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.

A ray of light passes in air is incident on the surface of a piece of  2 • 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 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.

During a physics experiment, you are measuring refractive indices of  3 • different transparent materials using a red helium-neon laser beam. F or 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.

A ray of light passes from air into water, striking the surface of the 4 • 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 ( λ water  = λ air  nwater  ), its speed changes ( v water  = c n water  ), and the

direction of its propagation changes in accordance with Snell’s law. Its frequency does not change, so

(a )

,

(c )

and

(d )

change.

Earth’s atmosphere decreases in density as the altitude increases. As a 5 • consequence, the index of refraction of the atmosphere also decreases as altitude increases. Explain how one can see the Sun when it 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. Atmosphere

Earth

A physics student playing pocket billiards wants to strike her cue ball 6 • 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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Chapter 31

Determine the Concept When light passes from air into water its wavelength changes ( λ water  = λ air  nwater  ), its speed changes ( v water  = c n water  ), and the

direction of its propagation changes in accordance with Snell’s law. Its frequency does not change, so

(a )

,

(c )

and

(d )

change.

Earth’s atmosphere decreases in density as the altitude increases. As a 5 • consequence, the index of refraction of the atmosphere also decreases as altitude increases. Explain how one can see the Sun when it 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. Atmosphere

Earth

A physics student playing pocket billiards wants to strike her cue ball 6 • 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. Material A has a higher index of refraction than material B. Which 8 • 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 appe ars 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.

Let θ   be the angle between the transmission axes of two polarizing 10 • 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, )/2, (c) (I cos2 θ )/4, )/4, (d ) I cos θ , (e) (I cos θ )/4, )/4, ( f   f ) None of the above Picture the Problem The intensity of the light transmitted by the second  polarizer is given by  I trans = I 0 cos 2 θ  , where  I 0 = 12 I . Therefore,  I trans = 12 I cos 2 θ 

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.

Light from the sun Polaroid sunglasses

θ P θ P Smooth surface

θ r 

Use the photon model of light to explain why, biologically, it is far  12 • 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.

Three energy states of an atom are A, B and C. State B is 2.0 eV above 13 • 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.

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

Properties of Light

2893

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 E 2,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 E 2,He.

When a beam of visible white light that passes through a gas of atomic 16 • 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 Estimate the time required for light to make the round trip during 18 • 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 : Substitute numerical values and evaluate Δt : Because human reaction is approximately 0.3 s:

d  = cΔt ⇒ Δt  =

Δt 

=

d  c

6 km 2.998 × 108 m/s

Δt reaction Δt 

=

or  Δt reaction

0 .3 s 2 × 10

−5

s

= 2 ×10−5 s ≈ 2 × 10 4

≈ (2 × 10 4 )Δt 

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. Estimate the time delay in receiving a light on your retina when you 19 • 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: If glass of thickness d and index of  refraction n is inserted in the path of  the photon, its travel time becomes:

Δt 0

Δt 

=

 L c

= t in air  + t in glass = =

 L + (n − 1)d  c

The time delay is the difference  between Δt and Δt 0:

t delay

= Δt − Δt 0 =

Substitute numerical values and evaluate t delay:

t delay

=

 L − d  c

= Δt 0 +

+

d  c n

(n − 1)d  c

(n − 1)d  c

(1.5 − 1)(2 mm ) 2.998 × 10 8 m/s

≈ 3 ps

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

Properties of Light

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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  E  = hf  = hc λ ) 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:

dN  dt 

Pincident

=

on pupil

(1)

 E  per  photon

Pincident

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

 I Sun

=

on pupil

 A pupil

=

PSun  Asphere at Earth's distance from the Sun

Solving for Pincident yields:

Pincident

on pupil

=

on pupil

 A pupil

PSun

 Asphere at Earth's

(2)

distance from the Sun

Substituting in equation (1) yields:

dN  dt 

=

 A pupil

PSun

 Asphere at Earth's

 E  per  photon

distance from the Sun

Substitute for the two areas and simplify to obtain:

dN  dt 

=

1 4

2 π d  pupil

PSun

2 4π  REarth -Sun  E   per  photon

⎛  d   ⎞ = ⎜⎜  pupil ⎟⎟ ⎝ 4 REarth-Sun  ⎠

2

PSun  E  per  photon

Substitute numerical values and evaluate dN /dt : 2

⎛   ⎞ 1 mm 4.2 × 10 26 W ⎟⎟ = ⎜⎜ = 3.237 × 1015 s −1 −19 11 dt  ⎝ 4(1.50 × 10 m ) ⎠ 1.602 × 10 J 2.25 eV ×

dN 

eV or, separating variables and integrating this expression,

(

 N  = 3.237 ×1015 s

−1

)t 

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Chapter 31

(

 N (0.1 s ) = 3.237 ×10 s −

Evaluating N for t = 0.1 s yields:

15

1

)(0.1 s)

≈ 3 ×1014  photons 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: Substituting for Pincident from

 E  = Pincident t  on pupil

 E =

on pupil

equation (2) yields:

 A pupil

PSun t 

 Asphere at Earth's distance from the Sun

Substitute for the two areas and simplify to obtain:

 E =

1 4

2 π d  pupil

2 4π  REarth -Sun

PSun t 

⎛  d   ⎞ = ⎜⎜  pupil ⎟⎟ ⎝ 4 REarth-Sun  ⎠

2

PSun t 

Substitute numerical values and evaluate E for t = 0.1 s: 2

⎛   ⎞ 1 mm ⎟⎟ (4.2 × 10 26 W )(0.1 s ) = 0.1167 mJ ≈ 0.1 mJ  E (0.1 s ) = ⎜⎜ 11 ⎝ 4(1.50 × 10 m ) ⎠ Römer was observing the eclipses of Jupiter’s moon Io with the hope 21 •• 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.

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

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:

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365.24 d 24 h × 4 4 d = 2191.4 h T earth

t  A→ B

=

 N  =

t  A→ B T Io

=

=

2191.4 h 42.5 h

= 51.56

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  N th eclipse occurs to N and the  period T Io of Io: Evaluate t (52) to obtain:

t ( N ) = NT Io

⎛  ⎝ 

t (52) = (52)⎜⎜ 42.5 h ×

1d  ⎞ ⎟ 24 h ⎠⎟

= 92.083 d Subtract the number of whole days to find the clock time t :

t  = t (52 ) − 92 d = 92.083 d − 92 d

= 0.083 d ×

24 h d

= 1.992 h

≈ 2 : 00 a.m. Because June, July, and August have 30, 31, and 31 d, respectively, the date is: (b) Express the time delay Δt in the arrival of light from Io due to Earth’s location at B: Substitute numerical values and evaluate Δt :

September 1

Δt =

r earth -sun c

1.5 × 1011 m Δt = 2.998 × 108 m/s

= 500 s ×

1 min 60 s

= 8.34 min Hence, the eclipse will actually occur at 2:08 a.m., September 1

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If the angle of incidence is small enough, the small angle 22 •• 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:

δ (θ ) =

θ  − sin θ 

sin θ 

=

θ 

sin θ 

−1

A spreadsheet program was used to plot the following graph of δ  (θ  ). 0.016 0.014 0.012

   ) 0.010   a    t   e    h    t 0.008    (   a    t    l   e    d 0.006 0.004 0.002 0.000 0.00

0.05

0.10

0.15

0.20

0.25

0.30

theta (radians)

From the graph, we can see that δ  (θ  ) < 1% for θ  ≤ 0.24 radians. In degree measure, θ  ≤ 14° Remarks: Using the Solver program on a TI-85 gave

= 0.244 radians.

The Speed of Light Mission Control sends a brief wake-up call to astronauts in a spaceship 23 • 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:  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 :

d  = cΔt 

(

)

d  = 2.998 × 10 m/s (2.5 s ) 8

= 7.5 × 10 8 m and (a) is correct.

