Chapter Physics

Chapter 31 Properties of Light Conceptual Problems 1 • [SSM] A ray of light reflects from a plane mirror. The angle between the incoming ray and the reflected ray is 70°. What is the angle of reflection? (a) 70°, (b) 140°, (c) 35°, (d) Not enough information is given to determine the reflection angle. Determine the Concept Because the angles of incidence and reflection are equal, their sum is 70° and the angle of reflection is 35°. ( )c is correct.

2 • A ray of light passes in air is incident on the surface of a piece of glass. The angle between the normal to the surface and the incident ray is 40°, and the angle between the normal and the refracted ray 28°. What is the angle between the incident ray and the refracted ray? (a) 12°, (b) 28°, (c) 40° ,(d) 68° Determine the Concept The angle between the incident ray and the refracted ray is the difference between the angle of incidence and the angle of refraction. ( )a is correct.

3 • During a physics experiment, you are measuring refractive indices of different transparent materials using a red helium-neon laser beam. For a given angle of incidence, the beam has an angle of refraction equal to 28° in material A and an angle of refraction equal to 26° in material B. Which material has the larger index of refraction? (a) A, (b) B, (c) The indices of refraction are the same. (d) You cannot determine the relative magnitudes of the indices of refraction from the data given. Determine the Concept The refractive index is a measure of the extent to which a material refracts light that passes through it. Because the angles of incidence are the same for materials A and B and the angle of refraction is smaller (more bending of the light) for material B, its index of refraction is larger than that of A. ( )b is correct.

4 • A ray of light passes from air into water, striking the surface of the water at an angle of incidence of 45º. Which, if any, of the following four quantities change as the light enters the water: (a) wavelength, (b) frequency, (c) speed of propagation, (d) direction of propagation, (e) none of the above?

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Determine the Concept When light passes from air into water its wavelength changes ( waterairwater nλλ = ), its speed changes ( waterwater ncv = ), and the direction of its propagation changes in accordance with Snell’s law. Its frequency does not change, so ( )a , ( )c and ( )d change. 5 • Earth’s atmosphere decreases in density as the altitude increases. As a consequence, the index of refraction of the atmosphere also decreases as altitude increases. Explain how one can see the Sun when it is below the horizon. (The horizon is the extension of a plane that is tangent to Earth’s surface.) Why does the setting Sun appear flattened? Determine the Concept The decrease in the index of refraction n of the atmosphere with altitude results in refraction of the light from the Sun, bending it toward the normal to the surface of Earth. Consequently, the Sun can be seen even after it is just below the horizon.

Earth

Atmosphere

6 • A physics student playing pocket billiards wants to strike her cue ball so that it hits a cushion and then hits the eight ball squarely. She chooses several points on the cushion and then measures the distances from each point to the cue ball and to the eight ball. She aims at the point for which the sum of these distances is least. (a) Will her cue ball hit the eight ball? (b) How is her method related to Fermat’s principle? Neglect any effects due to ball rotation. Determine the Concept (a) Yes. (b) Her procedure is based on Fermat’s principle of least time. The ball presumably bounces off the cushion with an angle of reflection equal to the angle of incidence, just as a light ray would do if the cushion were a mirror. The least time would also be the shortest distance of travel for the light ray.

7 • [SSM] A swimmer at point S in Figure 31-53 develops a leg cramp while swimming near the shore of a calm lake and calls for help. A lifeguard at point L hears the call. The lifeguard can run 9.0 m/s and swim 3.0 m/s. She knows physics and chooses a path that will take the least time to reach the swimmer. Which of the paths shown in the figure does the lifeguard take? Determine the Concept The path through point D is the path of least time. In analogy to the refraction of light, the ratio of the sine of the angle of incidence to the sine of the angle of refraction equals the ratio of the speeds of the lifeguard in

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each medium. Careful measurements from the figure show that path LDS is the path that best satisfies this criterion. 8 • Material A has a higher index of refraction than material B. Which material has the larger critical angle for total internal reflection when the material is in air? (a) A, (b) B, (c) The angles are the same. (d) You cannot compare the angles based on the data given. Determine the Concept Because the product of the index of refraction on the incident side of the interface and the sine of the critical angle is equal to one (the index of refraction of air), the material with the smaller index of refraction will have the larger critical angle. ( )b is correct.

9 • [SSM] A human eye perceives color using a structure which is called a cone that is is located on the retina. Three types of molecules compose these cones and each type of molecule absorbs either red, green, or blue light by resonance absorption. Use this fact to explain why the color of an object that appears blue in air appears blue underwater, in spite of the fact that the wavelength of the light is shortened in accordance with Equation 31-6. Determine the Concept In resonance absorption, the molecules respond to the frequency of the light through the Einstein photon relation E = hf. Neither the wavelength nor the frequency of the light within the eyeball depend on the index of refraction of the medium outside the eyeball. Thus, the color appears to be the same in spite of the fact that the wavelength has changed. 10 • Let θ be the angle between the transmission axes of two polarizing sheets. Unpolarized light of intensity I is incident upon the first sheet. What is the intensity of the light transmitted through both sheets? (a) I cos2 θ, (b) (I cos2 θ)/2, (c) (I cos2 θ)/4, (d) I cos θ, (e) (I cos θ)/4, (f) None of the above Picture the Problem The intensity of the light transmitted by the second polarizer is given by where ,II θ2

0trans cos= .21

0 II = Therefore, θ221

trans cosII =

and )(b is correct.

11 •• [SSM] Draw a diagram to explain how Polaroid sunglasses reduce glare from sunlight reflected from a smooth horizontal surface, such as the surface found on a pool of water. Your diagram should clearly indicate the direction of polarization of the light as it propagates from the Sun to the reflecting surface and then through the sunglasses into the eye. Determine the Concept The following diagram shows unpolarized light from the sun incident on the smooth surface at the polarizing angle for that particular

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surface. The reflected light is polarized perpendicular to the plane of incidence, i.e., in the horizontal direction. The sunglasses are shown in the correct orientation to pass vertically polarized light and block the reflected sunlight.

Smooth surface

Light fromthe sun

Polaroidsunglasses

θ θ

θr

PP

12 • Use the photon model of light to explain why, biologically, it is far less dangerous to stand in front of an intense beam of red light than a very weak beam of gamma radiation. HINT: Ionization of molecules in tissue can cause biological damage. Molecules absorb light in what form, wave or particle?

Determine the Concept Molecules require that a certain minimum energy be absorbed before they ionize. The red light photons contain considerably less energy than the gamma photons so, even though there are likely to be fewer photons in the gamma beam, each one is potentially dangerous. 13 • Three energy states of an atom are A, B and C. State B is 2.0 eV above state A and state C is 3.00 eV above state B. Which atomic transition results in the emission of the shortest wavelength of light? (a) B → A, (b) C → B, (c) C → A, (d) A→ C Determine the Concept Because the wavelength of the light emitted in an atomic transition is inversely proportional to the energy difference between the energy levels, the highest energy difference produces the shortest wavelength light. ( )c is correct.

14 • In Problem 13, if the atom is initially in state A, which transition results in the emission of the longest wavelength light? (a) A → B, (b) B → C, (c) A → C, (d) B→ A

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Determine the Concept Because the energy required to induce an atomic transition varies inversely with the wavelength of the light that must be absorbed to induce the transition and the transition from A to B is the lowest energy transition, the transition B→ A results in the longest wavelength light. ( )d is correct. 15 • [SSM] What role does the helium play in a helium–neon laser? Determine the Concept The population inversion between the state E2,Ne and the state 1.96 eV below it (see Figure 31-51) is achieved by inelastic collisions between neon atoms and helium atoms excited to the state E2,He. 16 • When a beam of visible white light that passes through a gas of atomic hydrogen at room temperature is viewed with a spectroscope, dark lines are observed at the wavelengths of the hydrogen atom emission series. The atoms that participate in the resonance absorption then emit light of the same wavelength as they return to the ground state. Explain why the observed spectrum nevertheless exhibits pronounced dark lines. Determine the Concept Although the excited atoms emit light of the same frequency on returning to the ground state, the light is emitted in a random direction, not exclusively in the direction of the incident beam. Consequently, the beam intensity is greatly diminished at this frequency. 17 • [SSM] Which of the following types of light would have the highest energy photons? (a) red (b) infrared (c) blue (d) ultraviolet Determine the Concept The energy of a photon is directly proportional to the frequency of the light and inversely proportional to its wavelength. Of the portions of the electromagnetic spectrum include in the list of answers, ultraviolet light has the highest frequency. ( )d is correct.

Estimation and Approximation 18 • Estimate the time required for light to make the round trip during Galileo’s experiment to measure the speed of light. Compare the time of the round trip to typical human response times. How accurate do you think this experiment is? Picture the Problem We can use the distance, rate, and time relationship to estimate the time required to travel 6 km.

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Express the distance d the light in Galileo’s experiment traveled in terms of its speed c and the elapsed time Δt:

tcd Δ= ⇒cdt =Δ

Substitute numerical values and evaluate Δt:

s 102m/s10998.2

km6Δ 58

−×=×

=t

Because human reaction is approximately 0.3 s:

45

reaction 102s102

s 3.0Δ

Δ×≈

×= −t

t

or ( ) tt Δ102Δ 4reaction ×≈

Because human reaction time is so much longer than the travel time for the light, there was no way that Galileo’s experiment could demonstrate that the speed of light was not infinite. 19 • Estimate the time delay in receiving a light on your retina when you are wearing eyeglasses compared to when you are not wearing your eyeglasses. Determine the Concept We’ll assume that the source of the photon is a distance L from your retina and express the difference in the photon’s travel time when you are wearing your glasses. Let the thickness of your glasses by 2 mm and the index of refraction of the material from which they are constructed by 1.5. When you are not wearing your glasses, the time required for a light photon, originating a distance L away, to reach your retina is given by:

cLt =0Δ

If glass of thickness d and index of refraction n is inserted in the path of the photon, its travel time becomes: ( ) ( )

cdnt

cdnL

ncd

cdLttt

1Δ1

Δ

0

glassin airin

−+=

−+=

+−

=+=

The time delay is the difference between Δt and Δt0:

( )c

dnttt 1ΔΔ 0delay−

=−=

Substitute numerical values and evaluate tdelay:

( )( ) ps 3m/s10998.2

mm 215.18delay ≈

×−

=t

20 •• Estimate the number of photons that enter your eye if you look for a tenth of a second at the Sun. What energy is absorbed by your eye during that

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time, assuming that all the photons are absorbed? The total power output of the Sun is 4.2 × 1026 W. Picture the Problem The rate at which photons enter your eye is the ratio of power incident on your pupil to the energy per photon. We’ll assume that the electromagnetic radiation from the Sun is at 550 nm and that, therefore, its photons have energy (given by λhchfE == ) of 2.25 eV. The rate at which photons enter your eye is the ratio of the rate at which energy is incident on your pupil to the energy carried by each photon:

photonper

pupilon incident

E

P

dtdN

= (1)

Sun thefrom distancesEarth'at sphere

Sun

pupil

pupilon incident

Sun AP

A

PI ==

The intensity of the radiation from the Sun is given by:

Sun


pupil

pupilon incident P

AA

P = (2)

Solving for yields: pupilon

incident P

Substituting in equation (1) yields:

photonper

Sun


pupil

EP

AA

dtdN

=

Substitute for the two areas and simplify to obtain:

photonper

Sun

2

-SunEarth

pupil

photonper

Sun2

-SunEarth

2pupil4

1

4

4

EP

Rd

EP

Rd

dtdN

⎟⎟⎠

⎞⎜⎜⎝

⎛=

=ππ

Substitute numerical values and evaluate dN/dt:

( )115

19

262

11 s 10237.3

eVJ 101.602eV 25.2

W102.4m 1050.14

mm 1 −− ×=

××

×⎟⎟⎠

⎞⎜⎜⎝

⎛×

=dtdN

or, separating variables and integrating this expression,

( )tN 115 s 10237.3 −×=

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2896

Evaluating N for t = 0.1 s yields:

( ) ( )( )photons 103

s 1.0s 10237.3s 1.014

115

×≈

×= −N

The energy deposited, assuming all the photons are absorbed, is the product of the rate at which energy is incident on the pupil and the time during which it is delivered:

tPEpupilon

incident =

Substituting for from

equation (2) yields: pupilon

incident PtP

AA

E Sun


pupil=

Substitute for the two areas and simplify to obtain:

tPRd

tPR

dE

Sun

2

-SunEarth

pupil

Sun2-SunEarth

2pupil4

1

4

4

⎟⎟⎠

⎞⎜⎜⎝

⎛=

=ππ

Substitute numerical values and evaluate E for t = 0.1 s:

( ) ( ) ( )( ) mJ 0.1mJ 1167.0s 1.0 W102.4m 1050.14

mm 1s 1.0 262

11 ≈=×⎟⎟⎠

⎞⎜⎜⎝

⎛×

=E

21 •• Römer was observing the eclipses of Jupiter’s moon Io with the hope that they would serve as a highly accurate clock that would be independent of longitude. (Prior to GPS, such a clock was needed for accurate navigation.) Io eclipses (enters the umbra of Jupiter’s shadow) every 42.5 h. Assuming an eclipse of Io is observed on Earth on June 1 at midnight when Earth is at location A (as shown in Figure 31-54), predict the expected time of observation of an eclipse one-quarter of a year later when Earth is at location B, assuming (a) the speed of light is infinite and (b) the speed of light is 2.998 × 108 m/s. Picture the Problem We can use the period of Io’s motion and the position of the earth at B to find the number of eclipses of Io during Earth’s movement and then use this information to find the number of days before a night-time eclipse. During the 42.5 h between eclipses of Jupiter’s moon, Earth moves from A to B, increasing the distance from Jupiter by approximately the distance from Earth to the Sun, making the path for the light longer and introducing a delay in the onset of the eclipse.

