OPTI 509Optical Design andInstrumentation II

Course overview andsyllabus

Instructor and notes informationInstructor: Kurt Thome

Email: kurt.thome@opt-sci.arizona.eduOffice: Meinel 605AOffice hours Tues. Wed. Phone: 621-4535

There is no assigned text for the class Reference texts:

Optical Radiation Measurements: Radiometry -Grum and Becherer

Radiometry and the Optical Detection of Radiation -BoydAberrations of Optical Systems -Welford

Class notes are available via web accesshttp://www.optics.arizona.edu/kurt/opti509/opti509.php

i

Discussion of optical systems via ray trace codes, ray fans, & spotdiagrams. The effects of optical aberrations and methods for balancing effectsof various aberrations. Radiometric concepts of projected area & solid angle, generation andpropagation of radiation, absorption, reflection, transmission andscattering, and radiometric laws. Application to laboratory & natural sources and measurement ofradiation

Why radiometry and aberrations?OPTI 502 gave optical design from the “optics” point of view

OPTI 509 shows tools needed to ensure enough light gets throughthe system (radiometry) and goes to the right place (aberrations)

Course descriptionCourse covers the use of radiometry and impact of

aberrations in optical design

ii

1. Class introduction, Radiometric and photometric terminology2. Areas and solid angles; projected solid angles, projected areas3. Radiance, throughput4. Invariance of throughput and radiance; Lagrange invariant,

radiative transfer, radiant flux5. E, I, M, reflectance, transmittance, absorptance, emissivity,

Planck's Law6. Lambert's law; isotropic vs. lambertian; M vs. L, inverse

square and cosine laws, examples7. Directional reflectance, simplifications, assumptions, view &

form factors, basic and simple radiometer8. The atmosphere and its effects on radiation, sources

Course overviewPropagation of radiation

iii

9. Radiometric systems, camera equation, spectralinstruments, demonstration of radiometric systems

10. Basic detection mechanisms and detector types11. Noise12. Figures of merit, Basic electronics; photodiodes and op-

amps13. Spectral selection terms; spectral selection methods14. Relative and absolute calibration15. Examples of detector calculations and vision

END OF TEST 1 MATERIAL

Course overviewDetectors

iv

16. Monochromatic aberrations; causes of aberrations,coordinate system; wave aberrations, tangential andsagittal rays, transverse and longitudinal ray aberrations

17. Ray fans, spot diagrams; RMS spot size18. Spherical aberration; variation with bending; high-order

spherical aberration19. Distortion20. Field curvature and astigmatism21. Coma; stop-shift effects22. Longitudinal chromatic aberrations of a thin lens, thin lens

achromatic doublet, spherochromatism, secondarychromatic aberration; lateral chromatic aberration

Course overviewAberrations

v

23. Defocus and tilt; balance with defocus and tilt24. Combined aberrations; aberration balancing; ray/wave fan

decoding25. Seidel aberration coefficients; wavefront expansion; wave

fans; wavefront variance, demonstration of optical designsoftware and optimization

26. Strehl ratio, calculations of PSFs and MTFs from raytracedata, influence of aberrations on MTFs

27. Imaging detectors - general characteristics28. Aspheric systems, conics; two mirror system, example

instrumentation: imaging systems; sensor examples

END OF TEST 2 MATERIAL

Course overviewImage quality and imaging systems

vi

Grading and homework policies

PHomework accounts for 25% of the final grade• Homework is due by the date and time listed on the assignment• Late homework has a 20% deduction if turned in prior to posting of

solutions• 40% deduction if turned in after solutions are given out (one week

after the due date)POne midterm, in-class exam is given worth 35% covers topics 1-15PExam 2 covers topics 16-28 plus one comprehensive question and

is worth 40% of the grade and given during final exam periodPExams are closed book, closed notesPEquation sheet will be given for use on the exam

vii

Propagation of RadiationOPTI 509

Lecture 1Radiometric and photometric terminology

1-1

Radiometry - DefinitionRadiometry is the science and technology of measuring

electromagnetic radiant energyPBasis for or part of many other fields

• Astronomy• Optical design• Remote sensing• Telecommunications

PField of study in and of itself• Metrology• Calibration• Detector development

PPhotometry is similar except that it is limited to the visible portion ofthe electromagnetic spectrum

1-2

Radiometric systemsAll radiometric systems have the same basic

componentsPSourcePObjectPTransmission mediumPOptical systemPDetectorPSignal processingPOutput

1-3

Output

SignalProcessing

Detector

OpticalSystem

TransmissionMedium

Object

Source

Radiometric sourcesSource illuminates the object

PAnything warmer than 0 K can be a sourcePSource partially determines the optical system

and detector• Source type defines the spectral

shape/output• Source output is combined with the object

properties to get the final spectral shape at the entrance of the optical system

• Source also defines the power available tothe optical system

Output

SignalProcessing

Detector

OpticalSystem

TransmissionMedium

Object

Source

1-4

Transmission mediaTransmission medium is what the radiation from the

source and object travels through to the optical systemPTransmission medium scatters, absorbs, and

emits radiation and confuses the object signal PCan be used to enhance the spectral shape

and power of the energy at the optical systemPTypically is viewed as a degradation of the

energy at the optical system

Output

SignalProcessing

Detector

OpticalSystem

TransmissionMedium

Object

Source

1-5

ObjectObject is the ultimate source of energy collected by the

optical systemPCan attempt to infer the properties of the

object from the measurements• Characterization of reflectance standards• Characterization of a calibration standard

PKnown properties of the object can be used to infer the properties of the optical system and detector (radiometric calibration)

Output

SignalProcessing

Detector

OpticalSystem

TransmissionMedium

Object

Source

1-6

Optical systemOptical system collects the radiation and sends it to the

detectorPOptical system is used to condition the

energy into a more useful form for the detetector and problem• Spectral filtering• Larger entrance pupil to collect more energy• Neutral density filter to reduce energy• Limit the field of view of the detector

P In radiometry, simpler is better

Output

SignalProcessing

Detector

OpticalSystem

TransmissionMedium

Object

Source

1-7

DetectorDetectors convert the energy of the photon into a form

that is easier to measure such as electric currentPAt this point in the system, the rest of

the work is selecting and characterizing the detector

PSelection of the detector is based on the spectral range of the problem and other issues• Operational requirements• Cost, size, availability

PCharacterize detector to understand relationship between input energy and ouput

Output

SignalProcessing

Detector

OpticalSystem

TransmissionMedium

Object

Source

1-8

Signal processingSignal processing is necessary to convert the detector

output to something useful to the userPSignal processing can improve the output

of the detector• Integration/averaging of multiple

measurements• Amplification of the detector output

(and noise)PMethods for signal processing can feed back

into choices for the other portions of the system

1-9

Output

SignalProcessing

Detector

OpticalSystem

TransmissionMedium

Object

Source

Variations in radiometric systemsEach radiometric system can have slightly different

configurations but the basic elements are always present

1-10

Radiometric systems - exampleImages below are of an Exotech radiometer being

absolutely calibrated in the RSG’s blacklabSpectralondiffuser(object)

Exotech radiometer(optical system anddetector)

Gain switches(signalprocessing)Output

DXW Lamp(source)

1-11

Radiometric systems - examplesImages below show an example of radiometric calibration

of using a spherical integrating sourcePRighthand image includes an ASD FieldSpec FR monitoring the

spectral characteristics of the SIS output for an airborne imagerPLefthand system is effectively identical to the radiometric system

shown on the previous viewgraph

1-12

Radiometric systemsOne goal of this course is to illustrate the importance of

viewing radiometry as the entire systemPEasy to concentrate on the equipment

• Optical system• Detectors• Signal processing

P Ignores the source, the transmission medium, and object• What is the spectral output of the source• How does the energy from the source interact with the object

PDesign of the optical system should optimize the outputPConsider the radiometer viewing the sphere source and the panel

• Both sources are spectrally similar• Sphere source has lower output• Optimizing the output of the radiometer for the sphere source could

make the output while viewing the panel unusable

1-13

Electromagnetic radiationRadiometry is the study of the measurement of

electromagnetic radiationPThus, radiometry is impacted by the ray, wave, and quantum nature

of electromagnetic radiation PMaterial in this course is dominated by the ray nature of light

• Geometric optics• Ignore diffraction in most cases• Not worry about the particle nature of light

PAlso ignore coherent light effects in most casesPRadiometric concepts will still apply in the coherent case and in the

presence of diffraction• Some modifications are required in the details• Basic philosophy is still the same

1-14

Electromagnetic spectrum1-15

1022 10

410

1910

1610

1310

1010

7Frequency (Hertz)

Cosmic Rays

Short Radio Waves

FM, TV AM

Long Radio Waves

Gamma Rays

X-rays

Ultraviolet InfraredVisible

Microwave and Radar

10-14 10

410

-1110

-810

-510

-210

1

Wavelength (meters)

Electromagnetic spectrumCommonly used labels

PThis class focuses on the 0.2 to 100 :m range• Shorter wavelengths are absorbed completely in the earth’s

atmosphere• Shorter wavelengths dominated by the particle nature of light• Longer wavelengths dominated by diffraction effects

PThis spectral range covers >99% of the total energy emitted byobjects warmer than 273 K

PSpectral regions are delineated by either detector, source, ortransmission medium effects

PAll of the wavelength ranges given are approximatePVisible part of the spectrum

• 0.35 to 0.7 :m (350 to 700 nm)• Wavelength range over which the human eye is sensitive

1-16

Electromagnetic spectrumMore labels

PUltraviolet (UV) - less than 0.35 :m PNear Infrared (NIR) - 0.7 to 1.1 :m

• Lower end is the upper end of the visible range• Upper end is the limit of which silicon-based detectors are sensitive

PVNIR - Visible and near infrared - 0.35 to 1.1 :m (region over whichsilicon detectors work well)

PShortwave Infrared (SWIR) - 1.1 to 2.5 :m• Lower limit is related to the upper limit of silicon detectors• Upper limit is the wavelength at which the sun’s output becomes

less dominant relative to emitted energy from the earthPMid-wave Infrared (MWIR) - 2.5 to 7.0 :mPThermal Infrared (TIR) - 8.0 to 16.0 :m

• Also called LWIR• Emissive portion of the spectrum for the earth-sun case

1-17

Equipment terminologyWill use several instrument types as examples in this

courseP Instruments are labeled according to their usagePMonochromator

• Produces quasi-monochromatic light• “One” wavelength as a source or in measurement

PSpectrometer• Measures a spectrum suitable for locating spectral features• Not an absolute device (cannot give accurate values of radiance)

PSpectroradiometer• Spectrometer for which the spectral response can be determined• Can give the radiance of a spectral feature

PRadiometer is the general term for an instrument used in radiometry

PSpectrophotometer• Relative instrument that can determine the reflectance

or transmittance spectrum of a sample• Strictly speaking only applies to visible light

1-18

Radiometric quantitiesList several key terms and units for radiometric quantities

for referencePMuch of what follows will discuss these parameters and how they

relate to each other in greater detailPRadiant Energy - Q

• Energy traveling as electromagnetic waves • Units of J [Joules] in radiometric units

PRadiant Flux (or power) - M• Rate at which radiant energy is transferred from a point on a

surface to another surface• Typical units of W [Watts or J/s]

PRadiant flux density is the power per unit area -[W/m2]• Radiant exitance - M• Irradiance - E

PRadiant intensity - I [W/sr]PRadiance - L [W/(m2 sr)]

1-19

Radiometric quantitiesThe terms just given are based on radiant energy

PKey to the usage of units is consistency• Using a given unit does not change the final conclusion• Often a matter of preference• There are some situations where choosing a given set of units will

simplify the problemPThis course will use radiometric units

• More general since it covers a wider portion of the EM spectrum• Instructor is most familiar with these units

P In addition to units, there are the labels as well• Note that radiant flux is oftened used interchangeably with power• Radiant flux density has two terms for the same units -irradiance

and radiant exitance• Intensity is sometimes used to refer to W/(m2 sr) {radiance}• Flux is sometimes used to refer to W/m2 {radiant flux density}

PTerms on the previous viewgraph are those used in this course

1-20

Phototmetric quantitiesWill not use photometric quantities in this course but they

are given here for referencePPhotometric units are based on the measurement of light

• Determined relative to a standard photometric observer• Visible part of the EM spectrum• Used in fields related to visual applications such as illumination

PBasic photometric units are the candela and lumen• Candela is luminous intensity of 1/60 of 1 cm2 of projected area of

a blackbody operating at the melting point of platinum• 1 lumen is equivalent to 1/685 Watts at 555 nm

PLuminous Energy - Q• Similar to radiant energy• Quantity of light • Units of lumen-seconds (lm s) or talbots

1-21

Photometric quantitiesOther units have analogies to radiometric quantities as

wellPLuminous Flux - M

• Similar to radiant flux• Rate light is transferred from point on a surface to another surface• Typical units of lumen (lm) where a lumen is equal to the flux

through a unit solid angle from a point source of unit candela PLuminous exitance (M), Illuminance (E)

• Simlar to radiant flux density, radiant exitance, and irradiance• Units of lumens/m2 or lux for both

PLuminous Intensity - I• Similar to radiant intensity• Luminous flux per unit solid angle• Units of candela

PLuminance - L• Similar to radiance• Units of candela/m2 or the nit

1-22

N dNh

Qd

hcQ dp p= = =∫∫ ∫

1 1ν λ

νν λ λ

Photon unitsCan use the number of photons as the base unit as well

and this leads to photon quantitiesPNot used in this course, but often easier to use for some detector

applicationsPPhoton number is similar to radiant energy

• <, 8, and c are the frequency, wavelength, and speed of light• Np is number of photons• Q< and Q8 are the radiant energies in frequency and wavelength• h is Planck’s constant

PPhoton flux is similar to power/radiant flux [photons/second]PPhoton exitance and photon irradiance [photons/m2]PPhoton intensity [photons/sr]PPhoton radiance [photons/(s sr m2)]

1-23

Radiant fluxRadiant flux is a key quantity when determining the

behavior of radiometric systemsPDetector of a radiometer collects photons over the field of view of the

systemPRadiant flux is a quantitative unit related to the number of photonsPRadiant energy is the quantity more related to the energy collected

by the detector• Some situations will warrant the use of radiant energy• Statistical analysis of detector output (and optical collection) is one

case where radiant energy is preferred• Remove the integration time effects by using radiant flux

PGoal in the design of almost all radiometric systems is to maximizethe radiant flux

PExamine how to maximize radiant flux by• Use of collection optics and detector area• Increasing field of view• Spectral integration

1-24

Radiant exitance and irradianceRadiant flux density has two labels to clarify the geometry

of the energy transferPRadiant exitance refers to energy leaving a surface

• Usually restricted to the case of emitted energy• Symbol is M

P Irradiance refers to energy incident on a a surface• Symbol is E• Irradiance is useful when considering the amount of energy

incident on a detector or similar surface

Radiant Exitance Irradiance

1-25

RadianceRadiance is a critical unit for understanding the source

and transferring source energy to the radiometerPWill show that radiance is conserved with distance

• Assumes that there is no absorption by the medium or optics• Makes it easier to compute the incident energy onto a radiometric

systemPUse of radiance removes sensor-related effects

• Area of the collection optics and/or detector• Field of view of the radiometer

PRadiant intensity will likewise allow us to readily transfer energy froma source to radiometer to compute radiant flux

PNote that both radiance and radiant intensity havean sr unit• Steradian• Takes into account solid angle which is covered in lecture 2

1-26

Propagationof Radiation

OPTI 509Lecture 2

Areas and solid angles,projected areas,

projected solid angles

2-1

Solid angleImage below is from a 2-D CCD array airborne framing

camera system (whole image taken at one time)PObjects that look larger in this image subtend a larger solid angle at

the sensorPMcKale Center subtends a larger solid angle than Meinel Building

2-2

Solid angleSteradian unit of solid angle shows up in radiance and

radiant intensityPSolid angle is a three-dimensional angle

• Unit is steradian [sr]• Mathematically, solid angle is unitless• Carry the unit for “clerical” purposes

PConcept of solid angle and steradian not that different from two- dimensional angle (What is a radian, anyway?)