The distance from a point on the surface of Earth to a point on the 24 • 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 Δ x is related to the uncertainty in the measured time Δt by Δ x = 12 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 Δ x is related to the uncertainty in the time Δt by Δ x = cΔt to evaluate Δ x.

(a) The uncertainty in the distance is:

Δ x

= 12 cΔt 

Substitute numerical values and evaluate Δ x:

Δ x

= 12 (2.998 ×10 8 m/s )(± 1.00 ns )

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

Δ x Earth to Moon

= ± 15.0 cm

 xEarth to Moon

=

15.0 cm 3.84 × 108 m

≈ 10 −8% Ole Römer discovered the finiteness of the speed of light by 25 •• [SSM] 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

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

Sensitivity = 101 Δt  where Δt is the time required for light to travel between the two positions of  the moon. d moon Δt 

=

at perigee

c

Substituting for Δt yields:

− d moon

d moon

Sensitivity =

Substitute numerical values and evaluate the required sensitivity:

− d moon

at apogee

Δt 

=

at apogee

at perigee

10c

4.06 × 105 km − 3.63 × 105 km (10)(2.998 ×108 m/s)

= 14 ms th

Remarks: Instruments with this sensitivity did not exist in the 17 century.

Reflection and Refraction Calculate the fraction of light energy reflected from an air–water  26 • 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. 2

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  I 0 of the incident light:

⎛ n − nwater  ⎞ ⎟⎟  I 0  I  = ⎜⎜ air  n + n ⎝  air  water  ⎠

Solve for the ratio I / I 0:

 I   I 0

⎛ n − n  ⎞ = ⎜⎜ air  water  ⎟⎟ ⎝ nair  + nwater  ⎠

2

Properties of Light Substitute numerical values and evaluate I / I 0:

 I   I 0

2901

2

1.00 − 1.33 ⎞ = ⎛  ⎜ ⎟ = 2.0% ⎝ 1.00 + 1.33 ⎠

A ray of light is incident on one of a pair of mirrors set at right angles 27 • 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 θ 1, reflected as ray 2, and incident on the horizontal surface at an angle of incidence θ 3. We’ll prove that rays 1 and 3 are  parallel by showing that θ 1 = θ 4, 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

θ 1 θ 1 3 2

θ 2

θ 3 θ 3

θ 4

θ 2 + 90° + (90° − θ 1 ) = 180°

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

or  θ 1 = θ 2

The sum of θ 2 and θ 3 is 90°:

θ 3

= 90° − θ 2

= θ 2 :

θ 3

= 90° − θ 1

The sum of θ 4 and θ 3 is 90°:

θ 3

+ θ 4 = 90°

Substitute for θ 3 to obtain:

(90° − θ 1 ) + θ 4 = 90° ⇒ θ 1 =

Because θ 1

θ 4

(a) A ray of light in air is incident on an air–water interface. Using a 28 •• 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)

(b)

θ 2

θ 1 n1

Air  Water 

Air 

n2

Water 

n1

n2

θ 1

θ 2

(a) Apply Snell’s law to the airwater interface to obtain:

Solving for θ 2 yields:

n1 sin θ 1

= n2 sin θ 2

where the angles of incidence and refraction are θ 1 and θ 2, respectively.

θ 2

⎛ n  ⎞ = sin −1 ⎜⎜ 1 sin θ 1 ⎟⎟ ⎝ n2  ⎠

A spreadsheet program to graph θ 2 as a function of θ 1 is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 B2 A6 A7 B6

Content/Formula 1 1.33333 0 A6 + 5 A6*PI()/180

C6

ASIN(($B$1/$B$2)*SIN(B6))

Algebraic Form n1 n2 θ 1 (deg) θ 1 + Δθ  π  θ 1 × 180

⎛ n1

sin −1 ⎜⎜

⎝ n2

D6

C6*180/PI()

1 2

A B n1= 1 n2= 1.33333

θ 2 ×

C

 ⎞

sin θ 1 ⎟⎟

180 π 

D

 ⎠

Properties of Light 3 4 5 6 7 8 9

θ 1 (deg) 0 1 2 3

θ 1 (rad) 0.00 0.02 0.03 0.05

θ 2 (rad) 0.000 0.013 0.026 0.039

θ 2 (deg) 0.00 0.75 1.50 2.25

21 22 23 24

87 88 89 90

1.52 1.54 1.55 1.57

0.847 0.847 0.848 0.848

48.50 48.55 48.58 48.59

2903

A graph of θ 2 as a function of θ 1 follows: 50   g 45   e    d 40  ,   n 35   o    i    t   c 30   a   r 25    f   e   r    f 20   o 15   e    l   g 10   n    A 5

0 0

10

20

30

40

50

60

70

80

90

Angle of incidence, deg

(b) Change the contents of cell B1 to 1.33333 and the contents of cell B2 to 1 to obtain the following graph: 90   g 80   e    d  , 70   n   o 60    i    t   c   a 50   r    f   e   r 40    f   o 30   e    l 20   g   n    A 10

0 0

10

20

30

40

50

Angle of incidence, 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°.

2904

Chapter 31

The red light from a helium-neon laser has a wavelength of 632.8 nm 29 •• 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: The wavelength of light in a medium whose index of refraction is n is given by: The frequency of the light is equal to its frequency in air independently of  the medium in which the light is  propagating: Substitute numerical values in equation (1) and evaluate vwater :

v

c

=

λ n

(1)

n

λ air 

=

 f  =

(2)

n

c

λ 

=

2.998 × 10 8 m/s 632.8 nm

= 4.74 ×1014 Hz

v water  =

2.998 × 10 8 m/s 1.33

= 2.25 × 10 8 m/s Substitute numerical values in equation (2) and evaluate λ water :

λ water 

=

632.8 nm 1.33

= 476 nm

The other speeds and wavelengths are found similarly and are summarized in the following table: (a) speed (m/s)

(b) wavelength (nm)

(c) frequency (Hz)

Air 

3.00 × 108

633

4.74 × 1014

Water 

2.25 × 108

476

4.74 × 1014

glass

2.00 × 108

422

4.74 × 1014

Properties of Light

2905

The index of refraction for silicate flint glass is 1.66 for violet light 30 •• 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.

= θ 1,violet − θ 1,red

(b) Express the difference in their  angles of incidence:

Δθ 

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

n1 sin θ 1

Solving for θ 1 yields:

θ 1

(1)

= n2 sin θ 2

⎛ n  ⎞ = sin −1 ⎜⎜ 2 sin θ 2 ⎟⎟ ⎝ n1  ⎠

Substitute for  θ 1, violet and θ 1, red in equation (1) to obtain:

Δθ 

For  θ 2, violet

Δθ 

 ⎞ ⎛ n  ⎞ ⎛ n = sin −1 ⎜⎜ violet sin θ 2, violet ⎟⎟ − sin −1 ⎜⎜ red sin θ 2, red ⎟⎟ ⎝  nair   ⎠ ⎝ nair   ⎠

= θ 2, red = 30° :

⎛ n  ⎞ ⎛ n  ⎞ ⎛  n  ⎞ ⎛ n  ⎞ = sin −1 ⎜⎜ violet sin 30° ⎟⎟ − sin −1 ⎜⎜ red sin 30° ⎟⎟ = sin −1 ⎜⎜ violet ⎟⎟ − sin −1 ⎜⎜ red ⎟⎟ ⎝  2n air   ⎠ ⎝ 2n air  ⎠ ⎝  n air   ⎠ ⎝ n air   ⎠

Substitute numerical values and evaluate Δθ : Δθ 

⎛  1.66  ⎞ ⎛  1.61  ⎞ = sin −1 ⎜⎜ ⎟⎟ − sin −1 ⎜⎜ ⎟⎟ = 56.10° − 53.61° = 2.49° ( ) ( ) 2 1 . 00 2 1 . 00 ⎝   ⎠ ⎝   ⎠

2906

Chapter 31

Remarks: Note that 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 waterglass interface to obtain: Solving for θ 2 yields:

(a) Evaluate θ 2 for θ 1 = 60°:

(b) Evaluate θ 2 for θ 1 = 45°:

n1 sin θ 1

= n2 sin θ 2

θ 2

⎛ n  ⎞ = sin −1 ⎜⎜ 1 sin θ 1 ⎟⎟ ⎝ n2  ⎠

θ 2

1.33  ⎞ = sin −1 ⎛  sin 60° ⎟ = 50° ⎜ ⎝ 1.50  ⎠

θ 2

1.33  ⎞ = sin −1 ⎛  sin 45° ⎟ = 39° ⎜ ⎝ 1.50  ⎠

(c) Evaluate θ 2 for θ 1 = 30°: 2

1.33  ⎞ sin 30° ⎟ = 26° = sin −1 ⎛  ⎜ ⎝ 1.50  ⎠

Repeat Problem 31 for a beam of light initially in the glass that is 32 •• 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 waterglass interface to obtain: Solving for θ 2 yields:

(a) Evaluate θ 2 for θ 1 = 60°:

n1 sin θ 1

= n2 sin θ 2

θ 2

⎛ n  ⎞ = sin −1 ⎜⎜ 1 sin θ 1 ⎟⎟ ⎝ n2  ⎠

θ 2

1.50  ⎞ = sin −1 ⎛  sin 60° ⎟ = 78° ⎜ ⎝ 1.33  ⎠

Properties of Light (b) Evaluate θ 2 for θ 1 = 45°:

θ 2

(c) Evaluate θ 2 for θ 1 = 30°: 2

2907

1.50  ⎞ sin 45° ⎟ = 53° = sin −1 ⎛  ⎜ ⎝ 1.33  ⎠ 1.50  ⎞ = sin −1 ⎛  sin 30° ⎟ = 34° ⎜ ⎝ 1.33  ⎠

A beam of light in air strikes a glass slab at normal incidence. The 33 •• 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 on e 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.

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

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

n1 = 1.00

 I r,1

 I 1

n2 = 1.50

= 1.00

n3

 I r, 2

 I 2

 I 3

2

 I 2

⎛ n − n  ⎞ =  I 1 − I r,1 = I 1 − ⎜⎜ 1 2 ⎟⎟  I 1 ⎝ n1 + n2  ⎠ ⎡ ⎛ n − n  ⎞ 2 ⎤ =  I 1 ⎢1 − ⎜⎜ 1 2 ⎟⎟ ⎥ ⎢⎣ ⎝ n1 + n2  ⎠ ⎥⎦ 2

 I 3

Substitute for  I 2 to obtain:  I 3

⎛ n − n  ⎞ =  I 2 − I r,2 =  I 2 − ⎜⎜ 2 3 ⎟⎟  I 2 ⎝ n2 + n3  ⎠ ⎡ ⎛ n − n  ⎞ 2 ⎤ =  I 2 ⎢1 − ⎜⎜ 2 3 ⎟⎟ ⎥ ⎢⎣ ⎝ n2 + n3  ⎠ ⎥⎦ ⎡ ⎛ n − n  ⎞ 2 ⎤ ⎡ ⎛ n2 − n3  ⎞ 2 ⎤ ⎟ ⎥ =  I 1 ⎢1 − ⎜⎜ 1 2 ⎟⎟ ⎥ ⎢1 − ⎜⎜ ⎢⎣ ⎝ n1 + n2  ⎠ ⎥⎦ ⎢⎣ ⎝ n2 + n3  ⎠⎟ ⎥⎦

2908

Chapter 31

Solve for the ratio I 3/ I 1 to obtain:

 I 3  I 1

⎡ ⎛ n − n  ⎞ 2 ⎤ ⎡ ⎛ n2 − n3  ⎞ 2 ⎤ ⎟⎟ ⎥ = ⎢1 − ⎜⎜ 1 2 ⎟⎟ ⎥ ⎢1 − ⎜⎜ n + n n + n ⎢⎣ ⎝  1 2  ⎠ ⎥⎦ ⎢⎣ ⎝  2 3  ⎠ ⎥⎦

Substitute numerical values and evaluate I 3/ I 1:  I 3  I 1

⎡ ⎛ 1.00 − 1.50 ⎞ 2 ⎤ ⎡ ⎛ 1.50 − 1.00 ⎞ 2 ⎤ = ⎢1 − ⎜ ⎟ ⎥ ⎢1 − ⎜ ⎟ ⎥ = 92% + + 1 . 00 1 . 50 1 . 50 1 . 00  ⎠ ⎥⎦ ⎢⎣ ⎝   ⎠ ⎥⎦ ⎢⎣ ⎝ 

(b) With n1 = n3 = 1.33:  I 3  I 1

⎡ ⎛ 1.33 − 1.50 ⎞ 2 ⎤ ⎡ ⎛ 1.50 − 1.33 ⎞ 2 ⎤ = ⎢1 − ⎜ ⎟ ⎥ ⎢1 − ⎜ ⎟ ⎥ = 99% + + 1 . 33 1 . 50 1 . 50 1 . 33 ⎝   ⎠ ⎝   ⎠ ⎣⎢ ⎦⎥ ⎣⎢ ⎦⎥

This problem is a refraction analogy. A band is marching down a 34 •• 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 1

θ 2

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

n1 sinθ 1

= n2 sinθ 2

Properties of Light Solving for θ 2 yields:

The ratio of the indices of refraction is the reciprocal of the ratio of the speeds of the marchers in the two media: Because the left and right sides of  the 30° angle and θ 1 are mutually  perpendicular, θ 1 = 30°. Substitute numerical values and evaluate θ 2:

θ 2

2909

⎡n ⎤ = sin −1 ⎢ 1 sin θ 1 ⎥ ⎣ n2 ⎦ v

n1 n2

=

v1 v

=

v2 v1

=

1 2

v1

v1

= 12

v2

θ 2

= sin −1 [12 sin 30°] = 14°

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 θ 2 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 θ 1 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 θ 3 is the angle of refraction in the glass, show that n1 sin θ 1 = n3 sin θ 3. 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:

n1 sin θ 1

= n2 sin θ 2

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

n2 sin θ 2

= n3 sin θ 3

Equate the two expressions for  n2 sin θ 2 to obtain:

n1 sin θ 1

= n3 sin θ 3

On a safari, you are spear fishing while wading in a river. You observe 36 •• 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

2910

Chapter 31

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 .   m    0    5  .    1     =        h

64.0° α

θ 1

Air 

n1

Water 

n2

  m    0    2  .    1

θ 2

    =

       d

 L

Use the pictorial representation to express the aiming angle α :

⎡ h + d ⎤ α  = tan −1 ⎢ ⎣ L + l ⎥⎦

Referring to the diagram, note that:

 L = h tan θ 1 and l = d tan θ 2

Substituting for  L and l yields:

Apply Snell’s law at the air-water  interface to obtain: Solving for θ 2 yields:

⎡

⎤ ⎥ ⎣ h tan θ 1 + d tan θ 2 ⎦

α  = tan −1 ⎢

n1 sin θ 1

θ 2

h + d 

= n 2 sin θ 2

⎡n ⎤ = sin −1 ⎢ 1 sin θ 1 ⎥ ⎣ n2 ⎦

Properties of Light

2911

Substitute for θ 2 to obtain:

⎡ ⎤ ⎢ ⎥ + h d  ⎢ ⎥ α  = tan −1 ⎢ ⎛  −1 ⎡ n1  ⎞ ⎥ ⎤ ⎢ h tan θ 1 + d tan⎜ sin ⎢ sin θ 1 ⎥ ⎟ ⎥ ⎜ ⎢⎣ ⎣ n2 ⎦ ⎠⎟ ⎥⎦ ⎝   Noting that θ 1 is the complement of 64.0°, substitute numerical values and evaluate α :