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(a) Find the time it takes Earth to travel from point A to point B:

h4.2191d

h244

d24.3654

earth

=

×==→

Tt BA

Because there are 42.5 h between eclipses of Io, the number of eclipses N occurring in the time it takes for the earth to move from A to B is:

56.51h42.5

h4.2191

Io

=== →

TtN BA

Hence, in one-fourth of a year, there will be 51.56 eclipses. Because we want to find the next occurrence that happens in the evening hours, we’ll use 52 as the number of eclipses. We’ll also assume that Jupiter is visible so that the eclipse of Io can be observed at the time we determine. Relate the time t(N) at which the Nth eclipse occurs to N and the period TIo of Io:

( ) IoNTNt =

Evaluate t(52) to obtain: ( ) ( )

d083.92h24

d1h5.425252

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×=t

Subtract the number of whole days to find the clock time t:

( )

a.m.00:2

h992.1d

h24d0.083

d92d92.083d9252

≈

=×=

−=−= tt

Because June, July, and August have 30, 31, and 31 d, respectively, the date is:

1September

(b) Express the time delay Δt in the arrival of light from Io due to Earth’s location at B:

crt sun-earth=Δ

Substitute numerical values and evaluate Δt: min34.8

s 60min 1s500

m/s10998.2m105.1

8

11

=

×=××

=Δt

Hence, the eclipse will actually occur at 2:08 a.m., September 1

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22 •• If the angle of incidence is small enough, the small angle approximation sin θ ≈ θ may be used to simplify Snell’s law of refraction. Determine the maximum value of the angle that would make the value for the angle differ by no more than one percent from the value for the sine of the angle. (This approximation will be used in connection with image formation by spherical surfaces in Chapter 32.) Picture the Problem We can express the relative error in using the small angle approximation and then either use 1) trial-and-error methods, 2) a spreadsheet program, or 3) the Solver capability of a scientific calculator to solve the transcendental equation that results from setting the error function equal to 0.01. Express the relative error δ in using the small angle approximation:

( ) 1sinsin

sin−=

−=

θθ

θθθθδ

A spreadsheet program was used to plot the following graph of δ (θ ).

0.000

0.002

0.004

0.006

0.008

0.010

0.012

0.014

0.016

0.00 0.05 0.10 0.15 0.20 0.25 0.30

theta (radians)

delta

(thet

a)

From the graph, we can see that δ (θ ) < 1% for θ ≤ 0.24 radians. In degree measure, °≤ 14θ

Remarks: Using the Solver program on a TI-85 gave θ = 0.244 radians. The Speed of Light 23 • Mission Control sends a brief wake-up call to astronauts in a spaceship that is far from Earth. 5.0 s after the call is sent, Mission Control can hear the groans of the astronauts. How far from Earth is the spaceship? (a) 7.5 × 108 m, (b) 15 × 108 m, (c) 30 × 108 m, (d) 45 × 108 m, (e) The spaceship is on the moon.

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Picture the Problem We can use the distance, rate, and time relationship to find the distance to the spaceship.

Relate the distance d to the spaceship to the speed of electromagnetic radiation in a vacuum and to the time for the message to reach the astronauts:

tcd Δ=

( )( )m105.7

s5.2m/s10998.28

8

×=

×=d

and )(a is correct.

Noting that the time for the message to reach the astronauts is half the time for Mission Control to hear their response, substitute numerical values and evaluate d: 24 • The distance from a point on the surface of Earth to a point on the surface of the moon is measured by aiming a laser light beam at a reflector on the surface of the moon and measuring the time required for the light to make a round trip. The uncertainty in the measured distance Δx is related to the uncertainty in the measured time Δt by Δx = 1

2 cΔt. If the time intervals can be measured to ±1.00 ns, (a) find the uncertainty of the distance. (b) Estimate the percentage uncertainty in the distance. Picture the Problem We can use the given information that the uncertainty in the measured distance Δx is related to the uncertainty in the time Δt by Δx = cΔt to evaluate Δx.

(a) The uncertainty in the distance is:

tcx ΔΔ 21=

Substitute numerical values and evaluate Δx:

( )( )cm0.15

ns00.1m/s10998.2Δ 821

±=

±×=x

(b)The percent uncertainty in the distance to the Moon is: %10

m 103.84cm 0.15Δ

8

8Moon Earth to

Moon Earth to

−≈

×=

xx

25 •• [SSM] Ole Römer discovered the finiteness of the speed of light by observing Jupiter’s moons. Approximately how sensitive would the timing apparatus need to be in order to detect a shift in the predicted time of the moon’s eclipses that occur when the moon happens to be at perigee (3.63×105 km ) and those that occur when the moon is at apogee ( 4.06 ×105 km )? Assume that an

Chapter 31

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instrument should be able to measure to at least one-tenth the magnitude of the effect it is to measure. Picture the Problem His timing apparatus would need to be sensitive enough to measure the difference in times for light to travel to Earth when the moon is at perigee and at apogee. The sensitivity of the timing apparatus would need to be one-tenth of the difference in time for light to reach Earth from the two positions of the moon of Jupiter:

tΔy Sensitivit 101=

where Δt is the time required for light to travel between the two positions of the moon.

The time required for light to travel between the two positions of the moon is given by:

c

ddt perigeeat

moonapogeeat

moon

Δ−

=

c

dd

10y Sensitivit perigeeat

moon apogeeat

moon −=

Substituting for Δt yields:

Substitute numerical values and evaluate the required sensitivity:

( )( )ms 14

m/s 10998.210km 1063.3km 1006.4Δ 8

55

=

××−×

=t

Remarks: Instruments with this sensitivity did not exist in the 17th century. Reflection and Refraction 26 • Calculate the fraction of light energy reflected from an air–water interface at normal incidence. Picture the Problem Use the equation relating the intensity of reflected light at normal incidence to the intensity of the incident light and the indices of refraction of the media on either side of the interface.

Express the intensity I of the light reflected from an air-water interface at normal incidence in terms of the indices of refraction and the intensity I0 of the incident light:

0

2

waterair

waterair InnnnI ⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

=

Solve for the ratio I/I0:

2

waterair

waterair

0⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=nnnn

II

Properties of Light

2901

Substitute numerical values and evaluate I/I0: %0.2

33.100.133.100.1 2

0

=⎟⎠⎞

⎜⎝⎛

+−

=II

27 • A ray of light is incident on one of a pair of mirrors set at right angles to each A ray of light is incident on one of two mirrors that are set at right angles to each other. The plane of incidence is perpendicular to both mirrors. Show that after reflecting from each mirror, the ray will emerge traveling in the direction opposite to the incident direction, regardless of the angle of incidence. Picture the Problem The diagram shows ray 1 incident on the vertical surface at an angle θ1, reflected as ray 2, and incident on the horizontal surface at an angle of incidence θ3. We’ll prove that rays 1 and 3 are parallel by showing that θ1 = θ4, i.e., by showing that they make equal angles with the horizontal. Note that the law of reflection has been used in identifying equal angles of incidence and reflection.

1

2

3

θ

θ

θθ θ

θ

1

1

2

33

4

We know that the angles of the right triangle formed by ray 2 and the two mirror surfaces add up to 180°:

( ) °=−°+°+ 1809090 12 θθ or

21 θθ =

The sum of θ2 and θ3 is 90°:

23 90 θθ −°=

Because 21 θθ = : 13 90 θθ −°=

The sum of θ4 and θ3 is 90°: °=+ 9043 θθ

Substitute for θ3 to obtain: ( ) °=+−° 9090 41 θθ ⇒ 41 θθ =

28 •• (a) A ray of light in air is incident on an air–water interface. Using a spreadsheet or graphing program, plot the angle of refraction as a function of the angle of incidence from 0º to 90º. (b) Repeat Part (a), but for a ray of light in water that is incident on a water–air interface. [For Part (b), there is no reflected ray for angles of incidence that are greater than the critical angle.]

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Picture the Problem Diagrams showing the light rays for the two cases are shown below. In (a) the light travels from air into water and in (b) it travels from water into air.

(a)

Air

Water

θ

θ

1

2

1n

2n

(b)

Air

Water1n

2n

θ

θ

1

2

2211 sinsin θθ nn = where the angles of incidence and refraction are θ1 and θ2, respectively.

(a) Apply Snell’s law to the air-water interface to obtain:

Solving for θ2 yields: ⎟⎟

⎠

⎞⎜⎜⎝

⎛= −

12

112 sinsin θθ

nn

A spreadsheet program to graph θ2 as a function of θ1 is shown below. The formulas used to calculate the quantities in the columns are as follows:

Cell Content/Formula Algebraic Form n1B1 1 n2B2 1.33333

A6 0 θ1 (deg) A7 A6 + 5 θ1 + Δθ B6 A6*PI()/180 π

1801θ ×

C6 ASIN(($B$1/$B$2)*SIN(B6))⎟⎟⎠

⎞⎜⎜⎝

⎛−1

2

11 sinsin θnn

D6 C6*180/PI() π

θ 1802 ×

A B C D

1 n1= 1 2 n2= 1.33333

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3 4 θ1 θ1 θ2 θ25 (deg) (rad) (rad) (deg) 6 0 0.00 0.000 0.00 7 1 0.02 0.013 0.75 8 2 0.03 0.026 1.50 9 3 0.05 0.039 2.25

21 87 1.52 0.847 48.50 22 88 1.54 0.847 48.55 23 89 1.55 0.848 48.58 24 90 1.57 0.848 48.59

A graph of θ2 as a function of θ1 follows:

05

101520253035404550

0 10 20 30 40 50 60 70 80 90

Angle of incidence, deg

Ang

le o

f ref

ract

ion,

deg

(b) Change the contents of cell B1 to 1.33333 and the contents of cell B2 to 1 to obtain the following graph:

0102030405060708090

0 10 20 30 40 50

Angle of incidence, deg

Ang

le o

f ref

ract

ion,

deg

Note that as the angle of incidence approaches the critical angle for a water-air interface (48.6°), the angle of refraction approaches 90°.

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29 •• The red light from a helium-neon laser has a wavelength of 632.8 nm in air. Find the (a) speed, (b) wavelength, and (c) frequency of helium-neon laser light in air, water, and glass. (The glass has an index of refraction equal to 1.50.) Picture the Problem We can use the definition of the index of refraction to find the speed of light in the three media. The wavelength of the light in each medium is its wavelength in air divided by the index of refraction of the medium. The frequency of the helium-neon laser light is the same in all media and is equal to its value in air. The wavelength of helium-neon laser light in air is 632.8 nm.