PSteradian unit is defined in terms of a unit sphere• Steradian is defined as the solid angle

subtended by an area that is equal tothe radius of the sphere squared

• Recall that the surface area of a sphere is 4BR2

• Thus, by extension, there are 4B sr in a sphere

R

S

2-3

d d dΩ = sinθ θ φ

Ω = ∫ ∫ϕ

ϕ

θ

θ

θ θ φmin

max

min

max

sin d d

Ω = − −( )(cos cos )max min min maxϕ ϕ θ θ

Solid angleSolid angle “can be shown” in spherical coordinates to be

PWhere the terms are as follows• S is the solid angle• 2 is the polar or zenith angle• N is the rotational or azimuth angle

P Integrating the above gives

and

where the cosine difference is “flipped” due to a negative sign in the integration

PAbove assumes that we are determining solidangle subtended on a circular surface of a sphere

2

N

2-4

Solid angle of a hemisphereFrom the previous formula, it should be clear that the

solid angle subtended by a hemisphere is 2B srPThe limits on azimuth, or rotation, are 0 to 2B

• Limits on zenith are 0 to B/2• Then S=(2B-0)(cos[0]-cos[B/2])=2B sr

PThe solid angle of a sphere is 4B sr which is what was seen earlierPThese calculations are rather simplistic

• Unfortunately, typical solid angle calculations not always this basic• Thus, need a method to allow computation of solid angle for more

complicated situations

B/22B

2-5

( )S R R+ −2 2

Solid angle of a sphereWhat about a spherical object as seen from the outside?

PThis is a straightforward case analogous to determiningthe solid angle of the entire earth as seen by a satellite in orbit

PDiagram at rightillustrates the problem

PThe “trick” to this problem is that the solid angle is determined by the tangent view• Limits on zenith are 0 to 2• Then S=(2B-0)(cos[0]-cos[2])

PCos2=(S2+2SR)1/2/(S+R)PThen S=2B(1-(S2+2SR)1/2/(S+R))

R

R

2

S

2-6

Ω =⋅

∫ ∫$n drrS

r

2

r x y zn dr dxdy

zr

zx y z

2 2 2 2

2 2 2

= + +⋅ =

= =+ +

$ cos

cos

r θ

θ

Solid angle of a squareA bit more difficult is something that is not circular in

shape such as the solid angle subtended by a squarePMake the problem a bit easier by considering that the square has

length 2S on each side and the distance to the square is SPThere is no rotational symmetry so cannot

do the azimuthal integration easilyPSwitch to Cartesian coordinates and the

solid angle is written in terms of distance and radial direction from the observation point

PThe following relationships allow the integral tobe put in Cartesian coordinates

2S

2S

z=S

y

x

2-7

Ω

Ω

=+ +

=+ +

∫ ∫

∫ ∫− −

zdxdyx y z

adxdyx y aa

a

a

a

( )

( )

/

/

2 2 2 3 2

2 2 2 3 2

Solid angle of a squareUse the previous relationships to rewrite the integral in

terms of x, y, and zPThis gives

PNext assume the square is oriented perpendicular to the z axis at adistance S from the observation point (z=S)

PSolving this integral gives 2B/3 sr• Makes sense from a qualitative sense• Assume a cube of side 2S• Each face of the cube is identical in size and equal distance from

the center of a reference sphere• Six faces into 4B sr of a sphere gives 4B/6=2B/3 sr

2-8

drdr

SΩ =

( cos )( ' )

22

π θ

( )Ω =+

∫22 2 3 2

0

πS rdr

r S

R

/

Solid angle of a diskWhat about the solid angle subtended by a flat disk (thus

rotationally symmetric)PConsider the geometry to the right

• S is the distance to the disk• R is the radius of the disk• dr is an incremental change in

the disk radius• S’ is the distance to edge of dr• r is the radius to edge of dr

PThen the differential solid angle subtended by athin ring of the disk is

P Integrating this realizing that cos2=S/r and S&=(r2+S2)-1/2 gives

dr

R

2S

SN

r

2-9

( )Ω =+

= −+∫

22

12 2 3 2

02 2

0

ππ

S rdr

r SS

r S

R R

/

Ω = −+

= −+

−⎡

⎣⎢

⎤

⎦⎥ = −

+

⎡

⎣⎢

⎤

⎦⎥2

12

1 12 1

2 20

2 2 2 2π π πS

r SS

R S SS

R S

R

Solid angle of a diskThe disk integral, while not trivial can be solved from a

standard table of integralsPThus,

PDoing a bit of algebra gives

PThis is the analytic solution for the solid angle subtended by a diskPThis is readily done, but still, there must be an easier way

2-10

Ω = − − + +⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ = + +

⎡

⎣⎢

⎤

⎦⎥2 1 1

12

38

34

2

2

4

4

2

2

4

4π πRS

RS

RS

RS

K K

Ω = + +⎡

⎣⎢

⎤

⎦⎥ = =π π

RS

RS

RS

AreaS

disk2

2

4

4

2

2 2

34

K

Solid angle of a diskAssuming large distances gives us a simplification

PBinomial expansion of the disk solution gives

PNext, assume that the distance D is large relative to R• Higher order terms go to zero• Left with

PThus, the solid angle for cases when the distances are large relativeto the object can be determined by the ratio of the area of the objectto the distance squared

0

2-11

Ω =AreaS 2

Solid angle - simplified formulaSolid angle can be adequately computed by the ratio of

the area of the object of interest to distance squaredPThen

PThis actually goes back to the original definition for the steradian• Area equal to the radius of the sphere squared• Could do all of our calculations this way• Trick is determining the area on the sphere that is enclosed by our

physical objectPThat is, need to project the physical area onto the spherePThis projection gets easier at small solid angles

(small objects at large distances)

2-12

Solid angle approximationThe next step is to determine when this area over

distance squared approximation “fails”PFailure depends on the application and the level of error that can be

tolerated PAssume that the area of the sphere that is intercepted by the solid

angle can be assumed to be a disk• Solid angle subtended by the circular portion of the sphere is

2B(1- cos2)• Solid angle of the disk for our simplified formula is BR2/D2=Btan22

Area, BR2

distance, Sdistance, SN

R

distance, S

2

2-13

Solid angle approximationThe error due to making the area approximation can be

determined as a function of 2PExact formulation is based on the spherical coordinates integrationPApproximate formula is the ratio of area of the disk to the distance

squaredPDifference is small for small 2 angles

• Alternatively, the approximation is accurate for small solid angles• < 2% error out to half angles of 10 degrees

0 20 40 60 9001234567

Tangent approximationTangent approximationTheta angles (degrees)

Exact method

0 5 10 15 200

2

4

6

8

10

Theta angles (degrees)

2-14

Solid angle approximationCould also use sin2 instead of tan2 if the distance S is to

the edge of diskPDifference is still smaller for small 2 angles

• Alternatively, the approximation is accurate for small solid angles• < 2% error out to half angles of 16 degrees

PSlightly better approximation than the tangent formulation

00 20 40 60 80

Theta Angle (degrees)Exact formulation Approximate formulation

0

1020

30

40

50

60

0 20 40 60 80Theta Angle (degrees)

2-15

Solid angle computationsSine approximation will work most of the time in this class

but application will determine accuracy requirementPMust have an angle less than 11.5 degrees for errors <1%PAngles less than 3.7 degrees are needed for 0.1% errorsPThe errors in terms of size are

• 2% error when the distance squared is about four times the area• 1% error when the distance squared is about eight times the area• 0.1% inaccuracy when the distance squared is 80 times the area

00.040.080.120.160.2

00.40.81.21.62

0 2 4 6 8 10 12 14 16Theta Angle (degrees)

Exact formulationApproximate formulation Percent Difference

2-16

Solid angle and opticsF-number (F/#) and numerical aperture (NA) are both

related to solid angle subtended by the collection opticsPRecall F/#=f/Dpupil where Dpupil is the entrance pupil diameter and f is

the effective focal length• Then if " is the half angle it follows that F/#=1/(2tan")• Likewise, if the system is well-designed (aplanatic and following the

Abbé sine condition) you can show that F/#=1/(2sin")PNumerical aperture for the case of a system operating in air is

NA=sin"PThen S=B/(4 F/#2) = B(NA)2

• Will see something similar to this later in the Camera Equation

• Hopefully make sense physically<Faster systems (small F/#) have

larger solid angles<Systems with a larger NA have

larger solid angles

"Dpupil f

2-17

Units and quantitiesKey to radiometry is the interplay between the quantities

given previouslyPRemind ourselves of the radiometric units and quantities

• Radiant Energy - Q in J [Joules]• Radiant Flux (or power) - M in W [Watts or J/s]• Radiant flux density - [W/m2]<Radiant exitance - M< Irradiance - E

• Radiant intensity - I [W/sr]• Radiance - L [W/(m2 sr)]

PRadiant flux strongly determines the quality of the signal outputPRadiance is critical for understanding the

physical nature of the sourcePRadiant intensity and radiant flux density are

useful in transferring energy from source to sensorPDetermine the relationships between each of

these quantities

2-18

Simplified relationshipsAt one level, the relationships between the energy

quantities follow logically from the units of eachPTrue when the areas are much smaller than the distance between

them and radiance does not vary much spatially• Object of interest is located far from a sensor• Change in radiance from object is small over the view of a sensor

PRadiant fluxM = L × A × S

M = E × AP Irradiance

E = L × SPNot quite this simple

• Solid angle calculation is not simple• Area also has subtle effects

Area

S of theObject asseen bythe area

2-19

Projected areaCannot claim that objects with the same solid angle are

the same sizePThe moon and the sun subtend about the same solid angle at the

earth’s surface (think about a solar eclipse) but they are not thesame size

PThe football field andparking garage at rightsubtend about same solid angle• Garage is a 3-D object

with three parking levelsbelow the roof

• Football field is a 2-Dobject

• Stadium seats are a 3-D object tilted relativeto the camera

2-20

Projected areaThe stadium seats and parking garage present an

interesting examplePThe camera sees the projected area of the stadium seatsPProjected area is the planar area seen from the observation pointPClear that the angle of the object relative to the observer plays a role

in the solid anglePNote that it is also important to get the right distance as shown by

the parking garage

2-21

Area Areaprojected = × cosθ

Projected areaIn the simplest case, the projected area and actual area

are related by a cosine factorPProjected area is the physical area projected onto the plane that is

orthogonal to the observation pointPThen

• A is the physical area of the object• 2 is the angle between the normal to the physical area and the

observation point

2-22

Area

Areaprojected

2

Area Areaoff axis on axis− −= / cos3 θ

S Soff axis− = / cosθ

Physical areaPhysical area encompassed by a given solid angle

depends upon the distance and the angle to the observerPFor the case at right, the solid angle between the on-axis and off-

axis cases is constantPThis would be the case for a well-designed radiometer pointing

off- axis PThe distance between the object on axis to the

same object off-axis is related by cosine

PThe area seen by the instrument off-axis is

PAssumes that thearea is much less than the distance squared

S

2S/cos2

2-23

Ω projected d d= ∫ ∫ϕ

ϕ

θ

θ

ψ θ θ φmin

max

min

max

cos sin

Projected solid angleProjected solid angle (or weighted solid angle) is related

in philosophy to the projected areaPThe angle R is the angle between the plane of the observer to that of

the plane of the object

P In most cases, R=2, but this is not always the casePUsing projected solid angle allows the geometry of the source or

detector to be treated separately when doing flux computationsPThe cosine weighting factor can also be added later when doing the

energy computations• Reduces confusion as to whether using projected solid angle or

solid angle• Fits in nicely with the cosine law that we will see later

2-24

Ω projected d d sr= =⎛⎝⎜

⎞⎠⎟ =∫ ∫

0

2

0

2 2

0

2

22

π π π

θ θ θ φ πθ

π/ /

cos sinsin

Projected solid angle - hemisphereClassic illustration of projected solid angle is the

hemispheric casePRecall that solid angle subtended by a hemisphere is 2B srPProject this solid angle to a flat plane gives

• R=2 in this case• Most cases the difficulty is determining what R is

2-25

Simplified formulasRewrite the simplified formulas including projected area

and solid angle effectsPWe’ll see more details on this later related to specifically what areas

and what solid angles to usePRadiant flux

M = L × A × cos2 × S

M = E × A× cos2P Irradiance

E = L × SPS=Aprojected/D2

Projected Area

S of theObject asseen bydetector

2-26

Propagationof Radiation

OPTI 509Lecture 3

Radiance, throughput

3-1

Why use radiance?Radiance can either be incident onto a surface or come

from a surface area in a specified directionPOne of the most important radiometric quantities

• It is conserved with distance in a non-scattering, non-absorbingmedium

• Removes all geometric dependenciesPUnits of radiance are W/(m2 sr)PDiagrams below attempt to illustrate the concept

Normal to differential area

Differential area, dA

Differential solid angle, dS

Normal to differential area

Differential area, dA

Differential solid angle, dS

3-2

L A Ad

dAd=

→ →⎡

⎣⎢

⎤

⎦⎥ =

lim limcos cosΔ Δ Ω

Δ ΦΔ Δ Ω

ΦΩ0 0

2

θ θ

RadianceRadiance is the radiant flux in a specified direction as the

differential area and solid angle go to zeroPPhysically, we are counting the number of photons that are travelling

through an infinitely small solid angle from an infinitely small area ina direction from the area given by 2

PRadiant energy is the basic energy unit• dM=dQ/dt• Will almost always exclusively go to radiant flux rather than radiant

energyPRadiance is the basic radiometric unit from which the other

quantities will be determinedP Ignore spectral aspect now for simplicity of notation

3-3

RadianceExamine radiance from a geometric viewpoint first by

examining the energy traveling between two areasPConsider a bundle of rays that passes through both areas

• Projected area of A1 that is seen by A2 is simply A1cos21• Likewise, the area of A2 that is seen by A1 is simply A2cos22

PSolid angles can be shown to be• S2=A2cos22/S2 is the solid angle subtended by area A2 at area A1• S1=A1cos21/S2 is the solid angle subtended by area A1 at area A2

A1 A2

distance, S21 22

3-4

A AA

S1 1 2 1 12 2

2cos coscos

θ θθ

Ω =

A AA

S2 2 1 2 21 1

2cos coscos

θ θθ

Ω =

RadianceProduct of projected area of one surface and solid angle

subtended by other is the entire bundlePThis product gives identical values for both directionsPCase of A1 looking at A2

PCase of A2 looking at A1

PThis AS product will show up again later

A1 A2

distance, S21 22

3-5

Radiance conservationCan now show that the radiance leaving area A1 is

identical to that incident on area A2P Implies radiance is conservedPNow consider the energy passing through the areas

• Concept is that our AS product encompasses all rays that passthrough both areas

• Consider the rays to be related to energyPRadiant flux from all points on area A1 that passes through all points

on area A2 is L1A2 cos22S1

PRadiant flux from all points on area A2 that passes through all pointson area A1 is L2A1cos21S2

PTotal number of photons collected is the same in either case• AS product is independent of direction• One concludes that M1=M2• Then L1=L2