⎡ ⎤ ⎢ ⎥ 1.50 m + 1.20 m −1 ⎢ ⎥ = 66.9° = tan ⎢ ⎛  −1 ⎡1.00 ⎤ ⎞ ⎥ ⎢ (1.50 m ) tan 26.0° + (1.20 m ) tan⎜⎜ sin ⎢1.33 sin 26.0°⎥ ⎟⎟ ⎥ ⎣ ⎦ ⎠ ⎦ ⎝  ⎣ 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. Sitting

Standing α hstanding

θ 1

Air  Water 

hsitting

n1 n2

θ 2

α'

θ 1'

Air  Water 

d 

n2

θ 2'

h

h  L'

 L

n1

' d 

2912

Chapter 31

Use the fact that angles α  and θ 1 are complementary, as are α ′ and θ 1′ to determine θ 1 and θ 1′:

θ 1

= 90° − α  = 90° − 28° = 62°

and θ 1 ' = 90° − α ' = 90° − 14° = 76°

Express the distances L and L′ in terms of θ 1 and θ 1′:

 L = hstanding tan θ 1

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′:

 L = (1.7 m ) tan 62° = 3.197 m

Referring to the pictorial representations, note that:

and  L' = hsitting tan θ 1 '

and  L' = (0.7 m) tan 76° = 2.808 m

tan θ 2

l

d − L

h

h

= =

(1)

and tan θ 2 ' =

Divide the first of these equations by the second to obtain: Solving for d yields:

tan θ 2 tan θ 2 ' d  =

=

l'

h

=

d  − L ' h

d − L d − L '

 L' tan θ 2 − L tan θ 2 '

tan θ 2 − tan θ 2 '

(2)

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

θ 2

⎡n ⎤ = sin −1 ⎢ 1 sin θ 1 ⎥ ⎣ n2 ⎦

Substitute numerical values and evaluate θ 2 :

θ 2

1.00 ⎤ = sin −1 ⎡⎢ sin 62°⎥ = 41.60° ⎣1.33 ⎦

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

θ 2 ' = sin −1 ⎢

Substitute numerical values and evaluate θ 2 ' :

⎡1.00 sin 76°⎤ = 46.85° θ 2 ' = sin −1 ⎢ ⎥⎦ ⎣1.33

⎡ n1 ⎤ sin θ 1 '⎥ ⎣ n2 ⎦

Properties of Light

2913

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

(2.808 m ) tan 41.60° − (3.197 m ) tan 46.85° tan 41.60° − tan 46.85°

Solving equation (1) for h yields:

Substitute numerical values and evaluate h:

h=

h

=

= 5.130 m = 5.1 m wide

d − L

tan θ 2 5.130 m − 3.197 m

= 2.2 m deep

tan 41.60°

be am of light incident on a glass plate of  38 ••• Figure 31-57 shows a beam 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? c m? l

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 , θ r r,  and θ i. Setting the derivative of the resulting equation equal to zero will yield the value of  θ i that maximizes x.

θ i θ i

 x Air

θ i θ r  d 

Glass

θ r 

Air

= 2d tan θ r 

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

l

Express x as a function of l, d , θ r r , and θ i:

 x = 2d tan θ r  cosθ i

Differentiate x with respect toθ i: dx d θ i

= 2d 

⎛  d θ   ⎞ (tan θ r  cosθ i ) = 2d ⎜⎜ − tan θ r  sin θ i + sec 2 θ r  cos θ i r  ⎟⎟ d θ i d θ i  ⎠ ⎝  d 

(1)

2914

Chapter 31 n1 sin θ i

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

= n2 sin θ r 

(2)

or, because n1 = 1 and n2 = n, sin θ i = n sin θ r 

cosθ i d θ i

Differentiate implicitly with respect toθ I to obtain:

or  d θ r  d θ i

=

= n cosθ r d θ r 

1 cos θ i n cos θ r 

Substitute in equation (1) to obtain: dx d θ i

⎛  sin θ r  ⎛ 1 cos 2 θ i sin θ r  sin θ i  ⎞ 1 cos θ i cos θ i  ⎞ ⎟⎟ = 2d ⎜⎜ ⎟⎟ = 2d ⎜⎜ − − sin θ i + 2 3 cos cos θ  n θ  n θ  cos cos cos θ  θ  r  r  ⎠ r  ⎝   ⎠ r  r  ⎝ 

Substitute 1 − sin 2 θ i for  cos 2 θ i 1 and sin θ i for  sin θ r  to obtain:

dx d θ i

⎛ 1 − sin 2 θ i sin 2 θ i  ⎞ ⎟⎟ = 2d ⎜⎜ − 3 cos n θ  cos n θ  r  ⎠ r  ⎝ 

n

Multiply the second term in parentheses by cos 2 θ r  cos 2 θ r  and simplify to obtain:

⎛ 1 − sin 2 θ i sin 2 θ i cos 2 θ r  ⎞ ⎟⎟ = 2d 3 (1 − sin 2 θ i − sin 2 θ i cos 2 θ r  ) = 2d ⎜⎜ − 3 3 d θ i n cos θ r   ⎠ n cos θ r  ⎝  n cos θ r  dx

Substitute 1 − sin 2 θ r  for cos 2 θ r  : dx d θ i

Substitute

=

2d  3

n cos θ r 

[1 − sin

2

θ i

− sin 2 θ i (1 − sin 2 θ r  )]

1 sin θ i for  sin θ r  to obtain: n dx d θ i

=

⎡ ⎛  1  ⎞⎤ 1 − sin 2 θ i − sin 2 θ i ⎜1 − 2 sin 2 θ i ⎟⎥ ⎢ n cos θ r  ⎣ ⎝  n  ⎠⎦ 2d  3

Factor out 1/n2, simplify, and set equal to zero to obtain: dx d θ i

=

2d  3

3

n cos θ r 

[sin

4

θ i

− 2n 2 sin 2 θ i + n 2 ] = 0 for extrema

Properties of Light If dx/d θ1  = 0, then it must be true that: Solve this quartic equation for θ i to obtain:

sin 4 θ i

θ i

− 2n 2 sin 2 θ i + n 2 = 0

⎛  1  ⎞ = sin − 1 ⎜⎜ n 1 − 1 − 2 ⎟⎟ n ⎝   ⎠

(b) Evaluate θ I for n = 1.60:

⎛  1  ⎞⎟ ⎜ θ i = sin 1.6 1 − 1 − 2 ⎟ ⎜ 1 . 60 ( ) ⎝   ⎠ = 48.5°

(c) In (a) we showed that:

 x = 2d tan θ r  cosθ i

Solve equation (2) for θ r :

2915

−1

⎛ n1  ⎞ sin θ i ⎟⎟ ⎝ n2  ⎠

θ r  = sin −1 ⎜⎜

⎛  1 sin 48.5° ⎞ = 27.9° ⎟ ⎝ 1.60  ⎠

Substitute numerical values and evaluate θ r: 

θ r  = sin −1 ⎜

Substitute numerical values and evaluate x:

 x  = 2(4.0 cm ) tan 27.9° cos 48.5°

= 2.8 cm

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:

n1 sin θ 1

When there is total internal reflection:

θ 1

Substitute to obtain:

n1 sin θ c

Solving for θ c yields:

θ c

= n2 sin θ 2

= θ c and θ 2 = 90° = n2 sin 90° = n2

⎛ n  ⎞ = sin −1 ⎜⎜ 2 ⎟⎟ ⎝ n1  ⎠

2916

Chapter 31

Substitute numerical values and evaluate θ c:

θ c

1.00 ⎞ = sin −1 ⎛  ⎜ ⎟ = 48.8° ⎝ 1.33 ⎠

A glass surface (n = 1.50) has a layer of water (n = 1.33) on it. Light in 40 • 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 glasswater interface under total internal reflection conditions.

n2 = 1.33

θ 2

Water Glass n1 = 1.50

θ 1

Apply Snell’s law to the glasswater interface:

n1 sin θ 1

= n2 sin θ 2

At the critical angle, θ 1 = θ c and θ 2 = 90°:

n1 sin θ c

= n2 sin 90°

Solve for θ c:

Substitute numerical values and evaluate θ c:

θ c

⎡n ⎤ = sin −1 ⎢ 2 sin 90°⎥ ⎣ n1 ⎦

θ c

1.33 ⎤ = sin −1 ⎡⎢ sin 90°⎥ = 62.5° ⎣1.50 ⎦

A point source of light is located 5.0 m below the surface of a large 41 •  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 θ c 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 θ c.

