Upload
others
View
5
Download
0
Embed Size (px)
Citation preview
ECE 5616Curtis
Foundations of Geometrical Optics
• Eikonal equation • optical path length
• Fermat’s Principle• Basic of Radiometry
•Inverse square law• Postulates of Geometrical Optics
• Sign convention
ECE 5616Curtis
Geometrical Optics - Introduction
What’s it good for ?1. Where is the image?2. How large is it?3. How bright is it?4. What is the image quality?
Useful when size of aperture is >100λWhy?
ECE 5616Curtis
Let there be light…
Hecht “Optics”
ECE 5616Curtis
Spectrum
659-769455-390Violet
610-659492-455Blue
520-610577-492Green
503-520597-577Yellow
482-503622-597Orange
384-482780-622Red
υ(Thz)λ (nm)Color
ECE 5616Curtis
193 nm (UV-C)Gas (excimer)ArF
351 nm (UV-A)Gas (excimer)XeF
364 nm (UV-A)Gas-IonArgon
416 nmGas-IonKrypton
442 nmGas-IonHeCd
457.9 nmGas-IonArgon
488.0 nmGas-IonArgon
514.5 nmGas-IonArgon
532 nmSemiconductorDPSS
568.2 nmGas-IonKrypton
633 nmGasHeNe
647.1 nmGas-IonKrypton
694 nmSolid StateRuby
700-1000 nm Solid StateTi:Sapphire
1064 nm Solid StateNd:YAG
WavelengthLaser TypeLasing Medium
Laser Lines
ECE 5616Curtis
Semiconductor Sources
ECE 5616Curtis
Major spectral lines usedFraunhofer lines, Telecom bands
Major Telecom wavelength windows for optical fiber 1.5um (C band, 1530-1565nm) and 1.3um (O band, 1260-1360nm). – 1.5 is dominant for long haul transmission as the erbium doped fiber
amplifiers are very effect at this wavelength and there is more chromatic dispersion that minimized non-linearity.
– 1.3 is more used for shorter connections or older lines.
These are historical visible lines use for camera, glass measurement, design etc.
ECE 5616Curtis
Basic History• 280BC Euclid's Catoptrics light goes in straight line. However light was emitted by eye
(not)– law of reflection was correctly formulated in Euclid's book
• Hero of Alexandria, in his Catoptrics (first century BC), also maintained that light travels with infinite speed.
– adopted the rule that light rays always travel between two points by the shortest path
• 1000 AD Arab philosopher ``Alhazan'' (Abu'ali al-hasan ibn al-haytham) first person to realize that light actually travels from the object seen to the eye
• 1676 Olaf Römer demonstrated that light must have a finite velocity, using his timings of the successive eclipses of the satellites of Jupiter
• 100-170AD Claudius Ptolemy rough law of refraction was studied experimentally (only close to normal incidence)
• 1621 Dutch mathematician Willebrord Snell comes up with correct law using sines. Does not publish
• 1637 French philosopher René Descartes was the first to publish, French countries call Snell’s law Descartes law of refraction
• 1658 French mathematician Pierre de Fermat demonstrated that all three of the laws of geometric optics can be accounted for on the assumption that light always travels between two points on the path which takes the least time
ECE 5616Curtis
The Ray Equation
Magnitude squared of change (gradient) in phase of wavefront equal index squared
2222
nzS
yS
xS
=⎟⎠⎞
⎜⎝⎛∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎠⎞
⎜⎝⎛∂∂
ECE 5616Curtis
RaysPhase fronts
A ray is a line drawn in space corresponding to the drive of flow of radiant energy
Rays are orthogonal trajectories of the wave-fronts
Light slows down bending wavefronts
ECE 5616Curtis
The eikonal equationAn equation for the evolution of a ray trajectory
ECE 5616Curtis
Eikonal equation continued
ECE 5616Curtis
Ray Trajectory with n constant ?
WOW !!!! Light travels in straight lines !!!!
ECE 5616Curtis
Example – Pinhole camera
ECE 5616Curtis
Pinhole Camera
ECE 5616Curtis
How small should pinhole be?A pinhole camera has no lens but uses a very small hole
some distance from the film/screen to produce an image. If we assume that light travels in straight lines, then the image of a distant point source will be a blur whose diameter is the same as the pinhole. However, diffraction will spread the beam into the Airy disk (remember last lecture). Assume that the sharpest picture is produced when the geometrical blur (hole size) is equal to central bright spot of the diffraction pattern – use diameter of first dark ring in Airy pattern. How large a hole should be used when the screen is 100mm from the hole, use λ=550nm ?