PRadiance is conserved in a lossless medium

3-6

Basic radiance and conservationBasic radiance takes into account changes in index of

refraction PAssume there are no other losses at the

boundary of two indexes of refraction.PThe amount of radiant energy through the

boundary does not changePChange in index of refraction causes the incident

radiance to alter direction• Change in direction leads to a change in

solid angle• Change in solid angle alters the radiance

(remember the per unit solid angle in radiance)PFrom a qualitative standpoint, expect the

radiance to be larger in medium 2• The same radiant flux is confined to a

smaller solid angle in medium 2• The radiance is larger for medium 2

n1

dA

n2>n1

22

21

3-7

LL

dd

dA ddA d

1

2

1

2

2 2

1 1=

ΦΦ

ΩΩ

coscos

θθ

LL

dd

dA d ddA d d

1

2

1

2

2 2 2 2

1 1 1 1=

ΦΦ

cos sincos sin

θ θ θ φθ θ θ φ

Basic radianceShow this more quantitatively using Snell’s Law and the

relationship between radiant flux and radiancePRecall

• dM=L dA dS cos2• Snell’s Law is n1sin21=n2sin22• Derivative of Snell’s law gives n1cos21d21=n2cos22d22

PDetermine the radiance onto the area of theinterface (L1) and the radiance from theinterface (L2)

PThe ratio of the radiance on either side of the boundary is

PSubstituting for the solid angle gives

n1

dA

n2>n1

22

21

3-8

LL nn21 2

2

12=

LL

dd

1

2

2 2 2

1 1 1=

cos sincos sin

θ θ θθ θ θ

LL

nn

n dnd

nn

1

2

1 1

2

1 1 1

2

1 1 1

12

22= =

sin cos

cos sin

θ θ θ

θ θ θ

Basic radianceThe radiant fluxes in both media must be identical since

there is no absorption (this will come up repeatedly)PDerivation of Snell’s law shows that the refracted beam and incident

beams are in the same plane and this means that dN1=dN2

PLikewise, the areas are identicalPThen

PSubstituting for Snell’s law such that sin22=(n1/n2)sin21 and forcos22d22=(n1/n2)cos21d21

POr and the radiance in medium 2 is larger

3-9

T A AAS

AAS

= = =Ω 212 1

22

Throughput, etendueThe AS product contains all of the geometric effects

related to the radiometric systemPProduct is so pervasive and important that it is studied in and of itselfPAS is the throughput of the systemPAlso called the geometric extentPThe situation we are working towards is the case of a detector

collecting photons• The entire area of the detector collects photons• Each elemental area on the detector has a finite field of view (solid

angle) over which to collect photonsPThroughput is not concerned with direction of energy flow

A1 A2

distance, S

3-10

T dAd dA d dA A

= =∫ ∫ ∫ ∫ ∫Ω

Ωφ θ

θ θ φsin

T dA d d AA

= ≈∫ ∫ ∫φ θ

θ θ θ φ θcos sin cosΩ

ThroughputPrevious formulation assumes large distances and small

areasPThe full formulation requires use of the integral form

PThis still assumes that the two areas are on axis• The more general case is below• This adds an additional cosine factor in the approximate formula

A’1

A’2

3-11

T A AAS

AAS

= = =Ω 212 1

22

ThroughputAll you are really trying to do is to convert the geometry

of any arbitrary case to that of the simple situation

A’1A’2

21

22

A1=A’1cos21

A2=A’2cos22

3-12

Why do we care?Throughput calculations allow for “quick” comparisons of

radiometric systemsPDetermine which system should collect more energyPAs an example consider the following

• Radiometer with a circular detector 0.1 mm in radius• Optical system is a simple tube that is 1 cm in radius and 10 cm

longPThus, S=10 cm, and both radii are more than a factor of 10 smaller

than the distance (thus we can justifiably use our approximation)PThen T=(B×0.12)(B×102)/1002=9.9×10-4 mm2 sr = 9.9×10-10 m2 sr

Raperture=1 cm = 10 mmRdetector=0.1 mm

Tube length = 10 cm =100 mm

3-13

Increasing etendueHopefully clear that increasing solid angle increases the

throughputP Increase the solid angle seen by the detector

• Decrease tube length (though if it is too short we can no longer usethe simplified formulation)

• Increase the aperture sizePCould increase the solid angle of the

detector but will do this as an increasein area

Raperture=1 cmRdetector=0.1 mm

Tube length = 5 cm

Raperture=2 cm

Rdetector=0.1 mm

Tube length = 10 cm =100 mm

3-14

Increasing etendueIncreasing the area of the detector is the moststraightforward way to increase the throughput

P Ignoring other effects such as vignetting that can limit the field viewPThat is, the example here assumes that entire detector can see the

entire entrance aperturePThe case shown here and those on previous viewgraph give a

throughput that is larger by a factor of four from the original case

Raperture=1 cm = 10 mmRdetector=0.2 mm

Tube length = 10 cm =100 mm

3-15

T A m sr= = × ×⎡

⎣⎢

⎤

⎦⎥ = ×− −Ω [ ( . ) ]

( . ).π

π2 0 10

4 1830 105 2

210 2

Magnitude of throughputIs the example shown a large value for throughput?

P It is difficult to say what is considered a large throughputPNeed to consider the source energy

• A small throughput will work fine when looking at the sun (do notneed a 36-inch telescope to count sunspots)

• Same throughput is not adequate to look for magnitude 10 starsPAlso need to consider the detector and signal processing packagePHowever, as an example that may be familiar consider a typical

digital camera• Large detector size would be around 20 micrometers• Fast camera would have an F/# of 1.8• Then the througput would be

3-16

Throughput cautionsCan use throughput for more complicated optical

systems but must not overextend the simplistic thinkingPAssume that we have a camera system with a manual lens

• Adjustable f-stop• We can vary the F/# of the lens• Recall that S=B/(4 F/#2) • Then changing the F/# by a factor of square root of 2 will change

the throughput by a factor of 2P Ignores other effects like

• Stray light• Reflections• Poorly designed optics• Variability in the grain size of the film

PStill allows us to design a camera lens so each click of the f-stop means about a factor of two more light (for smaller F/#)

3-17

T dA d d AA

proj= ≈∫ ∫ ∫φ θ

θ θ θ φcos sin Ω

Throughput calculationsThe subtlety in throughput calculations is to ensure that

the correct areas/solid angles are being usedPNeed to ensure that the tilts of the surfaces are taken into account

• Projected areas• Projected solid angles

PLess obvious aspect is to ensure that the limiting areas are used• This is where the source can play a role• Consider our earlier case• The original radiometer’s throughput is 9.9×10-10 m2sr• In this case the entrance aperture and detector are the limiting

areasRaperture=1 cm = 10 mmRdetector=0.1 mm

Tube length = 10 cm =100 mm

3-18

Throughput calculationsNow consider the two cases shown below of an extended

source and a small sourcePAssume in case 1 that the radiometer is viewing a large diffuser

screen• The diffuser screen fills the entire field of view of the sensor• The throughput is the same as computed before

PCan see how distance can play a role in this in that it is the solidangle subtended by the source not the physical size that is important

3-19

Throughput calculationsConsider the second case where the same radiometer

now views a “smaller” sourcePSmaller source can be something physically smaller or the same

source at a larger distancePDiffuser underfills the field of view of the sensor

• The throughput of the system is smaller than previously computed• The throughput of the radiometer is still the same• It is the throughput of the radiometric system that has changed

PShould always compute the throughput of the radiometric systemPStill have not included the source energy in the calculation

3-20

T n Tbasic = 2

Basic throughputBasic throughput accounts for refractive effects when the

the area and solid angle are not in the same mediumP In a similar fashion as basic radiance can show that basic

throughput (or optical extent) is

PConsider identical radiometers viewing an identical sourcePUpper radiometer is operating in airPLower radiometer is operating in water

• Throughput of the lower case is then (nair/nwater)2Tair• Can see that the throughput in water is smaller

3-21

nwater

Propagationof Radiation

OPTI 509Lecture 4

Invariance ofthroughput and

radiance, Lagrangeinvariant, Radiativetransfer, radiant flux

4-1

Throughput invarianceInteresting aspect to throughput is that it does not matter

which area/solid angle pair is usedPAssumes that the pairing of solid angle and area is done correctly

• One phrase used is “no snow cones”• Referring to the radiometer example, use the area of the detector

and the solid angle of the aperture as seen by the detectorPStraightforward when using simplified formulas

• Simply make sure both areas are used• Still have to be concerned about which angles to use for the

projected area and projected solid angle effectsPOften an issue when doing integral formulation

A1

A2 ?

Distance, SA1

A2

Distance, S

4-2

T A AAS

AAS

= = =Ω 212 1

22

Throughput invarianceOnce it is ensured that both areas are used then it should

be clear that either solid angle/area pair can be usedPRecall earlier simplified example repeated below

• Throughput can be computed using the solid angle subtended bythe area A1 as seen by area area A2

• Or, use the solid angle subtended by the area A2 seen by A1

PWhy do we care?• Goal is to compute the amount of energy collected by a detector

through an optical system or from a source of known size• Seems that the area of the detector and the solid angle of the

source (or optics) as seen by the detector would always be usedPSome cases are much easier when you “flip” the calculations

A1 A2

Distance, S

4-3

T dAdA

= ∫ ∫Ω

Ω1 2'

cosθ

T dAdA

= ∫ ∫Ω

Ω2 1'

cosθ

T dAdA

= ∫ ∫Ω

Ω1 1

cosθ

Throughput InvarianceThroughput is summing all of the solid angles subtended

by an area as seen by all points on a second area

A’1

A’2

Or

A’1

A’2BUT NEVER

A1A2

4-4

Throughput invariancePhysically, we are saying that we don’t care which

direction the photons are travelingPAssume the system is designed to operate left to right and collects

100 photons• Have to see the photons to collect them (field of view or solid angle

aspect)• Also need something for the photons to travel through (area)

PReverse the direction• Collecting area now becomes the source• Original source area becomes the collector• Still collect 100 photons

PRemember that this assumes the correct pairing of areas• Things become tricky when the source does not fill the field of view• Optical system drives the solid angle/area pair for the overfill case• Size of source limits the solid angle/area pair for the underfill case

4-5

Throughput invarianceThroughput invariance plays a big role when considering

an optical systemPUse the simple optical system below as an examplePEnergy from a source of area Asource enters a single lens system

• Area Alens • Exiting photons just fill the detector with area Adetector

POptical system (entrance pupil) subtends a solid angle, Sentrance, asseen from the source

POptical system (exit pupil) subtends a solid angle Sexit as seen fromthe detector

Sentrance

Asource AdetectorSexitSsource Sdetector

Alens

4-6

Throughput invarianceThen the throughput of the radiometric system can be

written in several waysP In terms of the lens collecting the source energy it is T=AsourceSentrance

PThe detector collecting energy from lens gives T=Adetector Sexit

PThen using throughput invariance can also write• Then T=AlensSsource• Then T=Alens Sdetector

PAll four are equivalentPCan use the entrance and exit pupils in the more complicated optical

case• This allows the designer to work in either image or object space• T=AsourceSentrance= AentranceSsource=Adetector Sexit=Aexit Sdetector

4-7

When invariance helpsConsider the case of the throughput for a system that

has two areas separated by distance HPAssume one area is much larger than the other area and HPFor example

• Circular area with radius 1000 m• Second circular with 1 m radius• Separation distance H=1 m

PSimplistic approach would be to use the areas and in this case thethroughput would be (B×12)((B×10002)/(12)=9.87 ×106 m 2 sr

PCan’t do this because the areas are not small relative to the distance

A2=3141590 m2

2

A1=3.14159 m2

H=1 m

4-8

T dAdA

= ∫ ∫Ω

Ω1 2

cosθ

When invariance helpsNeed to use the integral form of throughput in this case

of a large area relative to the distancePOne approach would be to use the area A2 and the solid angle

subtended by A1 at that distance• Then

• Not trivial because the cosine term depends on what part of thearea A2 you are using

P In other words, the smaller disk changes its appearance/solid angleas you move along the large area

4-9

T dAd dA

= ∫ ∫ ∫0

2

0

89 94

2

π

θ θ θ φ.

cos sino

When invariance helpsFlipping the problem around and using the solid angle ofthe large disk and area of the smaller simplifies things

PThus, use Area, A1 and the solid angle subtended by the larger areaP Integral formulation becomes

PReadily soluble, PFurther, in the limit of an infinitely large area, A2 (only an extra 0.06

degrees) we get a throughput of BA1

PWhen would we see this situation?• Bare detector viewing the ground or sky• Attempts to illuminate a large area using a lamp

4-10

Lagrange invariantAnother invariant is the Lagrange invariant and this too is

related to throughput and radiance invariancePRecall that the Lagrange invariant relates the chief and marginal

rays• The size of the image (chief ray) is related to the size of the optical

system (F/#)• Small angle approximation• N = n(Hmarginal"chief - Hchief"marginal) where n is the index of refraction

PSolid angle of the image is related to the solid angle of the object

Chief ray

Marginal ray Hchief

Hmarginal

"chief "marginal

4-11

( )N =n

A Achief m inal m inal chiefπ Ω Ωarg arg−

N =2 nAm inal chiefπ

⎛⎝⎜

⎞⎠⎟

2

Ω arg

( )N =n

Achief m inalπ Ω arg

Lagrange invariantAssume that the system is paraxial and objects are small

relative to distancesPThen sin"." giving S=B"2 PLarge distance versus object size allows us to use the area over

distance formula where A=BH2

PThen H"=(AS/B2)1/2 and

PAt the image plane, Hmarginal=0 and

PAt the pupil plane, Hchief=0 and

4-12

N =n

An

Tchief m inal image planeπ πΩ arg _=

N =n

Tπ

Lagrange invariantLook at the problem more generally in terms of the

relationship between throughput and Lagrange invariantPTake the pupil plane casePThen

PGeneralizing for throughput• Invariance of throughput• Recall example through a lens• Show that the Lagrange invariant and throughput are related via

PAnother way to think about this is that the Lagrange invariant isrelated to the light gathering ability of the radiometric system

4-13

Throughput and radianceCan relate three quantities to one another -Radiance, Throughput, and Radiant Flux

PUsually know two of these for a given radiometric system and canthen infer the third for future use

PFor example, consider a calibrated system• Know the relationship between radiant flux through the detector

and the output from the signal processing system• Have a known radiance source incident on the system• Measure the reported radiant flux• Infer the throughput

POnce throughput is known we can then look at an arbitrary source• Record the radiometer ouput and convert to radiant flux• Infer the radiance from the throughput and the radiant flux

PClearly, need to determine the radiant flux versus output relationship• Done theoretically at some level using known detector properties• This then relies on knowledge about the detector

4-14

Radiant flux and invarianceConsider the example of a radiometric system viewing a

source of known radiancePSingle lens system coupled to a fiber coupled to anther lens system

which then illuminates the detector• Illustrated schematically below• All of the light collected by the first lens system is captured by the

fiber optic• All of the light exiting the fiber optic is captured by the second lens

system• All of the light from the last optical element is collected by the

detectorPGoal is to compute the radiant flux on the detector

4-15

Radiant flux and invarianceThe fact that radiance is conserved and that throughput

is invariant simplifies the calculationPAll that is needed is to know

• Area of the entrance pupil• Area of the source as seen by the first len system• Distance from the entrance pupil to the source• Second two are equivalent to the field of view or NA

PThen the radiant flux through the pupil is

M=LsourceAsourceSentrance

PThis has to be same radiant flux as that incident on the detector• Photons are not lost anywhere along the way• If a photon is collected by the entrance pupil it must hit the detector

PAdmittedly, this is a simplified problem• Ignore losses at interfaces and in the fiber• Assumes perfect coupling• These factors can be readily calculated permitting

the radiant flux on the detector to be determined

4-16

dL d dA dA

S2 1 1 2 2

2Φ =( , , ) cos 'cos 'θ φ θ θ

Equation of radiative transferTake these examples a bit further and look more closely

at the relationship between radiant flux and radiancePEquation of radiative transfer