Express the area of the circle whose radius is r :

Air 

r 

n2 = 1.00

90º

Water 

θ c

n1 = 1.33

d  = 5.0 m

θ c

 A = π  r 

2

Properties of Light Relate the radius of the circle to the depth d of the point source and the critical angle θ c:

r  = d tan θ c

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

n1 sin θ c

Solving for θ c yields:

Substitute for r and θ c to obtain:

θ c

= n2 sin 90° = n2

⎛ n  ⎞ = sin −1 ⎜⎜ 2 ⎟⎟ ⎝ n1  ⎠

 A = π  [d tan θ c ]

2

⎡ ⎛  ⎧ n ⎫ ⎞⎤ = π  ⎢d tan⎜⎜ sin −1 ⎨ 2 ⎬ ⎟⎟ ⎥ ⎢⎣ ⎩ n1 ⎭ ⎠⎥⎦ ⎝  Substitute numerical values and evaluate A:

2917

2

⎡ ⎛  1 ⎧ 1 ⎫ ⎞⎤  A = π  ⎢(5.0 m ) tan⎜⎜ sin − ⎨ ⎬ ⎟⎟⎥ 1 . 33 ⎩ ⎭ ⎠⎦ ⎝  ⎣ = 1.0 ×10 2 m 2

2

Light traveling in air strikes the largest face of an isosceles-right42 •• 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-righttriangle prism to conclude that θ c = 45°.

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

45º

n1

θ c 45º n2 = 1.00

v

=

c n1

2918

Chapter 31

n1 sin θ c

Apply Snell’s law to the prism-air  interface to obtain: Solving for n1 yields:

n1

=

= n2 sin 90° = 1

1 sin θ c

Substitute for n1 and simplify to obtain:

v = c sin θ c

Substitute numerical values and evaluate v:

v = 2.998 ×10 m/s sin 45° 8

= 2.1×10 8 m/s

A point source of light is located at the bottom of a steel tank, and an 43 •• 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. r 

n2 n1

θ 2 θ c

d 

θ c

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

n1 sin θ 1

= n2 sin θ 2

Light is seen by the observer  when θ 1 = θ c and θ 2 = 90°:

n1 sin θ c

= n2 sin 90° = n2

Properties of Light Because the medium above the interface is air, n2 = 1. Solve for  n1 to obtain: From the geometry of the diagram: Substitute for  θ c to obtain:

n1

=

1 sin θ c

tan θ c

n1

2919

=

r  ⎞ ⇒ θ c = tan −1 ⎛  ⎜ ⎟ d  ⎝ d  ⎠ r 

1

=

⎡ ⎣

⎛  r  ⎞⎤ ⎟ ⎝ d  ⎠⎥⎦

sin ⎢ tan −1 ⎜ Substitute numerical values and evaluate n1:

n1

=

1

⎡ ⎛ 6.00 cm ⎞⎤ ⎟⎟⎥ sin ⎢ tan −1 ⎜⎜ 5 . 00 cm ⎝   ⎠⎦ ⎣

= 1.30

An optical fiber allows rays of light to propagate long distances by 44 •• 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 sinθ 1, where θ 1 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 corecladding interface just at the critical angle. Using the figure as a guide, show that the numerical aperture is given by sin θ 1 = n 22 − n 32 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 sin θ 1 Referring to the figure, note that:

Apply the Pythagorean theorem to the right triangle to obtain: Solving for 

b c

yields:

sin θ c

a

b c

2

=

n3

=

n2

a c

+ b = c or  2

= 1−

2

a2 c

2

=

2

n2

and sin θ 2

a

2

c2

+

b

2

c2

− n32 . =

=1

b c

2920

Chapter 31

Substitute for 

a c

and

b c

to obtain:

Use the law of refraction to relate θ 1 and θ 2: Substitute for sinθ 2, let n1 = 1 (air), and simplify to obtain:

2

sin θ 2

= 1−

n1 sin θ 1

n3

2

n2

= n2 sin θ 2 2

sin θ 1

= n2 1 −

n3

2 2

n

=

2

n2

− n32

45 •• [SSM] Find the maximum angle of incidence θ 1 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 μ m, 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:

sin θ 1

Solve for θ 1 to obtain:

θ 1

Substitute numerical values and evaluate θ 1:

θ 1

=

= sin −1

n22 − n32

(

n22 − n32

= sin −1 ⎛  ⎜ (1.492)2 − (1.489 )2  ⎞⎟ ⎝   ⎠ = 5°

Calculate the difference in time needed for two pulses of light to travel 46 •• 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 θ 1 is the ratio of  c to a multiplied by L.

The difference in the travel times Δt is given by:

Δt = t θ  − t normal 1

incidence

(1)

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

t θ 1

=

distance traveled speed in the medium

=

 L

2921

c

a c

n2

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

 L c

=

t normal incidence

n2

Substitute for t θ 1 and t normal

in

incidence

equation (1) to obtain:

Δt  =

 L

c

a c

−

n2

Simplify to obtain:

Δt  =

n2

n2 L ⎛ c c

 ⎞ ⎜ − 1⎟ ⎝ a  ⎠

sin θ c

=

a

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

sin θ c

=

n3

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

Δt  =

Substitute numerical values and evaluate Δt :

Δt 

Referring to the figure, note that:

=

 L c

(2)

c

n2

⇒

a c

=

n3 n2

n2 L ⎛ n2 c

 ⎞ ⎜⎜ − 1⎟⎟ ⎝ n3  ⎠

(1.492)(15 km ) ⎛ 1.492  ⎞ − 1⎟ ⎜ 2.998 ×108 m/s ⎝ 1.489  ⎠

= 150 ns 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 θ c 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 

2922

Chapter 31

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

n3 = 1.00

θ 3 θ  2

Water

n2 = 1.33

θ 2 Glass

n1 = 1.50

θ 

1

(a) Apply Snell’s law to the glasswater interface:

n1 sin θ 1

= n2 sin θ 2

At the critical angle, θ 1 = θ c and θ 2 = 90°:

n1 sin θ c

= n2 sin 90°

Solving for θ c yields:

Substitute numerical values and evaluate θ c: (b) Apply Snell’s law to a water-air  interface at the critical angle for a water-air interface: Solving for  θ c yields:

Substitute numerical values and evaluate θ c : Apply Snell’s law to a ray incident at the glass-water interface:

θ c

⎡n ⎤ = sin −1 ⎢ 2 sin 90°⎥ ⎣ n1 ⎦

θ c

1.33 ⎤ = sin −1 ⎡⎢ sin 90°⎥ = 62.5° ⎣1.50 ⎦

n2 sin θ c

= n3 sin 90°

θ c

⎛ n  ⎞ = sin −1 ⎜⎜ 3 ⎟⎟ ⎝ n2  ⎠

θ c

1.00 ⎞ = sin −1 ⎛  ⎜ ⎟ = 48.8° ⎝ 1.33 ⎠

n1 sin θ 1

= n2 sinθ 2

and θ 1

⎛ n  ⎞ = sin −1 ⎜⎜ 2 sin θ 2 ⎟⎟ ⎝  n1  ⎠

Properties of Light

For  θ 2

= θ c :

Substitute numerical values and evaluate θ 1:

θ 1

⎡⎛ n  ⎞ ⎛ n  ⎞⎤ ⎛ n  ⎞ = sin −1 ⎢⎜⎜ 2 ⎟⎟ ⎜⎜ 3 ⎟⎟⎥ sin −1 ⎜⎜ 3 ⎟⎟ ⎝ n1  ⎠ ⎣⎝  n1  ⎠ ⎝ n2  ⎠⎦