The speed of light in a medium whose index of refraction is n is given by:

ncv = (1)

The wavelength of light in a medium whose index of refraction is n is given by:

nair

nλ

λ = (2)

The frequency of the light is equal to its frequency in air independently of the medium in which the light is propagating:

Hz 1074.4nm 632.8

m/s 10998.2

14

8

×=

×==

λcf

Substitute numerical values in equation (1) and evaluate vwater:

m/s1025.233.1

m/s10998.2

8

8

water

×=

×=v

Substitute numerical values in equation (2) and evaluate λwater:

nm 47633.1

nm 8.632water ==λ

The other speeds and wavelengths are found similarly and are summarized in the following table:

(a) speed (b) wavelength (c) frequency (m/s) (nm) (Hz)

Air 81000.3 × 141074.4 ×633 Water 81025.2 × 141074.4 ×476 glass 81000.2 × 141074.4 ×422

Properties of Light

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30 •• The index of refraction for silicate flint glass is 1.66 for violet light that has a wavelength in air equal to 400 nm and 1.61 for red light that has a wavelength in air equal to 700 nm. A ray of 700-nm–wavelength red light and a ray of 400-nm-wavelength violet light both have angles of refraction equal to 30º upon entering the glass from air. (a) Which is greater, the angle of incidence of the ray of red light or the angle of incidence of the ray of violet light? Explain your answer. (b) What is the difference between the angles of incidence of the two rays? Picture the Problem Let the subscript 1 refer to the air and the subscript 2 to the silicate glass and apply Snell’s law to the air-glass interface.

(a) Because the index of refraction for violet light is larger than that of red light, for a given incident angle violet light would refract more than red light. Thus to exhibit the same refraction angle, violet light would require an angle of incidence larger than that of red light. (b) Express the difference in their angles of incidence:

red1,violet1,Δ θθθ −= (1)

Apply Snell’s law to the air-glass interface to obtain:

2211 sinsin θθ nn =

Solving for θ1 yields: ⎟⎟⎠

⎞⎜⎜⎝

⎛= −

21

211 sinsin θθ

nn

Substitute for violet1,θ and red1,θ in equation (1) to obtain:

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛= −−

red ,2air

red1 violet,2

air

violet1 sinsinsinsinΔ θθθnn

nn

For °== 30red 2, violet2, θθ :

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛°−⎟⎟

⎠

⎞⎜⎜⎝

⎛°= −−−−

air

red1

air

violet1

air

red1

air

violet1

2sin

2sin30sinsin30sinsinΔ

nn

nn

nn

nn

θ

Substitute numerical values and evaluate Δθ:

( ) ( ) °=°−°=⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛= −− 49.261.5310.56

00.1261.1sin

00.1266.1sinΔ 11θ

Chapter 31

2906

Remarks: Note that Δθ is positive. This means that the angle for violet light is greater than that for red light and confirms our answer in Part (a). 31 •• [SSM] A slab of glass that has an index of refraction of 1.50 is submerged in water that has an index of refraction of 1.33. Light in the water is incident on the glass. Find the angle of refraction if the angle of incidence is (a) 60º, (b) 45º, and (c) 30º. Picture the Problem Let the subscript 1 refer to the water and the subscript 2 to the glass and apply Snell’s law to the water-glass interface.

Apply Snell’s law to the water-glass interface to obtain:



⎞⎜⎜⎝

⎛= −

12

112 sinsin θθ

nn

(a) Evaluate θ2 for θ1 = 60°:

°=⎟⎠⎞

⎜⎝⎛ °= − 5060sin

50.133.1sin 1

2θ

(b) Evaluate θ2 for θ1 = 45°:

°=⎟⎠⎞

⎜⎝⎛ °= − 3945sin

50.133.1sin 1

2θ

(c) Evaluate θ2 for θ1 = 30°:

°=⎟⎠⎞

⎜⎝⎛ °= − 2630sin

50.133.1sin 1

2θ

32 •• Repeat Problem 31 for a beam of light initially in the glass that is incident on the glass–water interface at the same angles. Picture the Problem Let the subscript 1 refer to the glass and the subscript 2 to the water and apply Snell’s law to the glass-water interface.

Apply Snell’s law to the water-glass interface to obtain:



⎞⎜⎜⎝

⎛= −

12

112 sinsin θθ

nn

(a) Evaluate θ2 for θ1 = 60°:

°=⎟⎠⎞

⎜⎝⎛ °= − 7860sin

33.150.1sin 1

2θ

Properties of Light

2907

(b) Evaluate θ2 for θ1 = 45°:

°=⎟⎠⎞

⎜⎝⎛ °= − 5345sin

33.150.1sin 1

2θ

(c) Evaluate θ2 for θ1 = 30°:

°=⎟⎠⎞

⎜⎝⎛ °= − 3430sin

33.150.1sin 1

2θ

33 •• A beam of light in air strikes a glass slab at normal incidence. The glass slab has an index of refraction of 1.50. (a) Approximately what percentage of the incident light intensity is transmitted through the slab (in one side and out the other)? (b) Repeat Part (a) if the glass slab is immersed in water. Picture the Problem Let the subscript 1 refer to the medium to the left (air) of the first interface, the subscript 2 to glass, and the subscript 3 to the medium (air) to the right of the second interface. Apply the equation relating the intensity of reflected light at normal incidence to the intensity of the incident light and the indices of refraction of the media on either side of the interface to both interfaces. We’ll neglect multiple reflections at glass-air interfaces.

1I

1r,I 2r,I

2I 3I

00.11 =n 50.12 =n 00.13 =n

(a) Express the intensity of the transmitted light in the second medium:

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−=−=

2

21

211

1

2

21

211r,112

1nnnnI

Innnn

IIII

Express the intensity of the transmitted light in the third medium:

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−=−=

2

32

322

2

2

32

322r,223

1nnnn

I

Innnn

IIII

Substitute for I2 to obtain:

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−=2

32

322

21

2113 11

nnnn

nnnnII

Chapter 31

2908

Solve for the ratio I3/I1 to obtain:

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−=2

32

322

21

21

1

3 11nnnn

nnnn

II

Substitute numerical values and evaluate I3/I1:

%9200.150.100.150.11

50.100.150.100.11

22

1

3 =⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

+−

−⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

+−

−=II

(b) With n1 = n3 = 1.33:

%9933.150.133.150.11

50.133.150.133.11

22

1

3 =⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

+−

−⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

+−

−=II

34 •• This problem is a refraction analogy. A band is marching down a football field with a constant speed v1. About midfield, the band comes to a section of muddy ground that has a sharp boundary making an angle of 30º with the 50-yd line, as shown in Figure 31-55. In the mud, each marcher moves at a speed equal to

12 v1 in a direction perpendicular to the row of markers they are in.

(a) Diagram how each line of marchers is bent as it encounters the muddy section of the field so that the band is eventually marching in a different direction. Indicate the original direction by a ray and the final direction by a second ray. (b) Find the angles between these rays and the line normal to the boundary. Is their direction of motion ″bent″ toward the normal or away from it? Explain your answer in terms of refraction. Picture the Problem We can apply Snell’s law to find the angle of refraction of the line of marchers as they enter the muddy section of the field (a)

°30

θ

θ

1

2

(b) Apply Snell’s law at the interface to obtain:


Properties of Light

2909

Solving for θ2 yields: ⎥⎦

⎤⎢⎣

⎡= −

12

112 sinsin θθ

nn

The ratio of the indices of refraction is the reciprocal of the ratio of the speeds of the marchers in the two media:

21

1

121

1

2

2

1

2

1 ====vv

vv

vvvv

nn

Because the left and right sides of the 30° angle and θ1 are mutually perpendicular, θ1 = 30°. Substitute numerical values and evaluate θ2:

[ ] °=°= − 1430sinsin 211

2θ

As the line enters the muddy field, its speed is reduced by half and the direction of the forward motion of the line is changed. In this case, the forward motion in the muddy field makes an angle θ2 with respect to the normal to the boundary line. Note that the separation between successive lines in the muddy field is half that in the dry field. 35 •• [SSM] In Figure 31-56, light is initially in a medium that has an index of refraction n1. It is incident at angle θ1 on the surface of a liquid that has an index of refraction n2. The light passes through the layer of liquid and enters glass that has an index of refraction n3. If θ3 is the angle of refraction in the glass, show that n1 sin θ1 = n3 sin θ3. That is, show that the second medium can be neglected when finding the angle of refraction in the third medium. Picture the Problem We can apply Snell’s law consecutively, first to the n1-n2 interface and then to the n2-n3 interface. Apply Snell’s law to the n1-n2 interface:


Apply Snell’s law to the n2-n3 interface:


Equate the two expressions for 22 sinθn to obtain:


36 •• On a safari, you are spear fishing while wading in a river. You observe a fish gliding by you. If your line of sight to the fish is 64.0o degrees below the horizontal in air, and assuming the spear follows a straight-line path through the air and water after it is released, determine the angle below the horizontal that you

Chapter 31

2910

should aim your spear gun in order to catch dinner. Assume the spear gun barrel is 1.50 m above the water surface, the fish is 1.20 m below the surface, and the spear travels in a straight line all the way to the fish. Picture the Problem The following pictorial representation summarizes the information given in the problem statement. We can use the geometry of the diagram and apply Snell’s law at the air-water interface to find the aiming angle α.

θ

θ

1

2

α°0.64

m 50.1

=h

m 20.1

=d

lL

AirWater

1n

2n

Use the pictorial representation to express the aiming angle α:

⎥⎦⎤

⎢⎣⎡

++

= −

lLdh1tanα

Referring to the diagram, note that:

1tanθhL = and 2tanθd=l

Substituting for L and l yields: ⎥⎦

⎤⎢⎣

⎡++

= −

21

1

tantantan

θθα

dhdh

Apply Snell’s law at the air-water interface to obtain:


Solving for θ2 yields: ⎥

⎦

⎤⎢⎣

⎡= −

12

112 sinsin θθ

nn

Properties of Light

2911

Substitute for θ2 to obtain:

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛⎥⎦

⎤⎢⎣

⎡+

+=

−

−

12

111

1

sinsintantantan

θθα

nndh

dh

Noting that θ1 is the complement of 64.0°, substitute numerical values and evaluate α:

( ) ( )°=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛⎥⎦⎤

⎢⎣⎡ °+°

+=

−

− 9.660.26sin

33.100.1sintanm 20.10.26tanm 50.1

m 20.1m 50.1tan1

1α

That is, you should aim 66.9° below the horizontal. 37 ••• You are standing on the edge of a swimming pool and looking directly across at the opposite side. You notice that the bottom edge of the opposite side of the pool appears to be at an angle of 28° degrees below the horizontal. However, when you sit on the pool edge, the bottom edge of the opposite side of the pool appears to be at an angle of only 14o below the horizontal. Use these observations to determine the width and depth of the pool. Hint: You will need to estimate the height of your eyes above the surface of the water when standing and sitting. Picture the Problem The following diagrams represent the situations when you are standing on the edge of the pool (the diagram to the left) and when you are sitting on the edge of the pool (the diagram to the right). We can use Snell’s law and the geometry of the pool to determine the width and depth of the pool.