• Seen a portion of this is in the throughput calculations• Also shown in converting from one energy “type” to another

PNote thatradiance iswritten as a function of distance to account for transmission losses

PThe areas, angles, and distance are just as before in previous discussions

PThe first cosine factor allows for tilt of the source radiance area toaccount for the projected area normal to the second area

PThe second cosine factor takes into account projected solid angle

dA’1

dA’2

21

22

4-17

Φ = ∫ ∫A A

L dS

dA dA1 2

1 22 1 2

( , , ) cos cosθ φ θ θ

Equation of radiative transferRadiant flux depends on the apparent size of the source,

size of the collector, and the radiance from the sourcePThis becomes clearer when using the integral form of the equation

PThis problem is notstraightforward in general because• Distance can vary

across the areas we are using

• Radiance can vary as a function of angle

• Angles can vary across the areas we are using• Radiance can be attenuated as we travel through are transmission

media

dA’1

dA’2

21

22

L’

L’‘21'

4-18

Φ ΩΩ

= ∫ ∫A

L d d dA( , , ) cosθ φ θ

Equation of radiative transferAn alternative form for the equation of radiative transfer

can be written in terms of solid angle PSimpler in format, giving

PCan be more difficult to use• Distance effect is included in the solid angle• One of the cosine factors is taken into account in the solid angle

calculation• Second cosine factor leads to the projected solid angle

PThis was the form used earlierPWill see later that this form can make sense once the cosine law is

introduced

4-19

Radiance invarianceIn reality, this is conservation of radiance but the concept

of radiance invariance matches throughput invariancePThe primary reason that radiance plays such a key role in radiometry

is that it does not vary with locationPThat is, radiance is conserved in a lossless mediumPThis is a result of the fact that radiance was designed to be this wayPPhysically, as one moves away from a source of radiation nothing

changes geometrically from a radiance point of view• The differential area over which the radiation was emitted does not

change size as you move away from it• Differential solid angle in limit to zero solid angledoes not change• Think of it as following a single “ray” of photons leaving a surface

Normal to differential area

Differential area, dA

Differential solid angle, dS

4-20

Radiance conservationOptical system used cannot change the radiance (once

index of refraction effects are taken into account)PAssumes a lossless set of optics (kind of like the frictionless surface

and weightless string)• The radiance on the focal plane of a radiometer is IDENTICAL to

the radiance from the source if the focal plane is in the samemedium as the source

• Applies to both imaging and non-imaging systemsPOne advantage to this thinking is that now there is no need to worry

about the sensor• Can deal with the radiance on the entrance aperture• Helpful because models predicting the radiance incident on an

optical system have an easier time than optical design packages• For example, consider a 33-element fisheye lens with less than

perfect anti-reflection coatingPOne goal in radiometry is to determine the response of the system to

these input radiances

4-21

Φ Ω Ω ΩΩ Ω

= = ≈∫ ∫ ∫ ∫receiver source source receiverA

sourceA

sourceL dAd L dAd LAcos cos cosθ θ θ

Radiant flux is not conservedCan see from the equation of radiative transfer that

radiant flux is not conserved with distancePRecall throughput depends on the two areas being considered and

the distance between themPThroughput decreases with the distance between the areasPRadiant flux can be shown to be M=TL

• Thus, radiant flux decreases with distance as well• Using T=AS, then M=LAS• Once again, the actual formulation is an integral

P If 2.0, then cos2.1 and if radiance is relative constant over the areaand solid angle we get back to the simplified formulation of M=LAS• For solid angle and throughput only worry about geometry factors• Now, also need to ensure that the radiance behaves properly within

the integral in order to simplify

4-22

Examples of radiance and radiant fluxConsider case below where both radiometers are at the

same distance but different solid anglesPRadiant flux measured by the two radiometers will differ significantlyPRadiance is still identical for both sensors

Homogeneous surface - radiance same in all directions and positions

Sensor 1 Sensor 2

4-23

Radiance and radiant flux

Homogeneous surface

Area of the source that is seen by Sensor 1 is four times that of Sensor 2

Height=H

Solid angle same for both sensors

Height=1/2 H

Sensor 1

Sensor 2

4-24

Radiance and radiant flux

Homogeneous surface

Area seen by both radiometers is identical

Height=H

Height=1/2 H

Solid angle seen by Sensor 2 is twice that seen by sensor 1

Sensor 1

Sensor 2

4-25

Propagation of RadiationOPTI 509

Lecture 5E, I, M, reflectance, transmittance, absorptance,

emissivity, Planck’s Law

5-1

Summarize quantities so farAt this point we have the following radiometric quantities

excluding radiant energyPRadiant Flux - M

• Rate at which radiant energy is transferred from a point on asurface to another surface

• Typical units of W [Watts or J/s]PRadiant Flux Density - E and M

• Rate at which radiant energy is transferred per unit area• Radiant Exitance denoted by M and units of W/m2

• Irradiance denoted by E and units of W/m2

PRadiant Intensity - I• Radiant flux per unit solid angle• Units of W/sr

PRadiance - L• Radiant flux per unit solid angle per unit area• Units of W/(m2 sr)

5-2

Radiant flux densityHave examined radiant flux (W), radiance (W/[m2sr]), and

throughput (m2sr)PNow look at other quantities that can be formed by combining radiant

flux and other geometric factorsPFirst look at the ratio of radiant flux to the area -dM/dAPRadiant flux density is the general term applied to this quantityPUnits are W/m2

PThere is no directionality to radiant exitance (hence no solid angleunit)• Referenced relative to the normal to the area• More of an issue for incident energy

PMake a terminology distinction between the incident and exitingenergy• Irradiance refers to the radiant flux density incident on a surface• Radiant exitance refers to the radiant flux density leaving a surface

5-3

Radiant intensityWhat is left to examine is radiant flux per solid angle

P Intensity, I, is the radiant flux per unit solid• Units are W/sr• Intensity is not invariant

P Intensity is especially useful when dealing with the energy from a point source• One way to view a point source

is that it emits energy without an area

• Radiance in this case would not be defined since there is no area that is emitting

PSimplistic way to determine the intensity from a point source is to divide the radiant flux by 4B

I=N/4BI=N/4B I=N/4B

I=N/4B

I=N/4BI=N/4B

I=N/4B

I=N/4B

5-4

XX d

dλ

λλ

λ

λ

λ=∫

∫Δ

Δ

Spectral quantitiesAdd a further complication in that all of the quantities

shown vary with wavelengthPThe spectral variation is both a hindrance and an aidPNeed to account for the spectral nature of the quantities

• Spectral radiant flux, M8 [W/(:m)]• Spectral radiant exitance, M8 [W/(m2 :m)]• Spectral irradiance, E8 [W/(m2 :m)]• Spectral radiant intensity, I8 [W/(sr :m)]• Spectral radiance, L8 [W/(m2 sr :m)]

PThe idea is that these quantities are at a specific wavelength only• Can do this theoretically as will be seen shortly• In real life, the spectral quantities are averages over small

wavelength intervals

5-5

Φ ΩΩ

= ∫ ∫ ∫λ

θ φ θ λA

L d d dA d( , , ) cos

LddAd dλ θ λ

=3Φ

Ωcos

Spectral radianceSpectral radiance is the quantity that removes all of the

geometrically- and spectrally-related parametersPExamine the case of a radiometer collecting energyPRadiant flux is the more important parameter to ensure that we have

appropriate output quality

PSpectral radiance is more important when concerned about thetarget or object

• Removes all sensor related effects such as <Spectral width<Detector size<Solid angle of collection

• Can compare outputs from widely different sensors

5-6

Spectral radiance - exampleConsider the case of studying the spectral output from a

laboratory sample illuminated by a lamp sourcePTwo “identical” radiometers

• Radiometer 1 has a detector area that is two times that ofradiometer 2

• Radiometer 1 collects over twice the spectral interval ofradiometer 2

• Radiometer 1 has twice the solid angle of collection thanradiometer 2

PRadiant flux of radiometer 1 is eight times that of radiometer 2• This is typically a good thing• Science/application must still allow this to be done

PThe incident spectral radiance is identical for both radiometersPDifficulty is inferring the spectral radiance from the radiant flux

• More than one spectral radiance distribution can give the sameradiant flux

• Usually assume the band-averaged radiance is equivalent

5-7

More radiometric termsSince we’re defining terms, it’s worth bringing up four

more quantitiesPEmissivityPTransmittancePAbsorptancePReflectancePAll of these are unitless ratiosPAll of them can be determined from any of our energy quantities

• Differences between directional and non-directional• Ignore the directional aspects for now• Define all of the parameters in terms of radiant flux at this point• Examine later the reflectance from a directional standpoint as an

example• Numerator and denominator must have the same units

5-8

ρ =ΦΦ

refl

inc

ρλλ

λ=

ΦΦ

,

,

refl

inc

Reflection, reflectance, reflectivityReflection is the process in which radiant energy is

”thrown” back from a surfacePReflectance is the ratio of the reflected radiant flux to the incident

radiant flux

• Diffuse and specular (Fresnel)• More on this later

PReflectivity strictly refers to the reflectance of a layer of material• The layer is thick enough that there is no change in reflected

energy as the layer increases in thickness• In essence, it is the reflectance of an infinitely thick layer• Not used much anymore

PReflectance is also spectral in nature but still no units

Minc Mrefl

5-9

τ λλ

λ=

ΦΦ

,

,

trans

inc

Transmission, transmittance, transmissivity

Transmission is the process in which radiant energypasses through a surface or material

PSpectral transmittance is the ratio of the transmitted radiant flux tothe incident radiant flux

PDiffuse and regular transmission• Diffuse transmission is transmission that occurs independent of the

laws of refraction• Regular transmission is

transmission without diffusion• Total transmittance is the sum

of the diffuse and regular transmittance

PExamine transmittance morequantitatively later in terms of theproperties of the material

Minc

Mtrans

5-10

Transmission, transmittance, transmissivity

Other transmittance terms that are used in radiometryare included here for completeness

PSpectral internal transmittance is the transmittance within a layerand is the ratio of the radiant flux through the layer entry to thatexiting the layer

PSpectral transmissivity is the spectral internal transmittance througha layer of unit length under conditions when the boundary has noinfluence

PTransparent medium has a transmission that has a high regulartransmittance

PTranslucent medium transmits light by diffuse transmissionPOpaque medium transmits no radiant energyP4B transmittance

• Ratio of the forward and backward radiant fluxes leaving a surfaceto the incident radiant flux

• Typically measured in an integrating sphere

5-11

αλλ

λ=

ΦΦ

,

,

abs

inc

Absorption, absorptance, absorptivityAbsorption is the process in which incident energy is

retained without reflection or transmissionPSpectral absorptance is the ratio of absorbed spectral

radiant flux to the incident spectral radiant flux

PThe energy is converted to a different formPSpectral internal absorptance is absorptance within a layer and is

the ratio of the radiant flux absorbed in a layer to that exiting a layerPSpectral absorptivity is the spectral internal absorptance through a

layer of unit length under conditions when the boundary has noinfluence

PAbsorption coefficients• Characteristics of the material• Will use these later to define better the transmittance

P4B absorptance is the one’s complement to the 4B transmittance

MincMabs

5-12

ελλ

λ

λ

λ= =

ΦΦ

ΦΦ

,

,max

,

,

emitted

imum

emitted

blackbody

Emission, emissivity, emittanceEmissivity is the ratio of the amount of energy emitted by

an object to the maximum possiblePSpectral emissivity is then

PWill see later that an object emitting the maximum amount of energyis a blackbody

PAs before, emission refers to the process of emitting energyPEmittance is not used in practice, so emissivity is used with

reflectance, absorptance, and transmittance

MemittedMblackbody

5-13

Φ Φ Φ Φinc refl abs trans= + +

1 = + +ρ α τ

1 = + +ρ α τλ λ λ

Conserving energyPhotons are either absorbed, reflected, or transmitted in

this coursePThat is

PThen

POr in spectral terms

PKnowledge or measurement of any two parameters allows thederivation of the third• Directionality effects (radiance instead of radiant flux) complicates

this• Knowledge of the two is not always trivial

MincMrefl

Mtrans

Mabs

5-14

XX d

dλ

λλ

λ

λ

λ=∫

∫Δ

Δ

X X dλ λλ

λ= ∫Δ

Spectral ", D, J, ,Ideally, these quantities would be monochromatic values,

but there are also band-integrated and band-averagedPAs described previously, the monochromatic values we use will be

band-averaged values

PBand integrated values, or band-limited values are not divided by thewavelength interval

PAs has happened numerous times already, this is still simpler thanreal life• Weighting by the source energy• More on this later

5-15

BlackbodyA blackbody is one for which the absorptance is unity at

all wavelengthsPAlso a body for which the amount of energy emitted is a maximumPCan see this using the figure below

• Blackbody at some temperature T• Absorbs all energy incident on it• Inside an isothermal enclosure at temperature T

PLocal thermodynamic equilibrium (not lasers or gas discharges)• Object must have the same temperature as the enclosure• Object must emit as much as it absorbs< If it emits more it will have to cool< If it absorbs more it has to heat

5-16

[ ]Mhc

eW m mBB ch kT

c m s

h J s

k J K

λ λ

πλ

=−

= ×

= ×−

= ×−

21

2

52

2 99792 108

6 62607 1034

1 38065 1023

( / )

. [ / ]

. [ ] (Planck' s constant)

. [ / ] (Boltzmann' s constant)

[ / ( )]

Planck’s LawPlanck’s Law allows the spectral radiant exitance of an

object to be determined based on temperatureP In the case of a blackbody, the spectral radiant exitance is a function

only of the temperature (and the wavelength)PDerivation based on quantization of energy modes of oscillators

5-17

[ ]MC

e [W/(m m)]

c W m m

c m K

BB (C / T)λ λλμ

μμ

=−

= ×

= ×

15

2

18 4 2

24

2 1

374151 10

143879 10

. [( ) / ]

. [ ]

Planck’s LawCan substitute for the constants in Planck’s law and

rewrite it in a “simpler” fashionPConstants given here require that the input temperature and

wavelength have appropriate units• Temperature must be in Kelvin• Wavelength must be in :m

PConstants here and in previous formulation have a certain level ofuncertainty to them• Not an issue in this class• Can be an issue when attempting extremely high accuracy

measurements

5-18

[ ]MC

e [W/(m m)]

c W m mc m K

BB (C / T)λ λ

ελ

μ

μμ

=−

= ×

= ×

15

2

18 4 2

24

2 1

374151 10143879 10. [( ) / ]. [ ]

ε λ≠ <f ( ) .10

Planck’s Law - GraybodyGraybody is an object for which the emissivity is not a

function of wavelength but it is not unityPThat is,PThen

PThe shape of the Planck curve is identical, it is simply translateddownward• That is, the output at all wavelengths is reduced• The spectral radiant exitance is reduced by the same factor at all

wavelengths

5-19

Planck’s LawOnce you are given the temperature you can develop a

Planck curvePPlanck curves never

crossPCurves of warmer

bodies are above those of cooler bodies

PGiven a spectralradiant exitanceone can computean equivalenttemperature• Do this at one

wavelength• Can do it over a

spectral range

5-20

Planck’s LawOn linear scale it is difficult to display the broad range of

temperatures that can be seenPUsing a log-log scale shows dramatically the large range of exitance

that is seenPAlso note the

large differences in exitance between the two objects

PCurves do not cross each other as well

PFinally, the shapes of the curves do not change at all in this presentation

5-21

0.01 0.1 1 10 100Wavelength (micrometers)