θ 1

1.00 ⎞ = sin −1 ⎛  ⎜ ⎟ = 41.8° 1 . 53 ⎝   ⎠

2923

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 θ r .

b

θ i θ i θ i  x

t 

θ r 

 x = 2t tanθ r 

The separation of the reflected rays is:

b = x cosθ i

Substitute for  x to obtain:

b = 2t tan θ r  cosθ i

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

sin θ i

obtain:

Glass

n = 1.5

θ r 

Express the distance x in terms of t  and θ r :

Substitute for  θ r  in equation (1) to

Air

(1)

sin θ i  ⎞ = n sin θ r  ⇒ θ r  = sin −1 ⎛  ⎜ ⎟ n  ⎠ ⎝ 

⎡ ⎣

b = 2t tan ⎢sin

−1

⎛ sin θ i  ⎞⎤ cosθ  ⎜ ⎟⎥ i ⎝  n  ⎠⎦

2924

Chapter 31

Substitute numerical values and evaluate b:

⎡ ⎣

⎛ sin 40° ⎞⎤ cos 40° ⎟ ⎝  1.5  ⎠⎥⎦

b = 2(3.0 cm ) tan ⎢sin −1 ⎜

= 2.2 cm Dispersion A beam of light strikes the plane surface of silicate flint glass at an 49 • 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:

Δθ  = θ r,red − θ r,violet

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

sin 45° = n sin θ r  ⇒ θ r  = sin −1 ⎜

Substituting for  θ r in equation (1)

⎛  1  ⎞ ⎛   ⎞ ⎟⎟ − sin −1 ⎜⎜ 1 ⎟⎟ Δθ  = sin −1 ⎜⎜ ⎝  2 nred  ⎠ ⎝  2 nviolet  ⎠

yields: Substitute numerical values and evaluate Δθ  :

(1)

⎛  1  ⎞ ⎟ n 2 ⎝   ⎠

Δθ 

⎛  1  ⎞ ⎛  ⎟ − sin −1 ⎜ 1 ⎟ = sin −1 ⎜⎜ ⎟ ⎜ 2 (1.66 ) ⎟ ⎝  2 (1.60) ⎠ ⎝   ⎠ = 26.2° − 25.2° = 1.0°

In many transparent materials, dispersion causes different colors 50 •• (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

 L

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

Δt  =

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

Δt  =

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

n500

≈ 1.55

and n700

≈ 1.50

Substitute numerical values and evaluate Δt :

Δt 

=

v500

−

n500 L c

2925

 L v700

−

n700 L c

=

15.0 km 2.998 × 10 8 m/s

 L c

(n500 − n700 )

(1.55 − 1.50)

≈ 3 μ s 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: tan θ  p = n2 n1 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:

(a) For n1 = 1 and n2 = 1.33:

(b) For n1 = 1 and n2 = 1.50:

θ  p

⎛ n  ⎞ = tan −1 ⎜⎜ 2 ⎟⎟ ⎝ n1  ⎠

θ  p

1.33 ⎞ = tan −1 ⎛  ⎜ ⎟ = 53.1° ⎝ 1.00 ⎠

θ  p

1.50 ⎞ = tan −1 ⎛  ⎜ ⎟ = 56.3° ⎝ 1.00 ⎠

Light that is horizontally polarized is incident on a polarizing sheet. It 52 • 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?

2926

Chapter 31

Picture the Problem The intensity of the transmitted light I is related to the intensity of the incident light I 0 and the angle the transmission axis makes with the horizontal θ  according to  I  =  I 0 cos 2 θ .

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: Substitute numerical values and evaluate θ  :

⎛  ⎜ ⎝ 

 I  =  I 0 cos 2 θ  ⇒ θ  = cos −1 ⎜

θ  = cos −1

(

0.15

)=

 I   ⎞

⎟

 I 0  ⎠⎟

67°

Two polarizing sheets have their transmission axes crossed so that no 53 • 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 I 0 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 I n be the intensity after the nth polarizing sheet and use  I  =  I 0 cos 2 θ  to find the intensity of the light transmitted through all three sheets

for θ  = 45° and θ  = 30°. (a) The intensity of the light between the first and second sheets is:

 I 1

= 12 I 0

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

 I 2

=  I 1 cos2 θ 1, 2 =

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

 I 3

= I 2 cos2 θ 2,3 = 14 I 0 cos2 45° =

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

 I 1

=

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

 I 2

=  I 1 cos 2 θ 1, 2 = 12 I 0 cos 2 30° = 83 I 0

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

 I 3

= I 2 cos2 θ 2,3 = 38 I 0 cos2 60° =

1 2

 I 0 cos 45° = 14 I 0 2

1 8

 I 0

1 2

 I 0

3 32

 I 0

Properties of Light

2927

A horizontal 5.0 mW laser beam that is vertically polarized is incident 54 • 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. P

The intensity of the beam is the ratio of its power to cross-sectional area:

 I  =

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

 I 1 =  I 0 and P1

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

P  A

 A

=

P0  A

cos 2 θ 

= P0

⇒

P = P0 cos θ  2

Express the power of the beam after  the second transmission:

P2

= P1 cos 2 θ 1, 2 = P0 cos 2 θ 12

Substitute numerical values and evaluate P2:

P2

= (5.0 mW) cos 2 27° = 4.0 mW

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°:

θ  p + θ r  = 90° ⇒ θ r  = 90° − θ  p

Substitute for θ  p to obtain:

θ r  = 90° − 60° = 30°

2928

Chapter 31

(b) From Brewster’s law we have:

=

tan θ  p

n2 n1

or, because n1 = 1.00, n2 = tan θ  p Substitute for θ  p and evaluate n2:

n2

= tan 60° = 1.7

Two polarizing sheets have their transmission axes crossed so that no 56 •• 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 I n be the intensity after the nth polarizing sheet and use 2  I  =  I 0 cos θ  to find the intensity of the light transmitted through the three sheets.

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

 I 1

=

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

 I 2

=  I 1 cos 2 θ 1, 2 =

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

 I 3

=  I 2 cos 2 θ 2,3

1 2

 I 0

2  I 0 cos θ 

1 2

= 12 I 0 cos 2 θ  cos2 (90° − θ ) = 12 I 0 cos 2 θ sin 2 θ  = 18 I 0 (2 cosθ  sin θ )2 =

1 8

2

 I 0 sin 2θ 

(b) Because the sine function is a maximum when its argument is 90°, the maximum value of  I 3 occurs when θ  = 45°. If the middle polarizing sheet in Problem 56 is rotating at an angular  57 •• 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 I n be the intensity after the nth polarizing sheet, use 2  I  =  I 0 cos θ  to find the intensity of the light transmitted through each sheet, and

replace θ  with ω t .

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

 I 1

= 12 I 0

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

 I 2

=  I 1 cos 2 θ 1, 2 =

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

 I 3

= I 2 cos 2 θ 2,3

1 2

2929

2

 I 0 cos ω t 

= 12 I 0 cos 2 ω t cos2 (90° − ω t ) = 12 I 0 cos 2 ω t sin 2 ω t  = 18 I 0 (2 cos ω t sin ω t )2 =

 I 0 sin 2 2ω t 

1 8

A stack of  N + 1 ideal polarizing sheets is arranged so that each sheet 58 •• is rotated by an angle of π /(2 N ) rad with respect to the preceding sheet. A linearly  polarized light wave of intensity I 0 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 I 0 cos 2

 N 

⎡⎣π  (2 N )⎤⎦ . (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 I n be the intensity after the nth polarizing sheet and use 2  I  =  I 0 cos θ  to find the ratio of  I n+1 to I n.