D = 1.22 λ/sinθ ~ 1.22 λ (2*100)/D
D ~ 0.366mm
ECE 5616Curtis
Fermat’s Principle
ECE 5616Curtis
Fermat’s Principle
Why do we care ?At an image point, all OPL must be equal
See Born and Wolf for deviation from MESee next 2 examples.
ECE 5616Curtis
Concave Mirror Example
From M&M
ECE 5616Curtis
Lifeguard Example
You could write the equations for the time for the running part and the swimming part and then set the derivative of the time with respect to x1 to 0.
Problem: Life guard is outside pool at point A. Someone is drowning in pool at point B. On ground life guard can run at speed vr and can swim in the pool at speed vs where vr>vs. Where should life guard enter pool to get to the person as fast as possible?
Or you could recognize that this is a critical angle problem – angle onto ground is 90 degrees. With Fermat’s principle the answer will be the same.
Eliminate ic from both equations and solve for x1
ECE 5616Curtis
Snell’s LawFrom Fermat’s Principle
)sin()sin()sin()sin(2211
2
2
1
1 θθθθ nnvv
=→=
Can rewrite as:
ECE 5616Curtis
QuestionsWhat does stationary phase say about light reflecting off an elliptical mirror between 2 focii?
If minimum time was the constraint – what would the path that light would take if there was a high index brick between the source and the detector ?
It is stationary but it is an inflection point.
Light would go right along the surface of the brick rather than through it !
ECE 5616Curtis
RadiometryReview of terminology & inverse square law
Inverse Square Law
ECE 5616Curtis
Radiometry
ECE 5616Curtis
Radiometry via Rays
ECE 5616Curtis
Example
• 100 W light bulb – what is E (W/m2) when observing the light bulb 2 meters away ?
100/R2 = 25 W/m2
• An antenna puts out 100 W it reflects off a mirror/surface that is 100% reflective, 2 meters away. What does the antenna measure the reflectance E (W/m2) to be ?
100/R2 = 25 W/m2 -> 25/R2 = 6.25 W/m2
ECE 5616Curtis
Example
• The irradiance measured at a plane in a circular symmetric system (on-axis otherwise potentially cos4 law can apply) is given by
• Where T is transmission of the system, L is radiance of object in π steradians, θ is the half angle subtended by the exit pupil
• See Smith page 260.
θπ 2sinLTE =
ECE 5616Curtis
Example
• Source is 10 W ster-1 m-2, T = 80% and angle of collection of system is 60 degrees total angle.
θπ 2sinLTE =
E = .8 *3.14*10*(.5)2 = 7.85W/m2
Note; It the source is small and you are looking off-axis by 45 degrees the detected amount is decreased potentially by factor of cos4(45) = .25
ECE 5616Curtis
Postulates of Geometrical Optics
ECE 5616Curtis
Sign Convention
• Light travels left to right• Distances to the right are +• Heights above the axis are +• The radius of a surface is + if its center of curvature is to the right• Focal lengths of converging elements are +• Light traveling in a – direction (right to left after a mirror) uses distances and indices that have their signs reversed if the following surface is to the left.• Ray angles measured from axis to ray are positive if counterclockwise. Refraction measured from surface normal and if CC is positive.
ECE 5616Curtis
Sign Convention Graphically
ECE 5616Curtis
Sign Convention Example Lens
From M&M
Positivel’, u, R, PQ
Negativeu’, α, l
I suggest drawing arrows
ECE 5616Curtis
Sign Convention Example Mirror
R O
P
WA
B
u
NegativeR,u,OW,A, n for A
Positiveu’,B,OP,n
u’
ECE 5616Curtis
Basic Definitions
ECE 5616Curtis
Basic Definitions
Infinite Conjugate
Object or Image at infinity
Finite Conjugate
Object and Image at finite distances
F
Converging Diverging
ECE 5616Curtis
QuestionYou have a system with mirrors, stops, diffractive elements, lenses, source, and detectors. What are some potential source of throughput losses ?
Reflection lossesAbsorption from elements and mirrorsFluorescence from elementsScatter Limited NA (angles system accepts) to capture source outputQuantum efficiency of detectorFill Factor of detectorNoise in detector – thermal or shot (have to get above this level)Vignetting of image from stops or lens aperturesDiffraction efficiency of diffractive elementsPotentially aberrations can cause system throughput lossesDepolarizing the light in a polarization sensitive system
ECE 5616Curtis
Reading
Chapter 2 W. Smith, Modern Optical Engineering