1.00E+0

1.00E+1

1.00E+2

1.00E+3

1.00E+4

1.00E+5

1.00E+6

1.00E+7

1.00E+86000 K2000 K600 K

MckT

λπλ=

24

Mhc

eBBch kT

λλπ

λ= −2 2

5( / )

Planck’s LawPlanck’s Law was the first use of quantum physics

PQuantization of the energy was necessary to obtain satisfactoryagreement between measurements across the entire spectral range• Wien approximation was used for shorter wavelengths (short of the

peak)

• Rayleigh-Jean approximation works for longer wavelengths(beyond the peak) but fails at short wavelengths (ultravioletcatastrophe)

PNote that the goal was to match a set of measurements

5-22

[ ]Mh

c eBB h kTν ν

π ν=

−2

1

3

2 ( / )

Planck’s Law - frequency spaceCan also write Planck’s law in terms of frequency as well

as in terms of wavelengthPThen the spectral radiant flux is

PNote that the equation above does not simply replace<=c/8• Rather have to ensure that M8d8=M<d< (in absolute terms)• The problem is that d<=-c(d8/82)• These are not linearly related

PLeads to an interesting feature - the peak in frequency space doesnot match the peak in wavelength space converted to frequency• See this in Wien’s Law• Leads to the interesting question as to whether a system should be

optimized in wavelength or frequency space

5-23

[ ]Mhc

n eBB ch n kTλ λ

πλ

=−

21

2

2 5 ( / )

Planck’s Law - with indexStrictly speaking, Planck’s Law should include an effect

caused by the index of refraction of the mediumPOriginally assumed vacuum or material with index close to unityPPlanck’s Law was derived for frequency space

• Energy is linearly related to frequency in frequency space• Conversion to wavelength space from frequency depends upon the

index of refraction

PWe will ignore the index of refraction term because the index of air<1.0003• Thus, unless we are doing work requiring extremely high accuracy

we can ignore this factor• If you work in frequency space you can ignore the factor because

frequency does not change with index

5-24

ελλ

ε λ λ λ

λ λλλ

λ

= =∫

∫M T

M T

M T d

M T dBB

BB

BB

( , )( , )

( ) ( , )

( , )

ΔΔ

Δ

Δ

Band-averaged spectral emissivityRevisiting spectral emissivity, one should actually

determine it via a weighting by the blackbody curvePRecall emissivity was given as the ratio of the radiant fluxes emitted

by an object to that emitted by a blackbodyPNow write band-averaged spectral emissivity in terms of radiant

exitance

• Where M()8,T) is the band-integrated radiant exitance of an objectat temperature T over the wavelength interval )8 (at a given 8)

• MBB()8,T) is the band-integrated radiant exitance of the blackbodyPThus, there is the spectral, total or band-integrated, and band-

averaged emissivityPStill have not even considered the directionality aspect

5-25

Wien’s LawWien’s Law describes the relationship between the peaks

of Planck curves and the temperature of the object

5-26

[ ]λ μλmax ( ).

MT

m=×2 898 103

[ ]λ μνmax ( ).

MT

m=×5100 103

Wien’s LawObtain Wien’s Law by differentiating Planck’s Law and

setting equal to zeroPRecall that there are two versions of Planck’s Law -frequency and

wavelengthPThen the wavelength of maximum emission derived from the

wavelength version of Planck’s Law is

PWavelength of maximum emission as determined from thefrequency version of Planck’s Law is

PBrings up the question as to which is more relevant for energycalculations

5-27

Stefan-Boltzmann LawStefan-Boltzmann Law gives the radiant exitance (not

spectral radiant exitance) from the objectPDone empirically at first then with classical thermodynamicsPConceptually, it is the integral of Planck’s Law across all

wavelengthsPPlanck’s Law gives spectral radiant exitance -[W/(m2 :m)]PStefan Boltzmann Law gives radiant exitance - [ W/m2]

• M=,FT4 [W/m2] Where F=5.67×10-8 [W/(m2K4)]

• Emissivity factor assumes a gray bodyPWarmer objects emit more energy - by a power of fourPCorollary is that ,=M/FT4

5-28

1 = + +ρ ε τλ λ λ

Kirchhoff’s LawKirchoff’s law relates emissivity and absorptivity

,8="8PGood absorbers are good emitters lawPKirchoff’s Law says nothing about the amount of energy emitted

since this will also depend on the temperature and wavelengthPPlanck’s law gives the maximumum amount of energy that can be

emitted• A cold object that is an excellent absorber may not emit that much

energy• A piece of black paper absorbs very well but is too cold to emit a

significant amount of energy in the visiblePThermal equilibriumPFurther, this allows us to write

5-29

Propagation of RadiationOPTI 509

Lecture 6Lambert’s law; isotropic vs. lambertian,

M vs. L, inverse square and cosine laws, examples

6-1

Blackbody emission - directionalityIt can be shown that the radiance emitted by a blackbody

must be invariant with anglePConsider the small area dABB which is a blackbody inside a

blackbody enclosure, both at the same temperaturePRadiant flux collected by a small area dA located a distance R from

the blackbody, dABB at an angle (2,N)dM(2,N)=L8(2,N)dABBcos2dA/R2

PRadiant flux emitted by the area dA towardsthe blackbody area dABB is

dM(0,0)=L8(0,0)dA(dABBcos2/R2)PEmitted must equal absorbed thus the

conclusion is that L8(0,0)=L8(2,N)PMust also be true in the case when dABB

is not surrounded by a blackbody enclosuresince the enclosure would be a perfectabsorber

dA

dABB

R, 2, N

6-2

Lambert’s cosine lawA surface for which the radiance does not change with

angle follows Lambert’s cosine lawPLambert’s cosine law states the radiant intensity from a lambertian

surface varies with the cosine of the angle from the surface

I=I0cos2PRecall that radiance has a per unit area dependencePArea varies w/cosine of the angle thus radiance independent of

anglePThe variation of radiant intensity with cos2 offsets the change in

projected area so that radiance is constant

I=I0L=Llambertian I=I0cos2

L=Llambertian 2

6-3

Lambertian surfaceA surface that follows Lambert’s cosine law is a

lambertian surfacePLambertian surface plays a central role in radiometry and other fields

• Simplifies radiant flux calculations and other calculations as will beseen later

• Model applies to both emitted and reflected energyPBut, must keep in mind that it is a modelPRadiance from a lambertian surface does not vary with anglePBlackbodies are lambertian sources

• Know the spectral radiant flux from a blackbody -Planck’s Law• Know the wavelength of maximum emission -Wien’s Law• Know the radiant exitance - Stefan Boltzman Law• Radiance as a function of angle is a constant

6-4

Lambertian surfaceRadiance from a lambertian surface is independent of

view anglePAll of this knowledge can be applied to an arbitrary surface if we

assume that it is a blackbodyPApplies to gray bodies because we simply need to add a spectral

emissivityPFor surfaces that are not blackbodies, we can still make

assumptions that they are nearly blackbodies• Over small wavelength ranges• Over small angles of emission• Do this because we may not have any other information• Blackbody assumptions might be valid over a small range of

wavelengths and viewing anglesPThe one thing we don’t know yet is the amount of spectral radiance

in all directions (that is, Planck’s Law for spectral radiance)

6-5

M dhemisphere

lambertian lambertianL cos= ∫ θ Ω

Mlambertian lambertianhemisphere

L cos d= ∫ θ Ω

M vs. L for lambertianWant the relationship between radiant exitance and

radiance from a lambertian surfacePObtain the radiant exitance from a surface by integrating the

radiance through the entire hemisphere from the surfacePBecause of the large range of angles, we cannot simply use M=LS

PSimplifies because radiance does not vary with angle

MlambertianLlambertian

6-6

M

M

lambertian lambertian

/

lambertian lambertian

L cos sin d d

L

=

=

∫∫ θ θ θ φ

π

ππ

0

2

0

2

[ ]LBBBB

C T

M Ce

W sr m mλ λπ πλμ= =

−1

52

2 1( / ) [ / ( )]

Planck’s law for radianceHave seen that the integral over the hemisphere

weighted by cosine gives BP In spherical coordinates the integration is

PThen it can be shown that Planck’s Law can also be written in termsof radiance

PNote that the derivation of Planck’s law can be done either forradiance or radiant exitance and then the other determined fromlambertian aspect of the blackbody• Factor of B can be included in the constants thus may not appear

in the radiance version• The B factor then shows up in the numerator of the radiant exitance

6-7

Magic BNow have a relationship between radiance and irradiance

for the specific case of a lambertian surfacePMlambertian=BLlambertian

PSpectral radiant exitance given by Planck’s LawPWavelength of maximum emission described by Wien’s LawPRadiant exitance is described by Stefan Boltzman LawPSpectral radiance as a function of angle from a blackbody is a

constant and it’s value is gotten from Planck’s Law/BPRadiance from a black body is Stefan Boltzman Law/BPThis approach and methodology shows up repeatedly

• Look for a theoretical relationship or model that is convenient towork with (In this case we will assume a lambertian surface)

• Check that the model applies to our real life case (It won’t beperfect, but if it makes life much easier it still helps)

6-8

Irradiance on a surfaceNow want to flip the radiance around and see how the

irradiance on a surface behaves

Can probably guess whichside of the hill is to thesouth

Why is there snow in areasexposed to the sun?

Images here are of the same hill inNevada about 1 week after a snowevent

6-9

Cosine lawIrradiance on a surface is proportional to the cosine ofthe angle between the normal to the surface and the

incident irradianceP

PE = E0 cos2where E0 is the incident irradiance normal to the surface

PE0 is the maximum incident irradianceP Irradiance is zero for the tangent case

Normal to surface2

6-10

Radiometric Laws - Cosine LawCan also view the Cosine Law as a projected area effect

P Irradiance is a per unit area quantity• The same amount of radiant flux [Watts] spread out a larger area

gives less irradiance• Area increases as the angle of incidence increases

PThe beam of energy below is confined to the tube shown• The same amount of radiant flux hits both surfaces• In case B, the tilt of the second surface allows more surface area to

be illuminated thus giving lower irradiancePThe increase in area is inversely proportional to the cosine of the tilt

(area gets larger as angle increases)

6-11

Case B

Case A

Irradiance and directionalityIrradiance must always be defined relative to a given

normal to a surfacePThe useful amount of irradiance that a surface can “use” is only in

the normal directionPAttempt to illustrate this with the cases below

• The radiant exitance from the lamp is normal to a unit area of thelamp

• The irradiance from the lamp at the surface is relative to a normaldefined by the direction from the lamp to the reference surface

• The irradiance on the actual surface then includes the cosine factor

Lamp at temperatureT with radiantexitance FT4

Irradiance on areference normalsurface

Irradiance normal tosurface of interest

6-12

1/R2 lawIrradiance from a point source is inversely proportional to

the square of the distance from the source (1/R2 )PAlso called the distance squared lawPOnly true for a point sourcePCan still be used for cases where distance from the source is large

relative to the size of the source• Factor of five gives accuracy of 1%• Sun can be considered a point source at the earth

PCan understand how this law works by remembering that irradiancehas a 1/area unit and looking at the cases below

2DD

In both cases, the radiant fluxthrough the entire circle is same

Area of larger circleis 4 times that of thesmaller sphere andirradiance for a pointon the 2D circle is1/4 that of thesmaller circle

6-13

E cm E cm( ) (50 ) ( )10050

100100

14

2

2= = ⎡⎣⎢

⎤⎦⎥

1/R2 lawGiven the irradiance (or radiant exitance) at one location,

the irradiance can be computed at another locationPAssume the output of a lamp source is known to be 100 W/m2 at a

distance of 50 cmPThen the output of the lamp will be 25 W/m2 at a distance of 100 cm

• Assumes the lamp is a point source (typical lamp size <2 cm)• Still need the irradiance/radiant exitance at that initial distance

PDifficulties encountered when trying to use the radiant exitance froma point source and translating this over distance• Strictly cannot do this since a point source has no area, thus no

radiant exitance• In this case, need radiant intensity

6-14

Lens falloffCan combine distance-square law and cosine law to

determine incident irradiance on a planar surfaceP Important when looking at the irradiance reaching a detector in a

sensorPDiagram below shows the irradiance from a radiance source onto

the plane surfacePWant the irradiance at the off-axis point

in terms of the on-axis irradiance

E0

???

2

D

6-15

Lens falloffIt is “plainly” evident that the irradiance at the off-axis

location will be E0cos42PThe distance between the source and plane surface increases by a

cosine factor• Solid angle subtended by source depends on 1/R2 (if source small

enough)• Thus, the irradiance at the off-axis point decreases by cos2

PAlso have an angle of incidence onto the plane that gives anadditional cosine factor

PThe fourth cosine factor comesfrom the change in solidangle of the source as seen from the off-axis point

E0

2

D

D/cos2E0cos42

6-16

Radiometric Laws - Lens falloffThe impact is that images will appear darker at the edges

P Image at right is a fisheye lens image of the ground

PThe surface is reasonably close to lambertian• Bright spot at the top is

the hot spot• Bright spot at the bottom

is the specular reflection• These will be covered in

more detail laterPEdge of the scene is darker

primarily due to the lens falloff

6-17

Radiometric Laws - Lens falloffThe cosine4 factor is even more noticeable when the

effect is corrected as in the righthand image

6-18

Cosine4 falloffLens falloff is also known as cosine4 falloff and is true for

any extended sourcePExtended source can be a diffuse source of some finite size

• The ground outside for an imaging system• Could be a fully illuminated lens focusing light on the image plane

PExamples of “detectors” include what we normally would think ofsuch as CCD arrays and film

PSimilar falloff effects for sources illuminating a projector screen or adiffuser panel

PTechnically, the formulation is only good for small extended sources• If the source is too large, then our simplifying assumptions start to

breakdown and need the full integral• Still cos4 law is a useful tool

PJust keep in mind it is a theoretical idea - I have never seen a cos4

falloff in a real systemPFinally, it is straightforward to avoid this problem (Imax does this)

6-19

E Eoff axis− = 03cos θ

Point source falloffFor a point source we no longer have a solid subtended

by the sourcePThe solid angle was the source of three of our cosines

• The solid angle of the source changed with off-axis position• Area changed by a cosine and distance gave us two cosines

P In the case of the point source, there is no radiance from the source• Have irradiance, intensity, or radiant flux• Then it is possible to compute the irradiance for the on-axis point• Simply E0=Esource/D2

POff-axis you have the same formulation except thedistance is larger• Doff-axis=D/cos2• This gives a cos2 factor for Eoff-axis

P Including the cosine factor due tothe cosine incidence gives

E0

???