(a) Find the ratio of  I n+1 to I n:

 I n+1  I n

Because there are N such reductions of intensity:

 I  N +1  I 1

= cos 2

π 

2 N  π   ⎞ = cos 2 N ⎛  ⎜ ⎟ ⎝ 2 N  ⎠

=

 I  N +1

=

 I 0 cos 2 N ⎜

 I 0

and  I  N +1

⎛  π   ⎞ ⎟ ⎝ 2 N  ⎠

(b) A spreadsheet program to graph I  N +1/ I 0 as a function of  N is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell A2 A3

Content/Formula 2 A2 + 1

Algebraic Form  N   N + 1

2930

Chapter 31 B2

⎛  π   ⎞ ⎟ ⎝ 2 N  ⎠

(cos(PI()/(2*A2))^(2*A2)

cos 2 N ⎜

1 2 3 4 5

A  N  2 3 4 5

B  I / I 0 0.250 0.422 0.531 0.605

95 96 97 98 99 100

95 96 97 98 99 100

0.974 0.975 0.975 0.975 0.975 0.976

A graph of  I / I 0 as a function of  N follows. 1.0 0.9 0.8 0.7        0

       I 0.6    /        I

0.5 0.4 0.3 0.2 0

10

20

30

40

50

60

70

80

90

100

 N 

(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  I n be the intensity after the nth polarizing sheet and use 2  I  =  I 0 cos θ  to find the ratio of  I n+1 to  I n. Because each sheet introduces a 2%

loss of intensity, the net transmission after  N sheets (0.98) N .  I n+1

Find the ratio of  I n+1 to I n:

= (0.98)cos 2

 I n

Because there are N such reductions of intensity:

 I  N +1  I 0

π 

2 N 

π   ⎞ = (0.98) N  cos 2 N ⎛  ⎜ ⎟ ⎝ 2 N  ⎠

A spreadsheet program to graph I  N +1/ I 0 for an ideal polarizer as a function of  N, the percent transmission, and I  N +1/ I 0 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 A3 B2

Content/Formula 1 (cos(PI()/(2*A2))^(2*A2)

Algebraic Form  N  ⎛  π   ⎞ cos 2 N ⎜ ⎟ ⎝ 2 N  ⎠

C3

(0.98)^A3

(0.98) N 

D4

B3*C3

A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

 N 

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

B Ideal Polarizer 0.000 0.250 0.422 0.531 0.605 0.660 0.701 0.733 0.759 0.781 0.798 0.814 0.827 0.838 0.848 0.857 0.865

(0.98) N  cos 2 N ⎛  ⎜

π   ⎞

⎟ ⎝ 2 N  ⎠

C Percent Transmission 0.980 0.960 0.941 0.922 0.904 0.886 0.868 0.851 0.834 0.817 0.801 0.785 0.769 0.754 0.739 0.724 0.709

D Real Polarizer  0.000 0.240 0.397 0.490 0.547 0.584 0.608 0.624 0.633 0.638 0.639 0.638 0.636 0.632 0.626 0.620 0.613

2932

Chapter 31 20 21 22

18 19 20

0.872 0.878 0.884

0.695 0.681 0.668

0.606 0.598 0.590

A graph of  I / I 0 as a function of  N for the quantities described above follows: 1.0

0.8

0.6  I  / I  0

Ideal Polarizer

0.4

Percent Transmission 0.2

Real Polarizer

0.0 0

2

4

6

8

10

12

14

16

18

20

 Number of sheets ( N )

Inspection of the table, as well as of the graph, tells us that the optimum number  of sheets is 11. Show mathematically that a linearly polarized wave can be thought of  60 •• 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:

 E  x

= E 0 cos

t 

and  E  y =  E 0 sin t or  E  y

= − E 0 sin

t 

for left and right circular polarization, respectively. For a wave polarized along the x axis:

r

r

 E right + E left

= E 0 cos ω t i ˆ + E 0 cos ω t i ˆ = 2 E 0 cos ω t i ˆ

Suppose that the middle sheet in Problem 53 is replaced by two 61 ••  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 I n be the intensity after the nth polarizing sheet and use 2  I  =  I 0 cos θ  to find the intensity of the light transmitted by the four sheets.

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

 I 1

= 12 I 0

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

 I 2

=  I 1 cos 2 θ 1, 2 =

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

 I 3

=  I 2 cos 2 θ 2,3 = 83 I 0 cos 2 30° = 329 I 0

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

 I 4

27 =  I 3 cos 2 θ 3, 4 = 329 I 0 cos 2 30° = 128  I 0

 I 0 cos 30° = 83 I 0

1 2

2

= 0.211 I 0

(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.

Show that the electric field of a circularly polarized wave propagating 62 ••  parallel with the x axis can be expressed by r ˆ + E  cos(kx − ω t )k  ˆ.  E  = E  sin (kx − ω t )i  0

0

r

r

Picture the Problem We can use the components of  E  to show that  E is constant in time and rotates with angular frequency ω . r

Express the magnitude of  E in terms of its components:

 E  =  E  x + E  y 2

2

Substitute for  E  x and E  y to obtain:  E  =

[ E 0 sin(kx − ω t )]2 + [ E 0 cos(kx − ω t )] 2 =

= E 0

[

]

 E 02 sin 2 (kx − ω t ) + cos 2 (kx − ω t )

r

and the  E vector rotates in the yz plane with angular frequency ω .

2934

Chapter 31

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: r

ˆ + E  cos(kx − ω t )k  ˆ  E  =  E 0 sin (kx − ω t ) j  0

The corresponding magnetic field is: r

ˆ − B cos(kx − ω t ) j  ˆ  B =  B0 sin (kx − ω t )k  0

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: r

ˆ − E  cos(kx + ω t )k  ˆ  E  =  E 0 sin (kx + ω t ) j  0

Sources of Light A helium–neon laser emits light that has a wavelength equal to 632.8 64 • 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  E  photon : The energy of a photon is given by:

n=

P  E  photon

 E  photon

=

hc

λ 

Properties of Light Substitute for  E  photon to obtain:

n=

2935

λ P hc

Substitute numerical values and evaluate n: n=

1eV (632.8 nm )(4.00 mW ) = 1240 eV ⋅ nm 1.602 × 10 −19 J

1.27 × 1016 photons/s

The first excited state of an atom of a gas is 2.85 eV above the ground 65 • 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 E Raman = E inc − Δ E where Δ E is the energy for resonance absorption, to find the wavelength of the Raman scattered light. hc

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

λ  =

Substitute numerical values and evaluate λ :

λ  =

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

λ Raman

=

 E Raman

=  E inc − Δ E  1240 eV ⋅ nm = − 2.85 eV

Relate the energy of the Raman scattered light E Raman to the energy of the incident light E inc:

Substitute numerical values and evaluate λ Raman:

 E 

1240 eV ⋅ nm 2.85 eV

= 435 nm

1240 eV ⋅ nm  E Raman

320 nm = 1.025 eV

λ Raman

=

1240 eV ⋅ nm 1.025 eV

= 1210 nm

A gas is irradiated with monochromatic ultraviolet light that has a 66 •• 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

2936

Chapter 31

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 hc and the atomic state excited by the irradiation is given by Δ E  = hf  = . λ 

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

Δ E  = hf  =

Substitute 368 nm for λ  and evaluate Δ E :

Δ E  =

hc

λ 

=

1240 eV ⋅ nm λ 

1240 eV ⋅ nm 368 nm

= 3.37 eV

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/Δ E, longer wavelengths correspond to smaller energy differences.

(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:

λ max

=

3

4.35 eV

2

3.20 eV

1

2.11 eV

0

0

1240 eV ⋅ nm 3.20 eV

= 388 nm

Properties of Light

The fluorescence wavelengths are:

=

1240 eV ⋅ nm 3.20 eV − 2.11eV

λ 1→0

=

1240 eV ⋅ nm 2.11eV − 0

λ 0→3

=

λ 2→1

2937

= 1140 nm

and

(b) For excitation:

1240 eV ⋅ nm 4.35 eV

= 588 nm

= 285 nm

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

λ 3→2

=

1240 eV ⋅ nm 4.35 eV − 3.20 eV

= 1080 nm

(not in visible spectrum) λ 2→1

=

1240 eV ⋅ nm 3.20 eV − 2.11eV

= 1140 nm

(not in visible spectrum) λ 1→0

=

1240 eV ⋅ nm 2.11eV − 0

λ 3→1

=

1240 eV ⋅ nm 4.35 eV − 2.11eV

and λ 2→0

=

1240 eV ⋅ nm 3.20 eV − 0

= 588 nm

= 554 nm

= 388 nm

(not in visible spectrum) Singly ionized helium is a hydrogen-like atom that has a nuclear  68 •• charge of +2e. Its energy levels are given by E n = –4 E 0/n2, where n = 1, 2, … and  E 0 = 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 E 1 prior to irradiation.) Determine the Concept The energy difference between the ground state and the first excited state is 3 E 0 = 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.