θ

θ

1

2

α

h

L l

d

Standing

WaterAir

2n1nstandingh

θ

θ

1

2

α

h

L l

d

Sitting

'

'

' '

'Air

Water1n

2n

sittingh

Chapter 31

2912

°=°−°=−°= 622890901 αθ and

°=°−°=−°= 761490'90'1 αθ

Use the fact that angles α and θ1 are complementary, as are α′ and θ1′ to determine θ1 and θ1′:

1standing tanθhL =

and 'tan' 1sitting θhL =

Express the distances L and L′ in terms of θ1 and θ1′:

( ) m 197.362tanm 7.1 =°=L and

( ) m 808.276tanm 7.0' =°=L

Assuming that your eyes are 1.7 m above the level of the water when you are standing and 0.7 m above the water when you are sitting, evaluate L and L′: Referring to the pictorial representations, note that:

hLd

h−

==l

2tanθ (1)

and

hLd

h'''tan 2

−==

lθ

Divide the first of these equations by the second to obtain:

''tantan

2

2

LdLd

−−

=θθ

Solving for d yields: 'tantan

'tantan'

22

22

θθθθ

−−

=LLd (2)

Apply Snell’s law to the air-water interface when you are standing:

⎥⎦

⎤⎢⎣

⎡= −

12

112 sinsin θθ

nn

Substitute numerical values and evaluate 2θ :

°=⎥⎦⎤

⎢⎣⎡ °= − 60.4162sin

33.100.1sin 1

2θ

Apply Snell’s law to the air-water interface when you are sitting:

⎥⎦

⎤⎢⎣

⎡= − 'sinsin' 1

2

112 θθ

nn

°=⎥⎦⎤

⎢⎣⎡ °= − 85.4676sin

33.100.1sin' 1

2θ

Substitute numerical values and evaluate '2θ :

Properties of Light

2913

Substitute numerical values in equation (2) and evaluate d:

( ) ( ) widem 5.1m 130.585.46tan60.41tan

85.46tanm 197.360.41tanm 808.2==

°−°°−°

=d

Solving equation (1) for h yields:

2tanθLdh −

=

Substitute numerical values and evaluate h:

deep m 2.260.41tan

m 197.3m 130.5=

°−

=h

38 ••• Figure 31-57 shows a beam of light incident on a glass plate of thickness d and index of refraction n. (a) Find the angle of incidence so that the separation b between the ray reflected from the top surface and the ray reflected from the bottom surface and exiting the top surface is a maximum. (b) What is this angle of incidence if the index of refraction of the glass is 1.60? (c) What is the separation of the two beams if the thickness of the glass plate is 4.0 cm? Picture the Problem Let x be the perpendicular separation between the two rays and let l be the separation between the points of emergence of the two rays on the glass surface. We can use the geometry of the refracted and reflected rays to express x as a function of l, d, θr, and θi. Setting the derivative of the resulting equation equal to zero will yield the value of θi that maximizes x. Air

Glass

Airθ θ

θ

θ

θ

r

r

i ii

l

d

x

(a) Express l in terms of d and the angle of refraction θr:

rtan2 θd=l

Express x as a function of l, d, θr, and θi:

ir costan2 θθdx =

Differentiate x with respect toθi:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛+−==

i

rir

2irir

ii

cossecsintan2costan2θθ

θθθθθθθθ d

ddddd

ddx (1)

Chapter 31

2914

Apply Snell’s law to the air-glass interface:

r2i1 sinsin θθ nn = (2) or, because n1 = 1 and n2 = n,

ri sinsin θθ n=

rrii coscos θθθθ dnd = or

r

i

i

r

coscos1

θθ

θθ

ndd

=

Differentiate implicitly with respect toθI to obtain:

Substitute in equation (1) to obtain:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−=

r

ir

r3

i2

r

i

r2

ii

r

r

i cossinsin

coscos12

coscos

coscos1sin

cossin2

θθθ

θθ

θθ

θθ

θθθ

θ nd

nd

ddx

Substitute for

and

i2sin1 θ− i

2cos θ

isin1 θn

for rsinθ to obtain:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

r

i2

r3

i2

i cossin

cossin12

θθ

θθ

θ nnd

ddx

Multiply the second term in parentheses by r2

r2 coscos θθ and simplify to

obtain:

( )r2

i2

i2

r3

r3

r2

i2

r3

i2

i

cossinsin1cos

2cos

cossincossin1

2 θθθθθ

θθθθ

θ−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=

nd

nnd

ddx

Substitute for : r

2sin1 θ− r2cos θ

( )[ ]r2

i2

i2

r3

i

sin1sinsin1cos

2 θθθθθ

−−−=n

dddx

Substitute isin1 θn

for rsinθ to obtain:

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −−−= i

22i

2i

2

r3

i

sin11sinsin1cos

2 θθθθθ nn

dddx

Factor out 1/n2, simplify, and set equal to zero to obtain:

[ ] extremafor 0sin2sincos2 2

i22

i4

r33

i

=+−= nnn

dddx θθ

θθ

Properties of Light

2915

If dx/dθ1 = 0, then it must be true that:

0sin2sin 2i

22i

4 =+− nn θθ

Solve this quartic equation for θi to obtain: ⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛−−= −

21

i1

11sinn

nθ

(b) Evaluate θI for n = 1.60:

( )°=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−−= −

5.48

60.11116.1sin 2

1iθ

(c) In (a) we showed that:

ir costan2 θθdx =

Solve equation (2) for θr: ⎟⎟⎠

⎞⎜⎜⎝

⎛= −

i2

11r sinsin θθ

nn

Substitute numerical values and evaluate θr:

°=⎟⎠⎞

⎜⎝⎛ °= − 9.275.48sin

60.11sin 1

rθ

Substitute numerical values and evaluate x:

( )cm8.2

5.48cos9.27tancm0.42

=

°°=x

Total Internal Reflection 39 • [SSM] What is the critical angle for light traveling in water that is incident on a water–air interface? Picture the Problem Let the subscript 1 refer to the water and the subscript 2 to the air and use Snell’s law under total internal reflection conditions.

Use Snell’s law to obtain:


When there is total internal reflection:

c1 θθ = and °= 902θ

Substitute to obtain: 22c1 90sinsin nnn =°=θ

Solving for θc yields: ⎟⎟⎠

⎞⎜⎜⎝

⎛= −

1

21c sin

nnθ

Chapter 31

2916

Substitute numerical values and evaluate θc:

°=⎟⎠⎞

⎜⎝⎛= − 8.48

33.100.1sin 1

cθ

40 • A glass surface (n = 1.50) has a layer of water (n = 1.33) on it. Light in the glass is incident on the glass–water interface. Find the critical angle for total internal reflection. Picture the Problem Let the index of refraction of glass be represented by n1 and the index of refraction of water by n2 and apply Snell’s law to the glass-water interface under total internal reflection conditions.

GlassWater

50.11 =n

33.12 =n

θ

θ

1

2

Apply Snell’s law to the glass-water interface:


At the critical angle, θ1 = θc and θ2 = 90°:

°= 90sinsin 2c1 nn θ

Solve for θc: ⎥

⎦

⎤⎢⎣

⎡°= − 90sinsin

1

21c n

nθ


°=⎥⎦⎤

⎢⎣⎡ °= − 5.6290sin

50.133.1sin 1

cθ

41 • A point source of light is located 5.0 m below the surface of a large pool of water. Find the area of the largest circle on the pool’s surface through which light coming directly from the source can emerge. Picture the Problem We can apply Snell’s law to the water-air interface to express the critical angle θc in terms of the indices of refraction of water (n1) and air (n2) and then relate the radius of the circle to the depth d of the point source and θc.

90º

Air

Water33.11 =n

00.12 =n

m 0.5=d

r

θ

θ

c

c

Express the area of the circle whose radius is r:

2rA π=

Properties of Light

2917

Relate the radius of the circle to the depth d of the point source and the critical angle θc:

ctanθdr =

Apply Snell’s law to the water-air interface to obtain:

22c1 90sinsin nnn =°=θ

Solving for θc yields: ⎟⎟

⎠

⎞⎜⎜⎝

⎛= −

1

21c sin

nnθ

Substitute for r and θc to obtain:

[ ]2

1

21

2c

sintan

tan

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

⎭⎬⎫

⎩⎨⎧

=

=

−

nnd

dA

π

θπ

Substitute numerical values and evaluate A: ( )

22

2

1

m100.1

33.11sintanm0.5

×=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

⎭⎬⎫

⎩⎨⎧= −πA

42 •• Light traveling in air strikes the largest face of an isosceles-right-triangle prism at normal incidence. What is the speed of light in this prism if the prism is just barely able to produce total internal reflection? Picture the Problem We can use the definition of the index of refraction to express the speed of light in the prism in terms of the index of refraction n1 of the prism. The application of Snell’s law at the prism-air interface will allow us to relate the index of refraction of the prism to the critical angle for total internal reflection. Finally, we can use the geometry of the isosceles-right-triangle prism to conclude that θc = 45°.

45º

45º

θ

00.12 =n

1n

c

Express the speed of light v in the prism in terms of its index of refraction n1:

1ncv =

Chapter 31

2918

Apply Snell’s law to the prism-air interface to obtain:

190sinsin 2c1 =°= nn θ

Solving for n1 yields: c

1 sin1θ

=n

Substitute for n1 and simplify to obtain:

csinθcv =

Substitute numerical values and evaluate v:

( )m/s101.2

45sinm/s10998.28

8

×=

°×=v

43 •• A point source of light is located at the bottom of a steel tank, and an opaque circular card of radius 6.00 cm is placed horizontally over it. A transparent fluid is gently added to the tank so that the card floats on the fluid surface with its center directly above the light source. No light is seen by an observer above the surface until the fluid is 5.00 cm deep. What is the index of refraction of the fluid? Picture the Problem The observer above the surface of the fluid will not see any light until the angle of incidence of the light at the fluid-air interface is less than or equal to the critical angle for the two media. We can use Snell’s law to express the index of refraction of the fluid in terms of the critical angle and use the geometry of card and light source to express the critical angle.

θ

θ

c

c

θ2

r

d

n

n1

2

Apply Snell’s law to the fluid-air interface to obtain:


Light is seen by the observer when θ1 = θc and θ2 = 90°:


Properties of Light

2919

Because the medium above the interface is air, n2 = 1. Solve for n1 to obtain:

c1 sin

1θ

=n

From the geometry of the diagram:

dr

=ctanθ ⇒ ⎟⎠⎞

⎜⎝⎛= −

dr1

c tanθ

Substitute for cθ to obtain:

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

=−

dr

n1

1

tansin

1

Substitute numerical values and evaluate n1:

30.1

cm00.5cm00.6

tansin

1

11 =

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

n

44 •• An optical fiber allows rays of light to propagate long distances by using total internal reflection. Optical fibers are used extensively in medicine and in digital communications. As shown in Figure 31-58 the fiber consists of a core material that has an index of refraction n2 and radius b surrounded by a cladding material that has an index of refraction n3 < n2. The numerical aperture of the fiber is defined as sinθ1, where θ1 is the angle of incidence of a ray of light that impinges on the center of the end of the fiber and then reflects off the core-cladding interface just at the critical angle. Using the figure as a guide, show that the numerical aperture is given by sinθ1 = n2

2 − n32 assuming the ray is initially

in air. Hint: Use of the Pythagorean theorem may be required. Picture the Problem We can use the geometry of the figure, the law of refraction at the air-n1 interface, and the condition for total internal reflection at the n1-n2 interface to show that the numerical aperture is given by 2

3221sin nn −=θ .

Referring to the figure, note that: c

ann

==2

3csinθ and

cb

=2sinθ

Apply the Pythagorean theorem to the right triangle to obtain:

222 cba =+ or 12

2

2

2

=+cb

ca

Solving for cb yields:

2

2

1ca

cb

−=

Chapter 31

2920

Substitute for ca and

cb to obtain:

22

23

2 1sinnn

−=θ

Use the law of refraction to relate θ1 and θ2:


Substitute for sinθ2, let n1 = 1 (air), and simplify to obtain:

23

222

2

23

21 1sin nnnnn −=−=θ

45 •• [SSM] Find the maximum angle of incidence θ1 of a ray that would propagate through an optical fiber that has a core index of refraction of 1.492, a core radius of 50.00 μm, and a cladding index of 1.489. See Problem 44. Picture the Problem We can use the result of Problem 44 to find the maximum angle of incidence under the given conditions. From Problem 44: 2

3221sin nn −=θ

Solve for θ1 to obtain: ( )2

322

11 sin nn −= −θ

Substitute numerical values and evaluate θ1:

( ) ( )

°=

⎟⎠⎞⎜

⎝⎛ −= −

5

489.1492.1sin 2211θ

46 •• Calculate the difference in time needed for two pulses of light to travel down 15.0 km of the fiber that is described in Problem 44. Assume that one pulse enters the fiber at normal incidence, and the second pulse enters the fiber at the maximum angle of incidence calculated in Problem 45. In fiber optics, this effect is known as modal dispersion. Picture the Problem We can derive an expression for the difference in the travel times by expressing the travel time for a pulse that enters the fiber at the maximum angle of incidence and a pulse that enters the fiber at normal incidence. Examination of Figure 31-58 reveals that, if the length of the tube is L, the distance traveled by the pulse that enters at an angle θ1 is the ratio of c to a multiplied by L. The difference in the travel times Δt is given by:

incidence normal1

ttt −=Δ θ (1)

Properties of Light

2921

The travel time for the pulse that enters the fiber at the maximum angle of incidence is:

2

medium in the speed traveleddistance

1

ncacL

t ==θ

The travel time for the pulse that enters the fiber normally is:

2

incidence normal

ncLt =

Substitute for and in

equation (1) to obtain:

1θt

incidence normalt

22 ncL

ncacL

t −=Δ

Simplify to obtain: ⎟

⎠⎞

⎜⎝⎛ −=Δ 12

ac

cLnt (2)

Referring to the figure, note that: c

a=csinθ

From Snell’s law, the sine of the critical angle is also given by:

2

3csin

nn

=θ ⇒ 2

3

nn

ca=

Substitute for c/a in equation (2) to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−=Δ 1

3

22

nn

cLnt


( )( )

ns150

1489.1492.1

m/s10998.2km15492.1Δ 8

=

⎟⎠⎞

⎜⎝⎛ −

×=t

47 ••• Investigate how a thin film of water on a glass surface affects the critical angle for total reflection. Use n = 1.50 for glass and n = 1.33 for water. (a) What is the critical angle for total internal reflection at the glass–water interface? (b) Does a range of incident angles exist such that the angles are greater than θc for glass-to-air refraction and for which the light rays will leave the glass, travel through the water and then pass into the air? Picture the Problem Let the index of refraction of glass be represented by n1, the index of refraction of water by n2, and the index of refraction of air by n3. We can apply Snell’s law to the glass-water interface under total internal reflection conditions to find the critical angle for total internal reflection. The application of

Chapter 31

2922

Snell’s law to glass-air and glass-water interfaces will allow us to decide whether there are angles of incidence greater than θc for glass-to-air refraction for which light rays will leave the glass and the water and pass into the air.