2

D

6-20

Example of irradiance calculationCompute the irradiance from the sun at the top of the earth’s

atmosphere at the equator

Sun

Unit area onthe sun

Radiant exitancenormal to the unit area

AssumeT=6000 K

Since we are notcomputing the spectralirradiance, we start withStefan-Boltzman’s Law todetermine the radiantexitance from the sun atthe surface of the sun

Msun = FT4 = (5.67x10-8)(60004) = 7.348 X 107 W/m2

6-21

Irradiance calculation example, cont’d.Now have the radiant exitance from a small area on thesurface of the sun. To get irradiance, the next step is todetermine the total radiant flux from the sun as a wholeThinking from a logical view, wewant units of [W] based on aknown [W/m2]

Assuming that the radiantexitance is constant over theentire surface, then simplymultiplying by the area of thesun should get us what we want

Radius of the sun is 6.957x108mthus the radiant flux of the sun is

Sun

Rsun=6.957x108 m

Msun=Msun x Asun = Msun x 4B(Rsun)2

Msun=(7.348 X 107)4B(6.957x108)2=(4.469x1026) [W]

6-22

Irradiance calculation, cont’d.Knowing the radiant flux from the sun we can compute the

irradiance at the top of the earth’s atmosphere

Dearth-sun

Msun at sunMsun (at earth)

Assume that the solar radiant flux is equally distributed across theentire orbit of the earth

Then the irradiance on the earth isE=M/A = M/[4B(Dearth-sun)2]

E= (4.4693x1026)/[4B(1.496x1011)2]

E=1589 W/m2

Typically accepted value is measured to be 1367 W/m2

6-23

Irradiance calculation, alternate approach

Rather than use the radiant flux approach, we can also usethe radiance from the surface of the sun propagated to the

earth

Dearth-sun

Assume the sun is a blackbody and thus lambertian, thenLsun =Msun/B= FT4 /B

= (5.67x10-8)(60004)/B = 2.339 X 107 W/(m2sr)The advantage to this is that the radiance is constant withdistance, thus we know the radiance at the top of the earth’satmosphere

Then integrating over solid angle of the sun will give us theirradiance at earth orbit

Esun = ILsun cos1dSsun= I(Msun/B) cos1dSsun

6-24

Irradiance calculation, alternate approachBecause the sun is small relative to the distance and weare viewing at near normal, the integration simplifies to

E=LS

RsunSsun

Dearth-sun

Esun= LsunSsun = (Lsun)(Asun/D2

earth-sun) = (Lsun)(BR2

sun/D2earth-sun)

= (2.339 X 107 [W/(m2sr)]) B (6.957 x 108)2/(1.496 x 1011)2

= 1589 W/m2

And this matches the answer we obtained from the other approach

6-25

E I d A I A I DWmsun sun earth sun earth earth sun sun

earth

= ≈ = =∫Ω

Ω Ω/ / / 221589

Irradiance example, alternate approach 2Can also use Intensity

PThe sun is far enough away that it can be viewed as a point sourcePThen the radiant intensity from the sun is M/4B W/sr

PThen integrating over solid angle of the earth as seen by the sun andthen dividing by the area of the earth will give us the irradiance atearth orbit

• Note - we are assuming that the sun is a point source• Thus, there is no area of the sun• This may seem to conflict our throughput discussion of using the

two separate areas (collector and source)• However, we took into account the area of the sun in determining

the radiant flux

Isun =Msun /4B=(4.469x1026) /4B

= 3.5563 X 1025 W/sr

6-26

E I m I DWmsun sun unit area sun sun= = =Ω _ / /1 15892 2

2

Irradiance example, alternate approach 2Worthwhile to examine the geometry of the radiant

intensity approach a bitPNote, we don’t really need the area of the earthPWe could have just as easily used a generic unit area at the earth-

sun distance

PWhat is being done in this case is determining how much radiantintensity is being collected by the unit area, hence the use of thesolid angle of the receiver

PThe real trick in this approach is that we need to know the size of thesun to determine the radiant flux to get the radiant intensity

Sunit area

Dearth-sunSun

Unit areaIsun = 3.5563 X 1025 W/sr

6-27

Esun at earth MsunRsun

Dearth sun

W

m_ __

..

.=

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥ = ×

×

×

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=⎡

⎣⎢

⎤

⎦⎥

2

7 348 107 6 957 108

1496 1011

21589 2

Irradiance example, yet another wayCould always rely on the 1/R2 law to determine the

irradiance on the earth from the sunPWe know the radiant exitance at a given distance

PThen the irradiance is simply

PThis approach is the least aesthetically pleasing• The radiant exitance we started with is clearly not from a point

source• We can view the radius of the sun factor as taking us back to a

point source, but this is a bit disturbing as wellPHowever, the method works because we have accounted for all of

the relevant geometry factors that are important• Earth-sun distance• Size of the sun

Msun = 7.348 X 107 W/m2 at the surface of the sun

6-28

Temperature of the earth

Assuming that the earth is a blackbody, compute thetermperature it will have when in radiative balance with the sun

If the earth is in balance with the incoming solar energy then theemission from the earth equals the incident solar irradianceEarth absorbs energy as a disk Earth emits energy as a sphere

Mincident=EincidentB(Rearth)2 Memitted=Memitted4B(Rearth)2

EincidentB(Rearth)2 = Memitted4B(Rearth)2

T4=Eincident/4F=7.007x109

T=289.3 K

6-29

Propagationof Radiation

OPTI 509Lecture 7

Directionalreflectance,

simplifications,assumptions, view &

form factors, basic andsimple radiometer

7-1

ρλλ

λ=

ΦΦ

,

,

refl

inc

Spectral reflectanceRecall the earlier discussion of spectral reflectance as

the ratio of reflected energy to the incidentPReflectance in terms of the radiant flux

P Ignored the directionality of the reflectance in this earlier discussion• Diffuse• Specular or Fresnel

PThe problem with reflectance is that it can be highly dependent uponthe geometry of the situation• Whether the incident energy is hemispheric or directional• Whether the reflected energy is hemispheric or directional• Thus, have to know both the incident angles and view angles

PWill search for the holy grail of reflectance -lambertian surface

7-2

Specular and diffuse reflectorsThere are no perfect diffusers or perfect specular

reflectors PDiffuse surface something that is nearly lambertianPMirrors are examples of a surface that is nearly specular

Lambertian surfaceSpecular surface

IncidentReflected

Specular direction Specular direction

Near-Specular Near-

lambertian

7-3

Examples of specular reflection7-4

Examples of directional reflectanceBare soil

Backscatter Forward Scatter

7-5

Examples of directional reflectanceSpruce forest

Backscatter Forward Scatter

7-6

Specular or Fresnel reflectionReflection from a smooth surface between two differing

indexes of refractionPAir-glass interface for examplePReflectance can be computed based on Maxwell’s equations and

ensuring consistent boundary conditions• Reflectance depends only on the angle of incidence and refractive

index• No spectral dependence except related to index of refraction• Fresnel reflectance is polarized

PFresnel equations give the reflection and transmission coefficientsPReflectance and transmittance used for energy calculations are the

squares of these coefficientsPUnpolarized transmittance and reflectance determined as the

average of the two polarized components

7-7

τ ρ=+

⎡

⎣⎢

⎤

⎦⎥ =

−+

⎡

⎣⎢

⎤

⎦⎥

2 1

1 2

22 1

2 1

2n

n nn nn n

Fresnel reflectionReflection effect occurs even at normal incidence at the

index of refraction interfacePReflectance and transmittance for normal incidence can be shown to

be

• Where n1 is the index of refraction of the incident medium• n2 is the index of the transmitted medium

PGraph here shows the reflectance as a function of the ratio of n2/n1• Values around 1.5

correspond to an air/glass interface

• Large ratios relate to several detector materials in air

1 1.5 2 2.5 3 3.5 4

n2/n1 ratio

0

0.2

0.4

0.6

0.8

1Reflectance Transmittance

7-8

ρθ θθ θ

ρθ θθ θ

=−+

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=−+

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⊥

tan( )tan( )

sin( )sin( )

inc ref

inc ref

inc ref

inc ref

2 2

τθ θ

θ θ θ θ τθ θθ θ=

+ −⎡

⎣⎢⎢

⎤

⎦⎥⎥

=+

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⊥

2 22 2

sin cossin( ) cos( )

sin cossin( )

ref inc

inc ref inc ref

ref inc

inc ref

Fresnel reflection versus angleCan compute the reflected and transmitted componentsfrom the interface based on angle of incidence and indexPVariety of forms of Fresnel equations based on what parameters

have been substitudedPSimple mathematical form is in terms of the incident (2inc) and

refracted (2ref) angles• Index of refraction contained within the refracted angle• Equations here are for reflectance not reflection coefficient

• Above assume that the magnetic permeabilities are nearly unity (orat least similar)

7-9

Fresnel reflection versus angleExample here shows the reflectance and transmittance

for an air/glass interfacePNote that the parallel component of the reflected beam becomes

completely polarized around 55 degreesPTransmitted beam goes toward zero at 90-degree incident angle

• See this with a piece of paper• Illustrated witht he photograph shown on viewgraph 11-4

0 10 20 30 40 50 60 70 80 90

Incident Angle

0

0.2

0.4

0.6

0.8

1Parallel Perpendicular Total

0 10 20 30 40 50 60 70 80 90

Incident Angle

0

0.2

0.4

0.6

0.8

1Parallel Perpendicular Total

7-10

Fresnel reflection versus angleConsider the case where the index of refraction of the

transmitted media is much largerPGraphs here were generated with an n2=3PGreater loss at normal incidence than for the smaller index casePLarge indexes such as this are not uncommon at air-detector

interfaces

0 10 20 30 40 50 60 70 80 90

Incident Angle

0

0.2

0.4

0.6

0.8

1Parallel Perpendicular Total

0 10 20 30 40 50 60 70 80 90

Incident Angle

0

0.2

0.4

0.6

0.8

1Parallel Perpendicular Total

7-11

Specular reflectionThere are three aspects of Fresnel/specular reflection

that are critical in radiometryPThe first is that there is reflection of energy at all optical interfaces

• Energy is reflected from the desired direction leading to loss ofsignal

• The reflected energy may go in unanticipated directions leading tostray light

• Anti-reflection coatings help alleviate thisPThe second aspect in radiometry is when the object is a specular

surface• Design of the radiometric system must take this into account• Geometry of the source/optical system must ensure that sufficient

energy gets to the detector• Geometry of the source/optical system must compensate for the

case where too much energy gets to the detectorPPolarization sensitivity and effects

7-12

Diffuse reflectanceDiffuse reflectance is the situation that occurs in bulk

scatterers or rough surfacesPBulk scatterer is one in which the light penetrates the “surface” and

exits the material after multiple scattering events• Even if the light went in only one direction (except backscatter)

after an individual scatter the exiting light would be difuse• Bulk scatterer must be thick enough to limit the effect of the

substratePSurface scatter is an issue in mirror fabrication and leads to stray

light issues

7-13

Geometry of reflectanceThe geometry of the situation significantly affects the

outcome of the reflectance resultsPConsider the case where a radiometer does not view the specular

direction of a Fresnel reflectionPFigures below illustrate the basic geometries that are encountered in

determining reflectancePNot shown are the directional limits that are actually desired

Bi-hemispheric Conical-hemispherical

Hemispherical-conical Bi-conical

7-14

ρθ θ θ φ

θ θ θ φ

π π

π π= = =∫ ∫

∫ ∫

ΦΦ

reflected

incident

reflected

incident

reflected

incident

EE

L d d

L d d

0

2

0

2

0

2

0

2

/

/

cos sin

cos sin

ρπ

=LE

reflected

incident

Hemispheric reflectanceAlso referred to as bi-hemispheric reflectance & is when

incident & reflected radiance are over a hemisphere PTerm albedo is often used and this strictly refers to the hemispheric

reflectance averaged over all wavelengthsPThe problem is determining the reflected and incident radiances

P If the surface is lambertian then

7-15

E Einc = ( ) cosθ θ

E L d dinc = ∫∫ ( , ) cos sinθ φ θ θ θ φ

Lambertian reflectanceReflected radiance from a lambertian surface does not

vary with view anglePThe radiance does vary with the angle of incidence

• Cosine law effect on the incident irradiance

• Incident irradiance can be determined by integrating the incidentradiance over the hemisphere in this case

7-16

RFE

Esample

lambertian

sample

lambertian= =

ΦΦ

RFE

E

L d d

Lreflected

reflectedlambertian

reflected

reflectedlambertian

reflected

reflectedlambertian= = =

∫ ∫ΦΦ

0

2

0

2π π

θ θ θ φ

π

/

cos sin

Reflectance factorWant reflectance, but characterizing the incident energy

is troublesome. Reflectance factor avoids thisPReflectance factor is the ratio of the reflected energy from the

sample of interest to what would be reflected from a ideal lambertiansurface

PRewrite in terms of the incident radiance

PThe above is valid for the case where the reflected energy iscollected over a hemisphere

7-17

Reflectance factorReference reflected measurements to known lambertiansurface with the same incident and reflected geometry

PNo measurement of incident radiation is neededP Incident radiation is essentially inferred from the measurement of the

reflectance factorP If the surface under study is lambertian then the reflectance factor is

simply the reflectancePOne difficulty is ensuring that the geometries match between the

sample and the lambertianPStill need someone at some point to either determine the reflectance

factor of a repeatable standard or a lambertian surface

7-18

RF RFsample standardsample

standard=

ΦΦ

Reflectance factorHave one or two groups make the time and effort tocharacterize the reflectance of a standard reference

PNational Institute of Standards and Technology (NIST) in the US• NIST characterizes the reflectance of a powder

(polytetrafluoroethylene, PTFE, Halon) pressed into a sample ofspecified thickness and density

• NIST measures the incident energy and can compute what wouldbe reflected from a lambertian surface

PUsing the computed Mlambertian and the measured Msample gives thereflectance factor of the standard

PThen make a version of the standard in another laboratory• Take an arbitrary sample and measure the reflected energy from

the sample and the standard• The reflectance factor of the sample is

7-19

RFL d d

L d d

L d d

L

reflected

reflectedlambertian

reflected

reflectedlambertian

= =−

−

∫ ∫

∫ ∫

∫ ∫φ

φ

θ

θ

φ

φ

θ

θφ

φ

θ

θ

θ θ θ φ

θ θ θ φ

θ θ θ φ

φ φθ θ

1

2

1

2

1

2

1

2

1

2

1

2

2 1

22

21

2

cos sin

cos sin

cos sin

( )sin sin

Hemispheric-conical RFHemispheric incident irradiance and a conical reflected

radiancePThe integration is not over the hemisphere

• The incident geometry has no bearing on the form of the integralsabove

• What the incident geometry does is impact the value of thereflected radiance

PExample of a hemispheric-conical situation is an instrument with asmall field of view looking at a surface under cloudy skies

P If the surface under study is also lambertian, we obtain thereflectance

7-20

RFL d d

L

reflected

reflectedlambertian=

∫ ∫0

2

0

2π π

θ θ θ φ

π

/

cos sin

Conical-hemispheric RFConical incident irradiance and hemispheric reflected

radiance gives a form identical to the bi-hemispheric caseP Identical in form as the bi-hemispheric case

P In this case, however, the incident light is considered over only asmall cone of incident angles• As a thought problem consider an isotropic incident source and a

lambertian reflector• Conica-hemispheric geometry leads to a reflected radiance that is

much smaller than for the hemispheric casePReciprocity means that the conical-hemispheric and the

hemispheric-conical are the samePAn example is measuring the upwelling radiance with a full 180-

degree field of view where the incident beam is from a flashlight

7-21

Bi-directionalBi-directional reflectance is the holy grail of reflectance

but cannot be derived via measurementsPBi-directional case has infinitely small solid angle of reflected

radiation and incident radiation over an infinitely small solid anglePCan develop a reasonably good directional source (e.g., lasers)

• Tougher to obtain an output from a sensor with an infinitely smallsolid angle but we can get close (integrate for a long time)

• Still not the bi-directional reflectancePActually measure he bi-conical reflectance

• Incident source has a small solid angle and is reasonably constantover the solid angle

• Reflected radiance is reasonably constant over a small solid angleof a sensor

• Then, bi-conical reflectance and bi-directional are similar

7-22

RFL d d

L

reflected

reflectedlambertian

=−

−

∫ ∫φ

φ

θ

θ

θ θ θ φ

φ φθ θ

1

2

1

2

2 1

22

21

2

cos sin

( )sin sin

BRF (or BDRF)The bidirectional reflectance factor is what is reported in

the literaturePAs mentioned, the best that can be done is bi-conicalP In general, reference to reflectance, directional reflectance, and

reflectance factor typically means BRFPReciprocity is also assumed valid here as well

PNote that the form of the bi-conical case is identical to thehemispheric-conical