00.13 =n

33.12 =n

50.11 =n

Air

Water

Glass

θ

θ

θ

θ1

2

2

3

(a) Apply Snell’s law to the glass-water interface:


At the critical angle, θ1 = θc and θ2 = 90°:


Solving for θc yields: ⎥

⎦

⎤⎢⎣

⎡°= − 90sinsin

1

21c n

nθ


°=⎥⎦⎤

⎢⎣⎡ °= − 5.6290sin

50.133.1sin 1

cθ

(b) Apply Snell’s law to a water-air interface at the critical angle for a water-air interface:


Solving for cθ yields: ⎟⎟⎠

⎞⎜⎜⎝

⎛= −

2

31c sin

nn

θ

Substitute numerical values and evaluate cθ :

°=⎟⎠⎞

⎜⎝⎛= − 8.48

33.100.1sin 1

cθ

2211 sinsin θθ nn = and

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

21

211 sinsin θθ

nn

Apply Snell’s law to a ray incident at the glass-water interface:

Properties of Light

2923

For c2 θθ = :

⎟⎟⎠

⎞⎜⎜⎝

⎛⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛= −−

1

31

2

3

1

211 sinsin

nn

nn

nnθ


°=⎟⎠⎞

⎜⎝⎛= − 8.41

53.100.1sin 1

1θ

Yes, if θ ≥ 41.8°, where is the angle of incidence for the rays in glass that are incident on the glass-water boundary, the rays will leave the glass through the water and pass into the air. 48 ••• A laser beam is incident on a plate of glass that is 3.0-cm thick (Figure 31-57). The glass has an index of refraction of 1.5 and the angle of incidence is 40º. The top and bottom surfaces of the glass are parallel. What is the distance b between the beam formed by reflection off the top surface of the glass and the beam reflected off the bottom surface of the glass. Picture the Problem The situation is shown in the adjacent figure. We can use the geometry of the diagram and trigonometric relationships to derive an expression for d in terms of the angles of incidence and refraction. Applying Snell’s law will yield θr.

Glass

Air

θ

θ

θ

θ

i

i

r

r

θi

x

5.1=n

t

b

Express the distance x in terms of t and θr:

rtan2 θtx =

The separation of the reflected rays is:

icosθxb =

Substitute for x to obtain: ir costan2 θθtb = (1)

Apply Snell’s law at the air-glass interface to obtain: ri sinsin θθ n= ⇒ ⎟

⎠⎞

⎜⎝⎛= −

ni1

rsinsin θθ

Substitute for rθ in equation (1) to obtain:

ii1 cossinsintan2 θθ⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛= −

ntb

Chapter 31

2924

Substitute numerical values and evaluate b:

( )

cm2.2

40cos5.140sinsintancm0.32 1

=

°⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ °

= −b

Dispersion 49 • A beam of light strikes the plane surface of silicate flint glass at an angle of incidence of 45º. The index of refraction of the glass varies with wavelength (see Figure 31-59). How much smaller is the angle of refraction for violet light of wavelength 400 nm than the angle of refraction for red light of wavelength 700 nm? Picture the Problem We can apply Snell’s law of refraction to express the angles of refraction for red and violet light in silicate flint glass. Express the difference between the angle of refraction for violet light and for red light:

violetr,redr, θθθ −=Δ (1)

Apply Snell’s law of refraction to the interface to obtain:

rsin45sin θn=° ⇒ ⎟⎠⎞

⎜⎝⎛= −

n21sin 1

rθ

Substituting for rθ in equation (1) yields: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛=Δ −−

violet

1

red

1

21sin

21sin

nnθ

Substitute numerical values and evaluate Δθ : ( ) ( )

°=°−°=

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛= −−

0.12.252.26

66.121sin

60.121sinΔ 11θ

50 •• In many transparent materials, dispersion causes different colors (wavelengths) of light to travel at different speeds. This can cause problems in fiber-optic communications systems where pulses of light must travel very long distances in glass. Assuming a fiber is made of silicate crown glass (see Figure 31-19), calculate the difference in travel times that two short pulses of light take to travel 15.0 km in this fiber if the first pulse has a wavelength of 700 nm and the second pulse has a wavelength of 500 nm. Picture the Problem The transit times will be different because the speed with which light of various wavelengths propagates in silicate crown glass is dependent on the index of refraction. We can use Figure 31-19 to estimate the indices of refraction for pulses of wavelengths 500 and 700 nm.

Properties of Light

2925

Express the difference in time needed for two short pulses of light to travel a distance L in the fiber:

700500 vL

vLt −=Δ

Substitute for L, v500, and v700 and simplify to obtain:

( )700500700500 nn

cL

cLn

cLnt −=−=Δ

Use Figure 31-19 to find the indices of refraction of silicate crown glass for the two wavelengths:

55.1500 ≈n and

50.1700 ≈n


( )

s3

50.155.1m/s10998.2

km0.15Δ 8

μ≈

−×

=t

Polarization 51 • [SSM] What is the polarizing angle for light in air that is incident on (a) water (n = 1.33), and (b) glass (n = 1.50)? Picture the Problem The polarizing angle is given by Brewster’s law:

12ptan nn=θ where n1 and n2 are the indices of refraction on the near and far

sides of the interface, respectively.

Use Brewster’s law to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛= −

1

21p tan

nnθ

(a) For n1 = 1 and n2 = 1.33: °=⎟

⎠⎞

⎜⎝⎛= − 1.53

00.133.1tan 1

pθ

(b) For n1 = 1 and n2 = 1.50: °=⎟

⎠⎞

⎜⎝⎛= − 3.56

00.150.1tan 1

pθ

52 • Light that is horizontally polarized is incident on a polarizing sheet. It is observed that only 15 percent of the intensity of the incident light is transmitted through the sheet. What angle does the transmission axis of the sheet make with the horizontal?

Chapter 31

2926

Picture the Problem The intensity of the transmitted light I is related to the intensity of the incident light I0 and the angle the transmission axis makes with the horizontal θ according to .cos2

0 θII =

Express the intensity of the transmitted light in terms of the intensity of the incident light and the angle the transmission axis makes with the horizontal:

θ20 cosII = ⇒ ⎟

⎟⎠

⎞⎜⎜⎝

⎛= −

0

1cosIIθ

Substitute numerical values and evaluate θ :

( ) °== − 6715.0cos 1θ

53 • Two polarizing sheets have their transmission axes crossed so that no light gets through. A third sheet is inserted between the first two so that its transmission axis makes an angle θ with the transmission axis of the first sheet. Unpolarized light of intensity I0 is incident on the first sheet. Find the intensity of the light transmitted through all three sheets if (a) θ = 45º and (b) θ = 30º. Picture the Problem Let In be the intensity after the nth polarizing sheet and use

to find the intensity of the light transmitted through all three sheets for θ = 45° and θ = 30°.

θ20 cosII =

(a) The intensity of the light between the first and second sheets is:

021

1 II =

The intensity of the light between the second and third sheets is:

0412

021

2,12

12 45coscos IIII =°== θ

The intensity of the light that has passed through the third sheet is:

0812

041

3,22


(b) The intensity of the light between the first and second sheets is:

021

1 II =


0832

021

2,12


The intensity of the light that has passed through the third sheet is:

03232

083

3,22


Properties of Light

2927

54 • A horizontal 5.0 mW laser beam that is vertically polarized is incident on a sheet that is oriented with its transmission axis vertical. Behind the first sheet is a second sheet that is oriented so that its transmission axis makes an angle of 27º with respect to the vertical. What is the power of the beam transmitted through the second sheet? Picture the Problem Because the light is polarized in the vertical direction and the first polarizer is also vertically polarized, no loss of intensity results from the first transmission. We can use Malus’s law to find the intensity of the light after it has passed through the second polarizer. The intensity of the beam is the ratio of its power to cross-sectional area:

API =

Express the intensity of the light between the first and second polarizers:

01 II = and 01 PP =

Express the law of Malus in terms of the power of the beam:

θ20 cosAP

AP= ⇒ θ2

0 cosPP =

Express the power of the beam after the second transmission:

122

02,12

12 coscos θθ PPP ==

Substitute numerical values and evaluate P2:

( ) mW0.427cosmW0.5 22 =°=P

55 •• [SSM] The polarizing angle for light in air that is incident on a certain substance is 60º. (a) What is the angle of refraction of light incident at this angle? (b) What is the index of refraction of this substance? Picture the Problem Assume that light is incident in air (n1 = 1.00). We can use the relationship between the polarizing angle and the angle of refraction to determine the latter and Brewster’s law to find the index of refraction of the substance.

(a) At the polarizing angle, the sum of the angles of polarization and refraction is 90°:

°=+ 90rp θθ ⇒ pr 90 θθ −°=

Substitute for θp to obtain:

°=°−°= 306090rθ

Chapter 31

2928

(b) From Brewster’s law we have: 1

2ptan

nn

=θ

or, because n1 = 1.00, p2 tanθ=n

Substitute for θp and evaluate n2: 7.160tan2 =°=n

56 •• Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that its transmission axis makes an angle θ with the transmission axis of the first sheet. (a) Derive an expression for the intensity of the transmitted light as a function of θ. (b) Show that the intensity transmitted through all three sheets is maximum when θ = 45º. Picture the Problem Let In be the intensity after the nth polarizing sheet and use

to find the intensity of the light transmitted through the three sheets. θ20 cosII =


021

1 II =


θθ 202

12,1

212 coscos III ==

Express the intensity of the light that has passed through the third sheet and simplify to obtain:

( )

( )θ

θθ

θθ

θθ

θ

2sin

sincos2

sincos

90coscos

cos

208

1

208

1

2202

1

2202

1

3,22

23

I

I

I

I

II

=

=

=

−°=

=

(b) Because the sine function is a maximum when its argument is 90°, the maximum value of I3 occurs when θ = 45°. 57 •• If the middle polarizing sheet in Problem 56 is rotating at an angular speed ω about an axis parallel with the light beam, find an expression for the intensity transmitted through all three sheets as a function of time. Assume that θ = 0 at time t = 0. Picture the Problem Let In be the intensity after the nth polarizing sheet, use

to find the intensity of the light transmitted through each sheet, and replace θ with ωt.

θ20 cosII =

Properties of Light

2929

The intensity of the light between the first and second sheets is:

021

1 II =


tIII ωθ 202

12,1

212 coscos ==

Express the intensity of the light that has passed through the third sheet and simplify to obtain:

( )

( )tI

ttI

ttI

ttI

II

ω

ωω

ωω

ωω

θ

2sin

sincos2

sincos

90coscos

cos

208

1

208

1

2202

1

2202

1

3,22

23

=

=

=

−°=

=

58 •• A stack of N + 1 ideal polarizing sheets is arranged so that each sheet is rotated by an angle of π/(2N) rad with respect to the preceding sheet. A linearly polarized light wave of intensity I0 is incident normally on the stack. The incident light is polarized along the transmission axis of the first sheet and is therefore perpendicular to the transmission axis of the last sheet in the stack. (a) Show that the intensity of the light transmitted through the entire stack is given by

I0 cos2N π 2N( )⎡⎣ ⎤⎦ . (b) Using a spreadsheet or graphing program, plot the transmitted intensity as a function of N for values of N from 2 to 100. (c) What is the direction of polarization of the transmitted beam in each case? Picture the Problem Let In be the intensity after the nth polarizing sheet and use

to find the ratio of Iθ20 cosII = n+1 to In.

(a) Find the ratio of In+1 to In:

NII

n

n

2cos21 π

=+

Because there are N such reductions of intensity:

⎟⎠⎞

⎜⎝⎛== ++

NII

II NNN

2cos2

0

1

1

1 π

and

⎟⎠⎞

⎜⎝⎛=+ N

II NN 2

cos201

π

(b) A spreadsheet program to graph IN+1/I0 as a function of N is shown below. The formulas used to calculate the quantities in the columns are as follows:

Cell Content/Formula Algebraic Form N A2 2

A3 A2 + 1 N + 1

Chapter 31

2930

B2 (cos(PI()/(2*A2))^(2*A2) ⎟⎠⎞

⎜⎝⎛

NN

2cos2 π

A B 1 N I/I02 2 0.250 3 3 0.422 4 4 0.531 5 5 0.605

95 95 0.974 96 96 0.975 97 97 0.975 98 98 0.975 99 99 0.975 100 100 0.976

A graph of I/I0 as a function of N follows.

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 10 20 30 40 50 60 70 80 90 100

N

I/I 0

(c) The transmitted light, if any, is polarized parallel to the transmission axis of the last sheet. (For N = 2 there is no transmitted light.) 59 •• [SSM] The device described in Problem 58 could serve as a polarization rotator, which changes the linear plane of polarization from one direction to another. The efficiency of such a device is measured by taking the ratio of the output intensity at the desired polarization to the input intensity. The result of Problem 58 suggests that the highest efficiency is achieved by using a large value for the number N. A small amount of intensity is lost regardless of the input polarization when using a real polarizer. For each polarizer, assume the transmitted intensity is 98 percent of the amount predicted by the law of Malus and use a spreadsheet or graphing program to determine the optimum number of sheets you should use to rotate the polarization 90º.

Properties of Light

2931

Picture the Problem Let In be the intensity after the nth polarizing sheet and use to find the ratio of Iθ2

0 cosII = n+1 to In. Because each sheet introduces a 2% loss of intensity, the net transmission after N sheets (0.98)N.

Find the ratio of In+1 to In: ( )

NII

n

n

2cos98.0 21 π

=+

Because there are N such reductions of intensity:

( ) ⎟⎠⎞

⎜⎝⎛=+

NII NNN

2cos98.0 2

0

1 π

A spreadsheet program to graph IN+1/I0 for an ideal polarizer as a function of N, the percent transmission, and IN+1/I0 for a real polarizer as a function of N is shown below. The formulas used to calculate the quantities in the columns are as follows:

Cell Content/Formula Algebraic Form N A3 1

B2 (cos(PI()/(2*A2))^(2*A2) ⎟⎠⎞

⎜⎝⎛

NN

2cos2 π

( )N98.0C3 (0.98)^A3 D4 B3*C3 ( ) ⎟

⎠⎞

⎜⎝⎛

NNN

2cos98.0 2 π

A B C D 1 Ideal Percent Real 2 N Polarizer Transmission Polarizer 3 1 0.000 0.980 0.000 4 2 0.250 0.960 0.240 5 3 0.422 0.941 0.397 6 4 0.531 0.922 0.490 7 5 0.605 0.904 0.547 8 6 0.660 0.886 0.584 9 7 0.701 0.868 0.608 10 8 0.733 0.851 0.624 11 9 0.759 0.834 0.633 12 10 0.781 0.817 0.638 13 11 0.798 0.801 0.639 14 12 0.814 0.785 0.638 15 13 0.827 0.769 0.636 16 14 0.838 0.754 0.632 17 15 0.848 0.739 0.626 18 16 0.857 0.724 0.620 19 17 0.865 0.709 0.613

Chapter 31

2932

20 18 0.872 0.695 0.606 21 19 0.878 0.681 0.598 22 20 0.884 0.668 0.590

A graph of I/I0 as a function of N for the quantities described above follows:

0.0

0.2

0.4

0.6

0.8

1.0

0 2 4 6 8 10 12 14 16 18 20

Number of sheets (N )

I /I 0

Ideal PolarizerPercent TransmissionReal Polarizer

Inspection of the table, as well as of the graph, tells us that the optimum number of sheets is .11

60 •• Show mathematically that a linearly polarized wave can be thought of as a superposition of a right and a left circularly polarized wave. Picture the Problem A circularly polarized wave is said to be right circularly polarized if the electric and magnetic fields rotate clockwise when viewed along the direction of propagation and left circularly polarized if the fields rotate counterclockwise.

For a circularly polarized wave, the x and y components of the electric field are given by:

tEEx ωcos0= and

tEEy ωsin0= or tEEy ωsin0−=

for left and right circular polarization, respectively.

For a wave polarized along the x axis: i

iiEE

ˆcos2

ˆcosˆcos

0

00leftright

tE

tEtE

ω

ωω

=

+=+rr

61 •• Suppose that the middle sheet in Problem 53 is replaced by two polarizing sheets. If the angle between the transmission axes in the second, third, and fourth sheets in the stack make angles of 30º, 60º and 90º, respectively, with the transmission axis of the first sheet, (a) what is the intensity of the transmitted

Properties of Light

2933

light? (b) How does this intensity compare with the intensity obtained in Part (a) of Problem 53? Picture the Problem Let In be the intensity after the nth polarizing sheet and use

to find the intensity of the light transmitted by the four sheets. θ20 cosII =


021

1 II =


0832

021

2,12


The intensity of the light between the third and fourth sheets is:

03292

083

3,22


The intensity of the light to the right of the fourth sheet is:

0

0128272

0329

4,32

34

211.0

30coscos

I

IIII

=

=°== θ

(b) The intensity with four sheets at angles of 0°, 30°, 30° and 90° is greater the intensity of three sheets at angles of 0°, 45° and 90° by a factor of 1.69. Remarks: We could also apply the result obtained in Problem 58(a) to solve this problem. 62 •• Show that the electric field of a circularly polarized wave propagating parallel with the x axis can be expressed by

( ) ( )kiE ˆcosˆsin 00 tkxEtkxE ωω −+−=r

. Picture the Problem We can use the components of E

r to show that E

ris

constant in time and rotates with angular frequency ω.

Express the magnitude of Er

in terms of its components:

22yx EEE +=

Substitute for Ex and Ey to obtain:

( )[ ] ( )[ ] ( ) ( )[ ]0

2220

20

20 cossincossin

EtkxtkxEtkxEtkxEE

=

−+−=−+−= ωωωω

and the Er

vector rotates in the yz plane with angular frequency ω.

Chapter 31

2934

63 •• [SSM] A circularly polarized wave is said to be right circularly polarized if the electric and magnetic fields rotate clockwise when viewed along the direction of propagation and left circularly polarized if the fields rotate counterclockwise. (a) What is the sense of the circular polarization for the wave described by the expression in Problem 62? (b) What would be the expression for the electric field of a circularly polarized wave traveling in the same direction as the wave in Problem 60, but with the fields rotating in the opposite direction? Picture the Problem We can apply the given definitions of right and left circular polarization to the electric field and magnetic fields of the wave.

(a) The electric field of the wave in Problem 62 is:

( ) ( )kjE ˆcosˆsin 00 tkxEtkxE ωω −+−=r

The corresponding magnetic field is:

( ) ( ) jkB ˆcosˆsin 00 tkxBtkxB ωω −−−=r

Because these fields rotate clockwise when viewed along the direction of propagation, the wave is right circularly polarized. (b) For a left circularly polarized wave traveling in the opposite direction:

( ) ( )kjE ˆcosˆsin 00 tkxEtkxE ωω +−+=r

Sources of Light 64 • A helium–neon laser emits light that has a wavelength equal to 632.8 nm and has a power output of 4.00 mW. How many photons are emitted per second by this laser? Picture the Problem We can express the number of photons emitted per second as the ratio of the power output of the laser and energy of a single photon.

Relate the number of photons per second n to the power output of the pulse and the energy of a single photon : photonE

photonEPn =

The energy of a photon is given by: λ

hcE =photon

Properties of Light

2935

Substitute for to obtain: photonEhcPn λ

=

Substitute numerical values and evaluate n:

( )( ) photons/s1027.1J10602.1

eV1nmeV1240

mW00.4nm8.632 1619 ×=

×⋅= −n

65 • The first excited state of an atom of a gas is 2.85 eV above the ground state. (a) What is the maximum wavelength of radiation for resonance absorption by atoms of the gas that are in the ground state? (b) If the gas is irradiated with monochromatic light that has a wavelength of 320 nm, what is the wavelength of the Raman scattered light? Picture the Problem We can use the Einstein equation for photon energy to find the wavelength of the radiation for resonance absorption. We can use the same relationship, with ERaman = Einc − ΔE where ΔE is the energy for resonance absorption, to find the wavelength of the Raman scattered light.

(a) Use the Einstein equation for photon energy to relate the wavelength of the radiation to energy of the first excited state:

Ehc

=λ

Substitute numerical values and evaluate λ:

nm435eV85.2

nmeV1240=

⋅=λ

(b) The wavelength of the Raman scattered light is given by:

RamanRaman

nmeV1240E

⋅=λ

Relate the energy of the Raman scattered light ERaman to the energy of the incident light Einc: eV1.025

eV85.2nm 320

nmeV1240incRaman

=

−⋅

=

Δ−= EEE

Substitute numerical values and evaluate λRaman:

nm1210eV025.1

nmeV1240Raman =

⋅=λ

66 •• A gas is irradiated with monochromatic ultraviolet light that has a wavelength of 368 nm. Scattered light that has a wavelength equal to 368 nm is observed, and scattered light that has a wavelength of 658-nm is also observed. Assuming that the gas atoms were in their ground state prior to irradiation, find

Chapter 31

2936

the energy difference between the ground state and the excited state obtained by the irradiation. Picture the Problem The incident radiation will excite atoms of the gas to higher energy states. The scattered light that is observed is a consequence of these atoms returning to their ground state. The energy difference between the ground state

and the atomic state excited by the irradiation is given byλhchfE ==Δ .

The energy difference between the ground state and the atomic state excited by the irradiation is given by:

λλnmeV1240 ⋅

===ΔhchfE

Substitute 368 nm for λ and evaluate ΔE:

eV37.3nm368

nmeV1240=

⋅=ΔE

67 •• [SSM] Sodium has excited states 2.11 eV, 3.20 eV, and 4.35 eV above the ground state. Assume that the atoms of the gas are all in the ground state prior to irradiation. (a) What is the maximum wavelength of radiation that will result in resonance fluorescence? What is the wavelength of the fluorescent radiation? (b) What wavelength will result in excitation of the state 4.35 eV above the ground state? If that state is excited, what are the possible wavelengths of resonance fluorescence that might be observed? Picture the Problem The ground state and the three excited energy levels are shown in the diagram to the right. Because the wavelength is related to the energy of a photon by λ = hc/ΔE, longer wavelengths correspond to smaller energy differences.

3

2

1

0 0

3.20 eV

2.11 eV

4.35 eV

(a) The maximum wavelength of radiation that will result in resonance fluorescence corresponds to an excitation to the 3.20 eV level followed by decays to the 2.11 eV level and the ground state:

nm388eV3.20

nmeV1240max =

⋅=λ

Properties of Light

2937

nm1140eV2.11eV3.20

nmeV124012 =

−⋅

=→λ The fluorescence wavelengths are:

and

nm5880eV2.11

nmeV124001 =

−⋅

=→λ

nm285eV35.4

nmeV124030 =

⋅=→λ

(not in visible the spectrum)

(b) For excitation:

nm1080eV3.20eV35.4

nmeV124023 =

−⋅

=→λ The fluorescence wavelengths corresponding to the possible transitions are: (not in visible spectrum)

nm1140eV11.2eV20.3

nmeV124012 =

−⋅

=→λ

(not in visible spectrum)

nm5880eV11.2

nmeV124001 =

−⋅

=→λ

nm554eV11.2eV35.4

nmeV124013 =

−⋅

=→λ

and

nm3880eV20.3

nmeV124002 =

−⋅

=→λ

(not in visible spectrum) 68 •• Singly ionized helium is a hydrogen-like atom that has a nuclear charge of +2e. Its energy levels are given by En = –4E0/n2, where n = 1, 2, … and E0 = 13.6 eV. If a beam of visible white light is sent through a gas of singly ionized helium, at what wavelengths will dark lines be found in the spectrum of the transmitted radiation? (Assume that the ions of the gas are all in the same state with energy E1 prior to irradiation.) Determine the Concept The energy difference between the ground state and the first excited state is 3E0 = 40.8 eV, corresponding to a wavelength of 30.4 nm. This is in the far ultraviolet, well outside the visible range of wavelengths. There will be no dark lines in the transmitted radiation.