7-23

BRDFLE

srreflected

incident= −[ ]1

BRDFBi-directional reflectance distribution function allows for

the conversion from irradiance to radiancePCan be determined by the ratio of the reflected radiance to the

incident irradiance

PThus, has units of sr-1

• BRDF has infinite value for a specular surface• BRDF=1/B sr-1 for a perfect lambertian reflector (hemispheric

reflectance of unity)PBRF=(B)BRDF for all casesPMajor advantage to BRDF is that if it is truly known, then it is

possible to derive the reflectance of a surface regardless ofview/sensor effects

7-24

ρπ

=LE

reflected

incident

Lambertian surfaceIt should be clear by now that the use of a lambertian

surface makes life very simplePRecall

• This now becomes Lrefl=(BRDF)(Einc) where the BRDF=D/B sr-1

PSimple model that allows us to know the radiance in all directionsgiven an incident irradiance• Reflected radiance is independent of view angle• Reflected radiance varies as the cosine of the incident angle since

the incident irradiance is normal to the surfacePBRF of this surface is DPBRDF is D/B sr-1

7-25

Example of BRF usageCan illustrate the use of BRF through the measurement

of reflectance of a field samplePThis approach is used throughout remote sensingPShow below the measurement of a desert test site for calibration

• Compare measurements of the surface to those of a reference ofknown BRF

• Ratio of the two allows the BRF of the surface to be determinedPStep 1 is then to determine the BRF of the reference

7-26

Laboratory BRFGoniometric measurements in a “blacklab” are used to

characterize the BRF of the field referencePStart with a NIST-traceable standard of reflectance (pressed PTFE

powder)• NIST gives the 8-degree hemispheric BRF• Powder must be pressed to specific standards

PBRF is found only for the zero-degree-viewing case since thissimulates its use in the field

PMeasurements of field reference compared to the NIST standard

7-27

Φ = ∫ ∫A A

L dS

dA dA1 2

1 22 1 2

( , , ) cos cosθ φ θ θ

Φ = ∫ ∫A A

LS

dA dA1 2

1 22 1 2

( , ) cos cosθ φ θ θ

Φ = ∫ ∫LS

dA dAA A1 2

1 22 1 2

cos cosθ θ

Radiometric radiative transfer equationRemind ourselves again of the basic formulation for

computing the radiant fluxPUse the dual area form of the equation to allow form factors later

PFirst simplification is to ignore transmission losses• Not saying there will not be such losses• Just want to make the equation easier to write for now• Bring it back later but will decouple transmission effects from the

integration and simply include a transmittance term at the end• Then

P If the radiance is lambertian (or equivalently does not change muchover 2) then

7-28

Φ Ω= LA cosθ

Φ Ω Ω= =LA and E L

Φ =LA A

S1 2 1 2

2

cos cosθ θ

Further simplificationsCase where the areas are far apart simplifies the

integralsPAll we are left with is

P If we recognize that the combination of areas, cosines, and distancegive us solid angle we can write

PAnd, for normal incidence on the “collector” we get

PRemind ourselves that this assumes• Large distance relative to the areas (distance/linear dimension=20

gives an uncertainty < 0.1%)• Little change in radiance over the areas (lambertian would be best)

7-29

Φ = ∫ ∫LS

dA dAA A1 2

1 22 1 2

cos cosθ θ

Form factorsGo back to the integral formulation for the case of the

lambertian sourceP In this case there is the radiation portion (L) and geometric portion

(everything else)

PGeometric term also known as a configuration factor, view factor,and others• Use this idea to develop the concept of a form factor• Cluster all of the

geometric termstogether

• Precompute thisform factor for a setof commonly-usedsetups

A1

A221

22

7-30

Φ onA A

MD

dA dA_

cos cos2

1 1 22 1 2

1 2

= ∫ ∫πθ θ

Form factorsForm factor is not simply the geometry but is the ratio of

the radiant flux on surface 2 to that emitted from 1PF=Mon_2/Mfrom_1

PRemember that we can use area of detector and solid angle ofsource or solid angle of the detector and area of source for radiantflux

PForm factor in terms of the radiant fluxes allows for a parameter thatincludes both geometries

PRadiant flux from the first surface is simply Mfrom_1 =M1A1

PThe radiant flux through A2 is

7-31

F

MD

dA dA

M A A DdA dA

from

A A

from A A1 2

1 1 22 1 2

1 1 1

1 22 1 2

21 2

1 2

1− = = =

∫ ∫∫ ∫

_

_

cos cos

cos cosπθ θ

πθ θ

πΩ

A F A1 2 1 2 1Ω = −π

A F A2 1 2 1 2Ω = −π

Form factorsThen the ratio of the two radiant fluxes from the previous

viewgraph givesP

PNow bring back invariance of throughput (AS)• Throughput for the case of area 1 and solid angle of surface 2 is

• Throughput for the case of area 2 and solid angle of surface 1 is

P Invariance of throughput then means that F1-2A1=F2-1A2

7-32

Form factorsCan thus compute the form factor in terms of which ever

geometry is easierPForm factor effectively allows for the determination of the solid anglePThen M=LAS=Lfrom_1A1BFPRealizing that BL=M gives M=Mfrom_1A1FPAgain, in the case of a lambertian source

• Can precompute form factors for typical geometries• Use the form factor to give the radiant flux

PThis approach gives more accurate results than obtained using moresimplified geometric assumptions

PCan still use this for cases where radiance does not change muchover the solid angles and areas of interest

7-33

F X XSS

where SRH

and XS

S

1 22 2

2

12

1 2

22

12

12

4

11

− = − −⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= = ++

/

( )

Form factor - exampleGeometry below would represent the simple radiometer

and spherical integrating source problemPA1 can represent the exit port areaPA2 is the detector areaPBoth areas are those that would be seen by the other based on the

design of the system (takes into account projected area effects)PReverts back to the simplified formulation as the radius becomes

small relative to the distance• M=L(A1)(A2/H2)=MA1F1-2• Then F1-2=(1/B)(A2/H2)=R2

2/H2=S22

A1 =BR12

H

A2 =BR22

7-34

Form factor - exampleUse the previous viewgraph’s geometry for a “real”

problemPUse the approximate geometry of the RSG’s black laboratory

• 10-cm radius lambertian source• 1-cm radius collector• Distance between the two is 0.5 m• Assume that the radiance from the source 100 W/(m2sr)

PThen the parameters are• S1= Rsource/Distance= 0.2• S2=Rdetector/Distance = 0.02• X=26.01• Form factor = 0.0003845• M = MAF= (BL)(BR2

source)(F)=0.003795 WPApproximation approach givesM=LAsourceAdetector/(Distance)2 =0.003948 W

PAbout 4% difference

R1=Rsource=10 cm

H = Distance = 0.5 m

R2=Rdetector=1 cm

7-35

Form factor - example, continuedCompare the simplified approach versus form factor as a

function of distancePLarger difference

at smaller distancesPMakes sense since

the approximate formulations are no longer valid

P1% difference at approximately a distance of 1 m (about a factor of five in distance to source size)

POf course thisstill assumes thatthe source is uniform spatially and angularly

0 0.2 0.4 0.6 0.8 1

Distance (m)

-100

-80

-60

-40

-20

0

0 0.5 1 1.5 2

Distance (m)

-5

-4

-3

-2

-1

0

7-36

Form factor - example, continuedNormalize the distance to the size of source

P1% error at a factor of 5P0.1% at a factor of 16P Implication for RSG work is the more exact formulation is needed

• In reality, the source size is about 2 cm due to FOV limits• Example illustrates a typical laboratory problem that requires the

use of more accurate formulations than will be relied on this course

2 3 4 5 6 7 8 9 10

Ratio of Distance to Source Size

-8

-6

-4

-2

0

7-37

FRH1 2

2

− = ⎛⎝⎜

⎞⎠⎟

Form factor - another exampleThis case would relate to the situation of measuring the

output from a spherical blackbodyPAssume the area in this case is a small differential area centered on

the sphereP In our simplification approach, H becomes large relative to R

• Treat the sphere as a disk• Then M=L(A1)(A2/H2)=L(A1) (BR2/H2 )

PThe form factor should simplify to what we had for the previous case• Then F1-2=(1/B)(A2/H2)=R2/H2

• Note, that this matches the form factor below

H

dA1

A2=4BR2

R

7-38

FZ S

Z S

where SRD

and ZRD

HD

1 2

2

2 2 1 2

2 2

12

124

1

− = −−

−⎡

⎣⎢

⎤

⎦⎥

= = + ⎛⎝⎜

⎞⎠⎟ + ⎛

⎝⎜⎞⎠⎟

( ) /

Form factors, examplesThis case could relate to the situation of looking at the

effects of off-axis lightPStray light test or MTF analysisPThe theoretical value allows us to determine how close the sensor

output matches the theory• Expect differences because no instrument is perfect• Large differences would indicate a possible problem in the design

or our understanding of the sensorA2=BR2

H

A1D

7-39

Simple radiometer - small sourceSimple radiometer consists of an aperture and detector

separated by a given distancePMake our simplifying assumption that distances and sizes allow us to

use M=LASPAsk the following questions?

• What happens to the radiant flux when source area increases butstays smaller than the FOV?

• Detector area increases?• Aperture area increases?• Either distance decreases?

FOV

Detector witharea Ad

Source witharea As andoutput radiantflux of Ms

Aperture ofarea Aa

Distance, Ss Distance, Sd

7-40

Simple radiometer - large sourceNow examine the case of an extended source

PAsk the following questions?• What happens to the radiant flux when source area increases?• Detector area increases?• Aperture area increases?• Distance between source and aperture decreases?• Distance between aperture and detector decreases?

FOV

Detectorwitharea Ad

Extended sourcewith outputradiance of Ls

Apertureof area Aa

Distance, SsDistance, Sd

7-41

Basic radiometerBasic radiometer is similar to the simple radiometer

except now include a lens at the front aperturePPlace the detector at the focal distance of the lens for simplicity

• Focal plane• Can place the detector at other locations with similar results

PFocal length and detector size define the resolution anglesPLens size and focal length (equivalently, f/#) define the speed angles

Resolution angles

Speed angles

7-42

Basic radiometerProduces an image of the source while giving a better-

defined field of viewPField of view of the simple aperture radiometer is not well defined for

similar reasons as a pinhole can “image” the sunPChief ray defines the FOV of the basic radiometerPStill use our simple formulation for radiant flux except now we should

consider the transmittance of the lens and reflections from itPAddition of the lens

makes the lensthe limiting aperture and the detector size the field stop (assume no vignetting) Detector with

area Ad

Source witharea As andoutput radiantflux of Ms

Lens witharea Aa

Distance, SsDistance, Sd

7-43

Basic radiometer - small sourceExamine the case again of the small source as we did

before with the simple radiometerPAsk the following questions?

• What happens to the radiant flux when output area increases butstay smaller than the FOV?

• Detector area increases?• Aperture area increases?• Distance between source and aperture decreases?• Distance between aperture and detector decreases?

Detectorwitharea Ad

Source witharea As andoutput radiantflux of Ms

Lens witharea Aa

Distance, SsDistance, Sd

7-44

Basic radiometer - large sourceNow consider the extended source which is larger than

the FOV of the radiometerPAsk the following questions?

• What happens to the radiant flux when output area increases?• Detector area increases?• Aperture area increases?• Distance between source and aperture decreases?• Distance between aperture and detector decreases?

Detectorwitharea Ad

ExtendedSource withoutputradiance, Ls

Lens witharea Aa

Distance, SsDistance, Sd

7-45

Φ = L AASS d

a

d2

Φ = L AASS d

a

d2

Basic and simple radiometers - summaryConsider the two radiometer types and the radiant flux

through the detector for the extended source casePSimple radiometer

• Collection area is defined by the area of the detector• Solid angle is determined by the size of the aperture and the

distance between the detector and aperture

PBasic radiometer• Collection area is defined by the area of the detector• Solid angle is determined by the size of the aperture and the

distance between the detector and aperture

7-46

Φ =+

L AA

S Ss ds

s d( )2

Φ = L AASs a

s

s2

Basic and simple radiometers - summaryConsider the two radiometer types and the radiant flux

through the detector for the small source casePSimple radiometer

• Collection area is defined by the area of the detector• Solid angle is determined by the size of the source and the

distance between the detector and source

PBasic radiometer• Collection area is defined by the area of the aperture• Solid angle is determined by the size of the source and the

distance between the source and aperture

7-47

Φ = IA

Ss 2

Φ =+

IA

S Ssd

s d( )2

Φ = IASs

a

s( )2

Basic and simple radiometer - summaryCan take the small source to the extreme of the point

sourcePThere is no radiance in this casePUse the radiant intensity of the source and

P In the case of the simple radiometer

P In the case of the basic radiomter (with lens)

7-48

Basic and simple radiometer - summaryThe lens improves the output for the point source but has

no effect in the extended source caseP Include the transmittance of the lens we actually lose power in the

extended source casePThe lens is the limiting aperture in both the extended and point

source casePSize of source has an impact on the simple radiometer

• Detector defines the throughput when the FOV is underfilled• Size of the tube opening defines the throughput for the case when

the radiometer FOV is overfilledPRays that would strike the wall of the radiometer and not reach the

detector are redirected by the lens towards the detector

7-49

Propagationof Radiation

OPTI 509Lecture 8 - The

atmosphere and itseffects on radiation,

sources

8-1

Effect of the Atmosphere on RadiationAtmosphere can refract, scatter, and absorb energy

(transmit means nothing happened)PRefraction refers to the bending of radiation from its original path

• Only a major factor in long path lengths through the atmosphere• Weather radar and horizontal view remote sensing must consider it• Mirages are a refractive phenomenon• We will not be concerned with refraction in this course

PScattering and absorption together also referred to as attenuation orextinction

PScattering refers to the redirection of radiation after interaction with aparticle in the atmosphere• Molecular scattering• Aerosol scattering

PAbsorption refers to the incorporation of the radiation within theparticle to alter the energy state of the particle• Gaseous absorption• Aerosol absorption

8-2

Atmospheric compositionUnderstanding what is in the atmosphere and where it is

will be important for determining the impact PEffect of different gases in the atmosphere depends upon their

location and concentration -both horizontally and vertically• Will be a difference between lab and open air• Concepts still the same in both lab and open air

PComposition of the atmosphere has two components• Molecules• Aerosols

PMolecules make up what we typically think of as atmospheric gases• 78% Nitrogen• 21% Oxygen• Argon• Water vapor, carbon dioxide

PAerosols are everything else

8-3

ScatteringScattering by aerosols and molecules is radically different

-4 -2 0 2 4

-4

-2

0

2

4

ForwardScatterBack

Scatter

1=90 or 270 degrees

Curve below shows thescattering phase function of aparticle

Aerosols have strongly peakedscattering probability in the forwarddirection

Molecules have a peanut-shapedprobability

The larger the value at a given 1angle, the larger the probability

Molecular case has beenmultiplied by 10 (red line) tomake it visible relative to theaerosol curve

8-4

Phase function is effectively aprobability function of where thephoton will go after hitting anaerosol or molecule

Polarization exampleMolecules also strongly polarize when scattering as seenin the two images below taken by a digital camera with a

polarizing filter in front of the lens

8-5

Third image illustrates the brightness one obtains by including thesolar beam in the image as opposed to only the aureole

Example of forward scatteringImages below show the solar aureole due to forward

scattering by aerosols

8-6

AbsorptionGaseous absorption is highly dependent upon

wavelength, atmospheric composition, and concentrationPGases causing the largest impact are carbon dioxide and water

vaporPWavelengths < 0.3 :m are almost completely absorbedPGaseous absorption in VNIR is small relative to longer wavelengths

• Ozone from approximately 0.5 to 0.7 :m• Oxygen absorption at 0.76 :m• Water vapor bands at several band locations (largest at 0.94 :m)

PSWIR, MWIR, and TIR affected by strong absorption featurespreventing surface radiation from reaching the top of the atmosphere

P “Window” region is from 8-12 :mPOther gases also absorb but impact can be readily decreased

8-7

δλ λ= ∫ k dss

0

δ μ ρλ λ= ∫ dss

0

Extinction coefficient and optical depthExtinction coefficient is an inherent property of a givenmaterial related to how efficiently it scatters or absorbs

POptical depth includes the distance that a photon must travel todetermine the probability that a photon is scattered or absorbed• Extinction coefficient has units of per distance [m-1]• Optical depth is unitless

• Alternate form uses the volume extinction coefficient and density

PLarge optical depth can occur because• Material is an efficient scatterer or absorber at a given wavelength• The physical path through the material is large

k8, x, and *8 k8, 2x, and 2*8 2k8, x, and 2*8

8-8

δ δ δscatter molec aerosol= +

δ δ δtotal scatter absorption= +

δ δ δ δabsorption H O O CO= + + +2 3 2

.........