Chapter 31

2938

69 • [SSM] A pulse from a ruby laser has an average power of 10 MW and lasts 1.5 ns. (a) What is the total energy of the pulse? (b) How many photons are emitted in this pulse? Picture the Problem We can use the definition of power to find the total energy of the pulse. The ratio of the total energy to the energy per photon will yield the number of photons emitted in the pulse.

(a) Use the definition of power to obtain:

tPE Δ=

Substitute numerical values and evaluate E:

( )( ) mJ15ns5.1MW10 ==E

(b) Relate the number of photons N to the total energy in the pulse and the energy of a single photon : photonE

photonEEN =

The energy of a photon is given by: λ

hcE =photon

Substitute for to obtain: photonE

hcEN λ

=

Substitute numerical values (the wavelength of light emitted by a ruby laser is 694.3 nm) and evaluate N:

( )( ) 1619 102.5

J10602.1eV1

nmeV1240mJ15nm3.694

×=×⋅

= −N

General Problems 70 • A beam of red light that has a wavelength of 700 nm in air travels in water. (a) What is the wavelength in water? (b) Does a swimmer underwater observe the same color or a different color for this light? Picture the Problem We can use v = fλ and the definition of the index of refraction to relate the wavelength of light in a medium whose index of refraction is n to the wavelength of light in air.

Properties of Light

2939

(a) The wavelength λn of light in a medium whose index of refraction is n is given by:

nnfc

fv

n0λλ ===

Substitute numerical values and evaluate λwater:

nm5261.33

nm700nm700

waterwater ===

nλ

(b) Because the color observed depends on the frequency of the light, a swimmer observes the same color in air and in water. 71 •• [SSM] The critical angle for total internal reflection for a substance is 48º. What is the polarizing angle for this substance? Picture the Problem We can use Snell’s law, under critical angle and polarization conditions, to relate the polarizing angle of the substance to the critical angle for internal reflection.

Apply Snell’s law, under critical angle conditions, to the interface:

2c1 sin nn =θ (1)

( ) p2p2p1 cos90sinsin θθθ nnn =−°=

or

1

2ptan

nn

=θ ⇒ ⎟⎟⎠

⎞⎜⎜⎝

⎛= −

1

21p tan

nnθ (2)

Apply Snell’s law, under polarization conditions, to the interface:

Solve equation (1) for the ratio of n2 to n1:

c1

2 sinθ=nn

Substitute for n2/n1 in equation (2) to obtain:

( )c1

p sintan θθ −=

Substitute numerical values and evaluate θp:

( ) °=°= − 3748sintan 1pθ

72 •• Show that when a flat mirror is rotated through an angle θ about an axis in the plane of the mirror, a reflected beam of light (from a fixed incident beam) that is perpendicular to the rotation axis is rotated through 2θ. Picture the Problem Let φ be the initial angle of incidence. Since the angle of reflection with the normal to the mirror is alsoφ, the angle between incident and reflected rays is 2φ. If the mirror is now rotated by a further angle θ, the angle of

Chapter 31

2940

incidence is increased by θ to φ +θ, and so is the angle of reflection. Consequently, the reflected beam is rotated by 2θ relative to the incident beam.

φ θ+φ θ+

φ φ

θ

73 •• [SSM] Use Figure 31-59 to calculate the critical angles for light initially in silicate flint glass that is incident on a glass–air interface if the light is (a) violet light of wavelength 400 nm and (b) red light of wavelength 700 nm. Picture the Problem We can apply Snell’s law at the glass-air interface to express θc in terms of the index of refraction of the glass and use Figure 31-59 to find the index of refraction of the glass for the given wavelengths of light.

Apply Snell’s law at the glass-air interface:


190sinsin c1 =°=θn If θ1 = θc and n2 = 1: and

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

1

1c

1sinn

θ

(a) For violet light of wavelength 400 nm, n1 = 1.67:

°=⎟⎠⎞

⎜⎝⎛= − 8.36

67.11sin 1

cθ

(b) For red light of wavelength 700 nm, n1 = 1.60:

°=⎟⎠⎞

⎜⎝⎛= − 7.38

60.11sin 1

cθ

74 •• Light is incident on a slab of transparent material at an angle θ1, as shown in Figure 31-60. The slab has a thickness t and an index of refraction n. Show that

n = sin θ1( ) sin tan -1 d/ t( )⎡⎣ ⎤⎦ , where d is the distance shown in the figure.

Properties of Light

2941

Picture the Problem We can apply Snell’s law at the air-slab interface to express the index of refraction n in terms of θ1 and θ2 and then use the geometry of the figure to relate θ2 to t and d. t

d

θ

θ

1

2

n

Applying Snell’s law to the first interface yields:

21 sinsin θθ n= ⇒2

1

sinsin

θθ

=n

From the diagram: 2tanθtd = ⇒ ⎟

⎠⎞

⎜⎝⎛= −

td1

2 tanθ


⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

=−

td

n1

1

tansin

sinθ

75 •• A ray of light begins at the point (–2.00 m, 2.00 m, 0.00 m), strikes a mirror in the y = 0 plane at some point (x, 0, 0), and reflects through the point (2.00 m, 6.00 m, 0.00 m). (a) Find the value of x that makes the total distance traveled by the ray a minimum. (b) What is the angle of incidence on the reflecting plane? (c) What is the angle of reflection? Picture the Problem We can write an expression for the total distance traveled by the light as a function of x and set the derivative of this expression equal to zero to find the value of x that minimizes the distance traveled by the light. The adjacent figure shows the two points and the reflecting surface. The x and y coordinates are in meters.

d1

d2

2− 2

2

0

4

6

minx

m ,y

m ,x

(a) Express the total distance D traveled by the light: ( ) ( ) 36242 22

21

+−+++=

+=

xx

ddD

Chapter 31

2942

Differentiate D with respect to x:

( ) ( )

( )[ ] ( ) ( )[ ] ( )( 1223622242

36242

21

21

2212

21

22

−−+−++++=

⎥⎦⎤

⎢⎣⎡ +−+++=

−−xxxx

xxdxd

dxdD

)

Setting this expression equal to zero for extreme values yields:

( ) ( )0

362

2

42

22

min

min2

min

min =+−

−−

++

+

x

x

x

x

Solve for xmin to obtain:

m00.1min −=x

(b) Letting the coordinates of the point at which the ray originates be (x1, y1) and the coordinates of the terminal point be (x2, y2), the tangent of the angle of incidence is given by:

1

1itan

yyxx

−−

=θ ⇒ ⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

= −

1

11i tan

yyxxθ

where (x, y) = (−1.00 m, 0).

Substitute numerical values and evaluate θi:

( )°=⎥

⎦

⎤⎢⎣

⎡−

−−−= − 6.26

m 00.20m 00.2m 00.1tan 1

iθ

(c) Letting the coordinates of the point at which the ray originates be (x, y) and the coordinates of the terminal point be (x2, y2), the tangent of the angle of reflection is given by:

yyxx

−−

=2

2rtanθ ⇒ ⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

= −

yyxx

2

21r tanθ

where (x, y) = (−1.00 m, 0).

Substitute numerical values and evaluate θr:

( )°=⎥⎦

⎤⎢⎣⎡

−−−

= − 6.260m 00.6

m 00.1m 00.2tan 1rθ

76 •• To produce a polarized laser beam a plate of transparent material, (Figure 31-61) is placed in the laser cavity and oriented so the light strikes it at the polarizing angle. Such a plate is called a Brewster window. Show that if θP1 is the

Properties of Light

2943

polarizing angle for the n1 to n2 interface, then θP2 is the polarizing angle for the n2 to n1 interface. Picture the Problem Let the angle of refraction at the first interface by θ1 and the angle of refraction at the second interface be θ2. We can apply Snell’s law at each interface and eliminate θ1 and n2 to show that θ2 = θP2.

Apply Brewster’s law at the n1-n2 interface:

1

2P1tan

nn

=θ

Draw a reference triangle consistent with Brewster’s law:

22

2

1

nn+

1n

2n

θP1

Apply Snell’s law at the n1-n2 interface:

12P11 sinsin θθ nn =

Solve for θ1 to obtain: ⎟⎟⎠

⎞⎜⎜⎝

⎛= −

P12

111 sinsin θθ

nn

Referring to the reference triangle we note that:

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+=

−

−

22

21

11

22

21

2

2

111

sin

sin

nnn

nnn

nnθ

i.e., θ1 is the complement of θp1.

Apply Snell’s law at the n2-n1 interface:


Solve for θ2 to obtain: ⎟⎟⎠

⎞⎜⎜⎝

⎛= −

11

212 sinsin θθ

nn

Chapter 31

2944

Refer to the reference triangle again to obtain:

P222

21

21

22

21

1

1

212

sin

sin

θ

θ

=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+=

−

−

nnn

nnn

nn

Equate these expressions for 12 sinθn to obtain:

21P1 sinsin θθ nn = ⇒ P2 θθ =

77 •• [SSM] From the data provided in Figure 31-59, calculate the polarization angle for an air–glass interface, using light of wavelength 550 nm in each of the four types of glass shown. Picture the Problem We can use Brewster’s law in conjunction with index of refraction data from Figure 31-59 to calculate the polarization angles for the air-glass interface.

From Brewster’s law we have:

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

1

21p tan

nnθ

or, for n1 = 1, ( )2

1p tan n−=θ

For silicate flint glass, n2 ≈ 1.62 and:

( ) °== − 3.5862.1tan 1flint silicatep,θ

For borate flint glass, n2 ≈ 1.57 and:

( ) °== − 5.5757.1tan 1flint boratep,θ

For quartz glass, n2 ≈ 1.54 and:

( ) °== − 0.5754.1tan 1quartz p,θ

For silicate crown glass, n2 ≈ 1.51 and:

( ) °== − 5.5651.1tan 1crown silicatep,θ

78 •• A light ray passes through a prism with an apex angle of α, as shown in Figure 31-63. The ray and the bisector of the apex angle bisect at right angles. Show that the angle of deviation δ is related to the apex angle and the index of refraction of the prism material by sin 1

2 α +δ( )⎡⎣ ⎤⎦ = n sin 12α( ).

Properties of Light

2945

Picture the Problem The diagram to the right shows the angles of incidence, refraction, and deviation at the first interface. We can use the geometry of this symmetric passage of the light to express θr in terms of α and δ1 in terms of θr and α. We can then use a symmetry argument to express the deviation at the second interface and the total deviation δ. Finally, we can apply Snell’s law at the first interface to complete the derivation of the given expression.

α

δ

θ

θ

δ1

r

θi

r−°90

1n2n 1n

With respect to the normal to the left face of the prism, let the angle of incidence be θi and the angle of refraction be θr. From the geometry of the figure, it is evident that:

αθ 21

r =

Express the angle of deviation at the refracting surface:

αθθθδ 21

iri1 −=−=

Referring to triangle ABC, we see that:

α

δδ1

1n2n 1n

δ

δ1A

B

C

( ) αθαθδδ −=−== i21

11 222

Solving for θI yields: ( )δαθ += 21

i

Apply Snell’s law, with n1 = 1 and n2 = n, to the first interface:

αθ 21

i sinsin n=

Chapter 31

2946

Substitute for θI to obtain: 2

sin2

sin αδα n=+ (1)

79 •• [SSM] (a) For light rays inside a transparent medium that is surrounded by a vacuum, show that the polarizing angle and the critical angle for total internal reflection satisfy tan θp = sin θc. (b) Which angle is larger, the polarizing angle or the critical angle for total internal reflection? Picture the Problem We can apply Snell’s law at the critical angle and the polarizing angle to show that tan θp = sin θc.