Optical depthOptical depth can be divided into separate components

due to absorption and scatteringPExtinction coefficients can be treated in like fashionPOptical depths can be summed to give a total optical depth

PScattering optical depth can be split into molecular and aerosol

PAbsorption is the sum of individual gaseous components

PThis can then be used to understand how radiance changes as itpasses through a scattering or absorbing material

PThe radiance coming out of the box depends on• The amount going in• The amount created within the box• The material in the box and how it

interacts with light

8-9

L J e d L eλ λ

δδ

λδδ θ φ δ θ φ δ θ φ( , , ) ( , , ) ( , , )= ′ ′ +∫ − ′ −

0

0

L L eλ λδδ θ φ θ φ( , , ) ( , , )= −0

− = −dLd

L Jλ

λλ λδ

Radiative transferThe other radiative transfer equation describes howradiance varies in a scattering and absorbing media

PLaw of conservation of energyPStart with the differential form -Schwarzchild’s equation

PThe solution gives the integral form

P Ignore the source terms for the rest of this course for simplicity (or we will use very simplified arguments to determine it)

PThen we are left with Beer’s Law

8-10

τ δ δ δ= − + +e molec aerosol absorption( )

E E etransmitted incident= −δL L etransmitted incident= −δ Φ Φtransmitted incidente= −δ

ln( ) ln( )E Etransmitted incident= − δ

Optical Depth and Beer’s LawBeer’s Law relates incident energy to the transmitted

P Increase in optical depth means decrease in transmittancePThe exponential term of Beer’s law is the transmittance

• Then, in terms of optical depth we have

• Total optical depth has been substituted with the sum of thecomponent optical depths

PBeer’s Law is also used in a logarithmic form to help linearize theproblem

8-11

AbsorptionMODTRAN3 output for US Standard Atmosphere, 2.54

cm column water vapor, default ozone60-degree zenith angle and no scattering

O2

O3

H2O

H2O

H2O

8-12

0

0.2

0.4

0.6

0.8

1

0.3 0.6 0.9 1.2Wavelength (micrometers)

AbsorptionSame curve as previous page but includes molecular

scatter

1/84

dependence

8-13

0

0.2

0.4

0.6

0.8

1

0.3 0.6 0.9 1.2Wavelength (micrometers)

AbsorptionAt longer wavelengths, absorption plays a stronger rolewith some spectral regions having complete absorption

H2OH2O

8-14

0

0.2

0.4

0.6

0.8

1

1.1 1.4 1.7Wavelength (micrometers)

Absorption

H2O

CO2

CO2

H2OCO2

8-15

0

0.2

0.4

0.6

0.8

1

1.7 2 2.3Wavelength (micrometers)

AbsorptionThe MWIR is dominated by water vapor and carbon

dioxide absorption

CO2

CO2

H2O

H2O

8-16

0

0.2

0.4

0.6

0.8

1

2 3 4 5 6 7 8Wavelength (micrometers)

AbsorptionIn the TIR there is the “atmospheric window” from 8-12 :m

with a strong ozone band to consider

H2O

H2O

H2OCO2

O3

8-17

0

0.2

0.4

0.6

0.8

1

8 12 16Wavelength (micrometers)

Atmospheric compositionThe biggest problem in radiometry due to atmospheric

composition is the spatial and termporal variabilityPAmount of water vapor can change dramatically from measurement

to the nextPMultiple ways to avoid this problem

• Remove the atmosphere<Vacuum chambers<Nitrogen purge

• Operate at a wavelength that is not sensitive to water vaporinfluences

• Shorten path lengthsPNot trivial to avoid this effect if one must operate in open air

• Can still shorten path lengths and change wavelengths• Science application must permit this

8-18

Atmospheric effects, exampleMODIS Airborne Simulator data

8-19

Atmospheric effects, example8-20

Atmospheric effects, example8-21

Atmospheric effects, another exampleResults here show percent differences between

laboratory calibrations and field verificationsPNote the odd behavior of three instruments at 905 nmPOne idea is that this is a residual water vapor absorption effect in the

laboratory calibration

412 nm 469 555 645 858 905-15

-10

-5

0

5

10

15Terra MODIS ETM+Aqua MODIS ALIHyperion MISR

8-22

Radiometric sourcesRecall that the source illuminates the object in the

radiometric systemPAnything warmer than 0 K can be a sourcePSource partially determines the optical system

and detector• Source type defines the spectral

shape/output• Source output is combined with the object

properties to get the final spectral shape at the entrance of the optical system

• Source also defines the power available tothe optical system

Output

SignalProcessing

Detector

OpticalSystem

TransmissionMedium

Object

Source

8-23

Effects of sourcesSources play a key role in the characterization of the

optical system and objectPCharacterization of an instrument in the laboratory for use in sunlight

is one example• Laboratory source could be a lamp (2900 K blackbody)• Source used in operation of instrument could be sun (5800 K

blackbody)• Geometric effects reduce the solar output such that the laboratory

source has larger output at 8>1 :m• Solar output dominates at shorter wavelengths

PSpectral leak at short wavelengths will not be seen in the laboratoryPSame spectral leak at a short wavelength could dominate the signalPThus, the spectral shape of the source needs to be well matched to

the application, or well understood

8-24

Spectral effects of sourcesAs an example of the impact of the spectral nature of the

source consider the measurement of emissivityPAssume that the emissivity of a material is desired across the 7 to 15:m spectral range

PTwo sources are used to determine the emissivity• 290 K blackbody• 330 K blackbody

PEmissivity varieslinearly across thespectral range

PDerived emissivityfrom the twosources differsby more than1%

7 8 9 10 11 12 13 14 15Wavelength (micrometers)

20

30

40

50

0.5

0.6

0.7

0.8330 K

290 K

Emissivity

8-25

Natural sourcesNatural sources are a driving influence in the design of

laboratory and artificial sourcesPExamples of natural sources include the sun, skylight, thermal

emission from the ground and skyPRare to see a natural source in the laboratory due to the difficulty to

control the output level• Several groups are now using light pipes and heliostats to bring

natural light into the laboratory• Must figure out ways to offset absorption and scattering effects

PAdvantage to natural sources is that they simulate the source whichmight be used for the operation of the system

PA great deal of effort is made to attempt to simulate natural sources• Human eye and brain has adapted to natural conditions and these

are viewed as normal• Background skylight• Sunlight

8-26

Sources - Solar radiationStudying the sun as a radiometric source is helpful both

from a modeling and measurement standpointPThe sun is the primary source

of energy that drives the earth’s weather and climate

PPeak of solar radiant exitance is at approximately 0.45 :m

PDistance from the earth to sun varies from 0.983 to 1.0167 AU• AU is an astronomical unit and

is the ratio of the actual distance to the mean distance

• Using 1/R2 gives a difference of 7% in the irradiance between minimum and maximum distance

• Mean distance is 1.496×1011 m• Mean radius 6.957×108 m

8-27

Solar radiationSolar constant has been studied quantitatively since the

early 1900sPSamuel Langley’s

work for the Smithsonian Institute used ground- based measurements from mountain tops

PNext improvement was balloon-borne instruments

P1960s and later used rocketsondes and space-based instruments

PCurrently agreed upon value for solar constant is 1367 W/m2 (based on the work ofFrohlich)

8-28

Solar radiation - spectralSolar constant is not as well understood at the spectral

levelPSolar variability at very short wavelengthsPCalibration uncertainty at longer wavelengthsPKnowledge in the VNIR is better than the SWIRPTwo models shown

here are both usedin the remotesensing community

PMODTRAN-basedmodel is from Chance-Kurucz based on measurements atKitt Peak, Arizona

PWRC is based on theFrohlich work 0.70 0.75 0.80 0.85 0.90

Wavelength (micrometer)

800

900

1000

1100

1200

1300

1400 MODTRAN-basedWRC-Based

8-29

Solar radiation - spectralKnowledge has improved since the 1990s due to several

shuttle-based experimentsPThese shuttle-based results are still being evaluatedPWill require several historical works to be revisedPThe newer results should point to which of the current set of spectral

curves is the most accuratePCurve here shows the percent

difference between the WRCand MODTRAN-based models

PNote that in many cases theuncertainty requirements of some sensors that aremeasuring terrestrial energy are exceeded in some bands solely by the differences in solar models

0.30 0.70 1.10 1.50 1.90 2.30 2.70Wavelength (micrometer)

-10-5

0

5

10

15

20

25 10-nm averages50-nm averages

100-nm averages

8-30

Solar radiation - blackbody modelCould also use the Planck curve to simulate the output of

the sunPNote that one

single curvewould notbe sufficient

PEquivalenttemperatureto obtain thethe 1367 W/m2

would be5778 K

8-31

Terrestrial radiationBlackbody approach works a bit better with terrestrial

surfacesPPlanck curves below show typical terrestrial temperaturesPPeak is in the TIRPSpectral emissivity must be taken into account

8-32

Terrestrial radiationWould still need to modifythe Planck curve output to

take into account the spectralemissivity of the surface

PMost terrestrial surfaces can be adequately approximated as gray bodies

PPlots hear are measured emisivities of several materials

Asphalt

0.964Wavelength (:m)

4 13

0.982

Railroad Valley Playa0.80

Wavelength (:m)4 13

1.00

Sea water

8-33

Solar-terrestrial comparisonIt can be shown that solar energy dominates in

VNIR/SWIR and emitted terrestrial dominates in the TIRPTakes into account the earth-sun distancePSun emits more

energy than the earth at ALL wavelengths

P It is a geometry effect that allows the wavelength regions to betreated separately

8-34

0.1 1 10 100Wavelength (micrometers)

1.00E-3

1.00E+0

1.00E+3

1.00E+6

1.00E+9

1.00E+12Solar exitance (5800 K)Solar irradianceTerrestrial exitance (300 K)

Solar-terrestrial comparisonThe same plot in linear space shows how the peakspectral output of the sun exceeds that of the earth

8-35

0 5 10 15Wavelength (micrometers)

0

500

1000

1500

2000

2500Solar irradianceTerrestrial exitance

Solar/Terrestrial comparisonPutting the two curves on a relative scale highlights thedifference between the solar reflective and the emissive

parts of the spectrum

8-36

0.1 1 10 100Wavelength (micrometers)

0

500

1000

1500

2000

2500

0

10

20

30

40

50Solar irradianceTerrestrial exitance

SkylightPlot here illustrates a modeled output from the sky for a

given view-sun geometryPPlot will also vary

depending upon atmospheric conditions

PCurve is clearly dominated by the solar output

PNow have two parts of the three part puzzle why the sky is blue

0.3 0.5 0.7 0.9 1.1Wavelength (micrometers)

0

20

40

60

80

100 Zenith angle = 45Solar zenith = 30

Sea LevelModel output

8-37

Artificial sourcesCommon sources used in the laboratory include lasers,

incandescent bulbs, LEDs, and discharge lampsPJust as there are large differences in the output of the sun and earth

with wavelength there are differences with artificial sourcesPThe source used depends heavily on the applicationPLasers are often used because of their spectral purity

• Downside is often the variability of the output• Also need to find a source at the appropriate wavelength

PDischarge lamps or fluorescent sources are used at shorterwavelengths• Require approximately a 50,000 K blackbody to simulate blue sky• Discharge lamps can give higher output at short wavelengths

relative to a blackbodyPLEDs have broader spectral output than lasers

• Can be more stable but still better to have controlled output• Mixtures of colors can be used to get the appropriate spectrum

8-38

Broadband versus spectralCurves below show typical spectral outputs from both a

fluorescent tube (left) and incandescent bulb (right)PFluorescent tube output is characterized by several spiked featuresP Incandescent bulb has a peak at a wavelength longer than the peak

output of the sunPThese two factors can lead to interesting color effects in clothing and

paints

8-39

Artificial sourcesArtificial sources have received widespread attention

outside the laboratory in reference to illumination studiesPComputer displaysPNighttime illumination

• Inside the home• Streetlights, traffic signals, vehicular lighting

PLarge amount of research in developing energy efficient lighting thatcan simulate the spectral output of normal daylight conditions• Incandescent bulbs with appropriate coatings• Halogen lamps that are hotter than standard tungsten filament

bulbs• Flourescent lamps with different mixtures of gases and internal

coatings• Mixtures of LEDs

PDevelopment of artificial lighting is a good example of whereradiometry (and photometry) is critical to understanding the problem

8-40

“Standard” LampsFull name is a standard source of spectral irradiance or

spectral irradiance standardPFilament lamp sources work on the

simple concept of heating the filament to the temperature needed to obtain the desired spectral output

P Image at right shows an FEL secondary standard• FEL just refers to an ANSI designation• Obtained from commercial vendor

PTypical operation parameters are120 V at 8 A current

PTungsten filamentPFilament operates at around 3000 K

8-41

Radiometric sourcesA major issue in radiometry is developing sources that satisfy

needed outputs both spectrally and absolutelyPExample here compares the output of the sun at the top of the earth’s

atmosphere to that of anFEL lamp standard• FEL lamp is at the

standard distance of50 cm

• Solar values are basedon the WRC model

PNote the low lamp outputin the blue part of thespectrum

PLarge output of lamprelative to the sun atlong wavelengths alsocauses problems 0.30 0.70 1.10 1.50 1.90 2.30 2.70

Wavelength (micrometer)

0

500

1000

1500

2000

2500 WRC-BasedFEL-lamp based

8-42

Lamp sourcesIn addition to FEL lamps are other tungsten lamps for

laboratory usagePPhoto here shows a DXW lamp that is ot absolutely calibrated and is

much less expensive than an FEL standardPDifficulties with lamp

sources are• Degradation• Changes in tungsten

spectral emissivity• Change in tungsten

resistance duringusage

PUse of halogen improves the lampstability over time

8-43

Spherical integrating sourceSIS relies on the fact that light from a lamp source goesthrough multiple reflections prior to exiting the sphere

P Images below show the RSG’s 40-inch SIS PAttempt to coat sphere with near-lambertian, spectrally-flat materialPSIS provides a near-lambertian source with some extended sizePThe lamp alone provides a near-point source

8-44

Blackbody sourcesLaboratory blackbodies are designed to have emissivities

that are near unityPSIS provides a uniform

illumination for radiance calibration at short wavelengths

PLaboratory blackbody provides a good source for longer wavelengths with much the same principles

PMemitted=,FT4

• Trick is to have knowledge ofthe temperature

• Original traceability to nationalstandards was based on thetemperature sensors for aspecific conical design

8-45

Blackbody sourcesTypical design of a blackbody incorporates some type of

cavity that enhances the absorption of lightPConical cavity is one of the easiest to produce

• Spherical design also works well (just like the SIS)• Other designs are also used

PThe idea is to have a large surface area to emit the radiant exitance but requiringmultiple bounces for the photons to exit the port

8-46