(a) Apply Snell’s law at the medium-vacuum interface:

r211 sinsin θθ nn =

For θ1 = θc, n1 = n, and n2 = 1: 190sinsin c =°=θn

For θ1 = θp, n1 = n, and n2 = 1: nn

n 1tan1

2p ==θ ⇒ 1tan p =θn

Because both expressions equal one:

cp sintan θθ =

(b) For any value of θ :

θθ sintan > ⇒ cp θθ >

80 •• Light in air is incident on the surface of a transparent substance at an angle of 58º with the normal. The reflected and refracted rays are observed to be mutually perpendicular. (a) What is the index of refraction of the transparent substance? (b) What is the critical angle for total internal reflection in this substance? Picture the Problem Let the numeral 1 refer to the side of the interface from which the light is incident and the numeral 2 to the refraction side of the interface. We can apply Snell’s law, under the conditions described in the problem statement, at the interface to derive an expression for n as a function of the angle of incidence (also the polarizing angle).

(a) Apply Snell’s law at the air-medium interface:

21 sinsin θθ n=

Because the reflected and refracted rays are mutually perpendicular:

°=+ 9021 θθ ⇒ 12 90 θθ −°=

Properties of Light

2947

( ) 111 cos90sinsin θθθ nn =−°= or

p1 tantan θθ ==n


Substitute for θp and evaluate n:

6.158tan =°=n

(b) Apply Snell’s law at the interface under conditions of total internal reflection:


Because n1 = 1: ⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

nn1sin1sin 1

2

1cθ

Substitute for n and evaluate θc: °=⎟

⎠⎞

⎜⎝⎛= − 39

6.11sin 1

cθ

81 •• A light ray in dense flint glass that has an index of refraction of 1.655 is incident on the glass surface. An unknown liquid condenses on the surface of the glass. Total internal reflection on the glass–liquid interface occurs for a minimum angle of incidence on the glass–liquid interface of 53.7º. (a) What is the refractive index of the unknown liquid? (b) If the liquid is removed, what is the minimum angle of incidence for total internal reflection? (c) For the angle of incidence found in Part (b), what is the angle of refraction of the ray into the liquid film? Does a ray emerge from the liquid film into the air above? Assume the glass and liquid have parallel planar surfaces. Picture the Problem We can apply Snell’s law at the glass–liquid and liquid–air interfaces to find the refractive index of the unknown liquid, the minimum angle of incidence (glass-air interface) for total internal reflection, and the angle of refraction of a ray into the liquid film.

(a) Apply Snell’s law, under critical-angle conditions, at the glass–liquid interface:

glass

liquidcsin

nn

=θ ⇒ cglassliquid sinθnn =

Substitute numerical values and evaluate : liquidn

( ) 33.17.53sin655.1liquid =°=n

(b) With the liquid removed: ⎟⎟⎠

⎞⎜⎜⎝

⎛= −

glass

1c

1sinn

θ

Chapter 31

2948


°=⎟⎠⎞

⎜⎝⎛= − 2.37

655.11sin 1

cθ

(c) Apply Snell’s law at the glass−liquid interface:

2liquid1glass sinsin θθ nn =

Solve for θ2:

⎥⎥⎦

⎤

⎢⎢⎣

⎡= −

1liquid

glass12 sinsin θθ

nn

Letting θ1 be the angle of incidence found in Part (b) yields:

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

liquid

1

glassliquid

glass12

1sin

1sin

n

nnn

θ


°=⎟⎠⎞

⎜⎝⎛= − 6.48

333.11sin 1

2θ

Because 48.6° is also the angle of incidence at the liquid-air interface and because it is equal to the critical angle for total internal reflection at this interface, no light will emerge into the air. 82 ••• (a) Show that for normally incident light, the intensity transmitted through a glass slab that has an index of refraction of n and is surrounded by air is

approximately given by IT = I0 4n n +1( )2⎡

⎣⎤⎦

2. (b) Use the Part (a) result to find

the ratio of the transmitted intensity to the incident intensity through N parallel slabs of glass for light of normal incidence. (c) How many slabs of a glass that has an index of refraction of 1.5 are required to reduce the intensity to 10 percent of the incident intensity?

Properties of Light

2949

Picture the Problem We’ll neglect multiple reflections at the glass-air interfaces. We can use the expression (Equation 31-7) for the reflected intensity at an interface to express the intensity of the light in the glass slab as the difference between the intensity of the incident beam and the reflected beam. Repeating this analysis at the glass-air interface will lead to the desired result.

Air Glass

0I glassITI

R,2IR,1I

n

(a) Express the intensity of the light transmitted into the glass:

R,10glass III −=

where IR,1 is the intensity of the light reflected at the air-glass interface.

The intensity of the light reflected at the air-glass interface is given by Equation 31-7:

0

2

R,1 11 I

nnI ⎟⎠⎞

⎜⎝⎛+−

=

Substitute and simplify to obtain:

( ) ⎥⎦⎤

⎢⎣

⎡

+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+−

−=

⎟⎠⎞

⎜⎝⎛+−

−=

20

2

0

0

2

0glass

14

111

11

nnI

nnI

InnII

Express the intensity of the light transmitted at the glass-air interface:

R,2glassT III −=

where IR,2 is the intensity of the light reflected at the glass-air interface.

The intensity of the light reflected at the glass-air interface is:

( ) 02

2

glass

2

R,2

14

11

11

Inn

nn

InnI

⎥⎦

⎤⎢⎣

⎡

+⎟⎠⎞

⎜⎝⎛+−

=

⎟⎠⎞

⎜⎝⎛+−

=

Chapter 31

2950

Substitute and simplify to obtain:

( ) ( )

( )

( ) ( )

( )

2

20

220

2

2

0

02

2

20T

14

14

14

14

111

14

11

14

⎥⎦

⎤⎢⎣

⎡

+=

⎥⎦

⎤⎢⎣

⎡

+⎥⎦

⎤⎢⎣

⎡

+=

⎥⎦

⎤⎢⎣

⎡

+⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+−

−=

⎥⎦

⎤⎢⎣

⎡

+⎟⎠⎞

⎜⎝⎛+−

−⎥⎦

⎤⎢⎣

⎡

+=

nnI

nn

nnI

nn

nnI

Inn

nn

nnII

(b) From Part (a), each slab reduces the intensity by the factor: ( )

2

214

⎥⎦

⎤⎢⎣

⎡

+nn

( )

N

nnII

2

20t 14

⎥⎦

⎤⎢⎣

⎡

+=

For N slabs:

and

( )

N

nn

II

2

20

t

14

⎥⎦

⎤⎢⎣

⎡

+= (1)

(c) Begin the solution of equation (1) for N by taking the logarithm (arbitrarily to base 10) of both sides of the equation:

( )

( ) ⎥⎦⎤

⎢⎣

⎡

+=

⎥⎦

⎤⎢⎣

⎡

+=⎟⎟

⎠

⎞⎜⎜⎝

⎛

2

2

20

t

14log2

14loglog

nnN

nn

II

N

Solve for N:

( ) ⎥⎦⎤

⎢⎣

⎡

+

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

2

0

t

14log2

log

nn

II

N

Substitute numerical values and evaluate N:

( )( )

( )

282.28

15.15.14log2

1.0log

2

≈=

⎥⎦

⎤⎢⎣

⎡

+

=N

Properties of Light

2951

83 ••• Equation 31-14 gives the relation between the angle of deviation φd of a light ray incident on a spherical drop of water in terms of the incident angle θ1 and the index of refraction of water. (a) Assume that nair = 1, and derive an expression for dφd/dθ1. Hint: If y = sin–1 x, then dy/dx = (1 – x2)–1/2. (b) Use this result to show that the angle of incidence for minimum deviation θ1m is given by

cosθ1m = 1

3 n 2 −1( ). (c) The index of refraction for a certain red light in water is

1.3318 and that the index of refraction for a certain blue light in water is 1.3435. Use the result of Part (a) to find the angular separation of these colors in the primary rainbow. Picture the Problem (a) We can follow the directions given in the problem statement and use the hint to establish the given result. (b) Treating the result of Part (a) as an extreme-value problem will lead to the given result. (c) We can use Equation 31-14 and the result of Part (b) to find the angular separation of these colors in the primary rainbow.

(a) Equation 31-14 is: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−+= −

water

1air11d

sinsin42n

n θθπφ

For nair = 1and nwater = n: ⎟

⎠⎞

⎜⎝⎛−+= −

n11

1dsinsin42 θθπφ

Use the hint to differentiate φd with respect to θ1:

1221

111

11

d

sincos42

sinsin42

θθ

θθπθθ

φ

−−=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−+= −

n

ndd

dd

(b) Set dφd/dθ1 = 0:

extremafor 0sin

cos421

221 =

−−

θθ

n

Simplify to obtain:

( )122

12 sin4cos16 θθ −= n

Replace sin2θ1 with 1 − cos2θ1 and simplify to obtain:

44cos12 21

2 −= nθ

Solve for cosθ1 = cosθ1m:

31cos

2

1m−

=nθ

Chapter 31

2952

(c) Express the angular separation Δφ of blue and red:

redd,blued, φφφ −=Δ (1)

From Equation 31-18, with nair = 1 and nwater = n:

⎟⎠⎞

⎜⎝⎛−+= −

n11

1dsinsin42 θθπφ

From Part (b):

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −= −

31cos

21

1mnθ

Substitute to obtain:

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −

−⎥⎥⎦

⎤

⎢⎢⎣

⎡ −+=

−

−−

n

n

n 31cossin

sin43

1cos2

21

12

1d πφ

Evaluate φd for blue light in water:

( )( )

°=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −

−⎥⎥⎦

⎤

⎢⎢⎣

⎡ −+=

−

−−

42.139

3435.1

313435.1cossin

sin43

13435.1cos2

21

12

1blued, πφ

Evaluate φd for red light in water:

( )( )

°=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −

−⎥⎥⎦

⎤

⎢⎢⎣

⎡ −+=

−

−−

75.137

3318.1

313318.1cossin

sin43

13318.1cos2

21

12

1redd, πφ

Properties of Light

2953

Substitute numerical values in equation (1) and evaluate Δφ:

°=°−=Δ 67.175.13742.139φ

84 ••• Show that the angle of deviation δ is a minimum if the angle of incidence is such that the ray and the bisector of the apex angle α (Figure 31-62) intersect at right angles. Picture the Problem The figure below shows the prism and the path of the ray through it. The dashed lines are the normals to the prism faces. The triangle formed by the interior ray and the prism faces has interior angles of α, 90° − θ2, and 90° − θ3. Consequently, .32 αθθ =+ We can apply Snell’s law at both interfaces to express the angle of deviation δ as a function of θ3 and then set the derivative of this function equal to zero to find the conditions on θ3 and θ2 that result in δ being a minimum.

a

d

n

u2

u3

α

θθ

θ

θ

δ

1

2 3

4

Express the angle of deviation:

αθθδ −+= 21 (1)

21 sinsin θθ n= (2) Apply Snell’s law to relate θ1 to θ2 and θ3 to θ4: and

43 sinsin θθ =n (3)

( )21

1 sinsin θθ n−= Solve equation (2) for θ1 and equation (3) for θ4: and

( )31

4 sinsin θθ n−= Substitute in equation (1) to obtain:

( ) ( ) ( )[ ] ( ) αθαθαθθδ −+−=−+= −−−−3

13

13

12

1 sinsinsinsinsinsinsinsin nnnn

Chapter 31

2954

Note that the only variable in this expression is θ3. To determine the condition that minimizesδ, take the derivative of δ with respect to θ3 and set it equal to zero.

( )[ ] ( ){ }( )( )[ ] ( )

extremafor 0sin1

cos

sin1

cos

sinsinsinsin

23

32

3

3

31

31

33

=−

+−−

−−=

−+−= −−

θ

θ

θα

θα

αθαθθθ

δ

n

n

n

n

nndd

dd

This equation is satisfied provided:

33 θθα =− ⇒ αθ 21

3 =

Because 32 θαθ −= : αααθ 21

21

2 =−=

Because 32 θθ = , we can conclude that the deviation angle is a minimum if the ray pthrough the prism symmetrically. Remarks: Setting dδ/dθ3 = 0 establishes the condition on θ3 that δ is either a maximum or a minimum. To establish that δ is indeed a minimum when

α,θθ 21

23 == we can either show that ,23

2 dθδd evaluated at αθ 21

3 = , is positive or, alternatively, plot a graph of δ (θ3) to show that it is concave upward at αθ 2

13 = .

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