2938

Chapter 31

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:

 E  = PΔt 

Substitute numerical values and evaluate E :

 E = (10 MW )(1.5 ns ) = 15 mJ

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

 N  =

The energy of a photon is given by:

Substitute for  E  photon to obtain:

 E   E  photon

 E  photon

 N  =

=

hc

λ 

λ  E  hc

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

1eV (694.3 nm )(15 mJ ) = 1240 eV ⋅ nm 1.602 × 10 −19 J

5.2 × 1016

General Problems A beam of red light that has a wavelength of 700 nm in air travels in 70 • 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 =  f λ  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 (a) The wavelength λ n of light in a medium whose index of refraction is n is given by: Substitute numerical values and evaluate λ water :

λ n

=

=

λ 0

700 nm nwater 

=

v

=

 f 

λ water  =

c nf 

2939

n

700 nm 1.33

= 526 nm

(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:

n1 sin θ c

= n2

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

n1 sin θ  p

= n2 sin 90° − θ  p = n2 cosθ  p

(1)

or  tan θ  p

=

⎛ n  ⎞ ⇒ θ  p = tan −1⎜⎜ 2 ⎟⎟ (2) n1 ⎝ n1  ⎠

n2

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

n2

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

θ  p

= tan −1 (sin θ c )

Substitute numerical values and evaluate θ  p:

θ  p

= tan −1 (sin 48°) = 37°

n1

= sin θ c

Show that when a flat mirror is rotated through an angle θ  about an 72 •• 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 

2940

Chapter 31

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  φ 

φ +θ 

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 θ c 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:

n1 sin θ 1

= n2 sin θ 2

If θ 1 = θ c and n2 = 1:

n1 sin θ c

= sin 90° = 1

and θ c

⎛  1  ⎞ = sin −1 ⎜⎜ ⎟⎟ ⎝ n1 ⎠

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

θ c

1  ⎞ = sin −1 ⎛  ⎜ ⎟ = 36.8° ⎝ 1.67 ⎠

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

θ c

1  ⎞ = sin −1 ⎛  ⎜ ⎟ = 38.7° ⎝ 1.60 ⎠

Light is incident on a slab of transparent material at an angle θ 1, as 74 •• shown in Figure 31-60. The slab has a thickness t and an index of refraction n.

( ) sin ⎡⎣tan (d / t )⎤⎦ , where d is the distance shown in the

Show that n = sin θ 1 figure.

-1

Properties of Light Picture the Problem We can apply Snell’s law at the air-slab interface to express the index of refraction n in terms of θ 1 and θ 2 and then use the geometry of the figure to relate θ 2 to t and d .

2941

θ 1

n

θ 2

t 

d 

Applying Snell’s law to the first interface yields: From the diagram:

Substitute for θ 2 to obtain:

sin θ 1

= n sin θ 2 ⇒ n =

d  = t tan θ 2

n=

sin θ 1 sin θ 2

d  ⎞ ⇒ θ 2 = tan −1⎛  ⎜ ⎟ ⎝  t  ⎠

sin θ 1

⎡ ⎣

⎛ d  ⎞⎤ ⎟ ⎝  t  ⎠⎥⎦

sin ⎢ tan −1 ⎜

A ray of light begins at the point (–2.00 m, 2.00 m, 0.00 m), strikes a 75 •• 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.

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

 y , m 6 4 d 2

2 d 1 −2

 x , m 0

 x min

2

 D = d 1 + d 2

= ( x + 2 )2 + 4 + (2 − x )2 + 36

2942

Chapter 31

Differentiate D with respect to x: dD dx

=

d 

⎡ ( x + 2)2 + 4 + (2 − x )2 + 36 ⎤ ⎥⎦ dx ⎢⎣ − 12

− 12

= [( x + 2) + 4] 2( x + 2) + [(2 − x ) + 36] 2(2 − x )(− 1) 2

1 2

2

1 2

Setting this expression equal to zero for extreme values yields: 2 − x min +2 − =0 2 2 2 4 2 36 ( x min + ) + ( − x min ) +  x min

= − 1.00 m

Solve for  xmin to obtain:

 x min

(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:

⎛  x − x1  ⎞ ⎟ ⇒ θ i = tan −1 ⎜⎜ ⎟  y − y1 −  y  y 1  ⎠ ⎝  where ( x, y) = (−1.00 m, 0). tan θ i

=

 x − x1

Substitute numerical values and evaluate θ i: θ i

⎡ − 1.00 m − (− 2.00 m) ⎤ = tan −1 ⎢ ⎥ = 26.6° 0 − 2.00 m ⎣ ⎦

(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:

⎛  x − x ⎞ ⇒ θ r  = tan −1 ⎜⎜ 2 ⎟⎟  y2 − y ⎝  y2 − y ⎠ where ( x, y) = (−1.00 m, 0). tan θ r  =

 x2 − x

Substitute numerical values and evaluate θ r :

⎡ 2.00 m − (− 1.00 m )⎤ = 26.6° ⎥⎦ 6.00 m − 0 ⎣

θ r  = tan −1 ⎢

To produce a polarized laser beam a plate of transparent material, 76 •• (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 θ P1 is the

Properties of Light

2943

 polarizing angle for the n1 to n2 interface, then θ P2 is the polarizing angle for the n2 to n1 interface. Picture the Problem Let the angle of refraction at the first interface by θ 1 and the angle of refraction at the second interface be θ 2. We can apply Snell’s law at each interface and eliminate θ 1 and n2 to show that θ 2 = θ P2.

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

tan θ P1

=

n2 n1

Draw a reference triangle consistent with Brewster’s law:

  2

  n  2

  2   +

  n  1

n2

θ 

P1

n1

Apply Snell’s law at the n1-n2 interface: Solve for θ 1 to obtain:

Referring to the reference triangle we note that:

n1 sin θ P1

θ 1

θ 1

= n2 sin θ 1

⎛ n  ⎞ = sin −1 ⎜⎜ 1 sin θ P1 ⎟⎟ ⎝ n2  ⎠ ⎛ n  ⎞ n = sin −1 ⎜ 1 2 2 2 ⎟ ⎜ n2 n + n ⎟ 1 2  ⎠ ⎝  ⎛  n  ⎞ = sin −1 ⎜ 2 1 2 ⎟ ⎜ n +n ⎟ ⎝  1 2  ⎠

i.e., θ 1 is the complement of θ  p1. Apply Snell’s law at the n2-n1 interface: Solve for θ 2 to obtain:

n2 sin θ 1

θ 2

= n1 sin θ 2

⎛ n  ⎞ = sin −1 ⎜⎜ 2 sin θ 1 ⎟⎟ ⎝ n1  ⎠

2944

Chapter 31

⎛ n  ⎞ n = sin −1 ⎜ 2 2 1 2 ⎟ ⎜ n1 n + n ⎟ 1 2  ⎠ ⎝  ⎛  n  ⎞ = sin −1 ⎜ 2 2 2 ⎟ = θ P2 ⎜ n +n ⎟ ⎝  1 2  ⎠

Refer to the reference triangle again to obtain:

θ 2

Equate these expressions for  n2 sin θ 1

n1 sin θ P

= n1 sin θ 2 ⇒ θ 2 =

θ P

to obtain: 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 airglass interface.

From Brewster’s law we have:

θ  p

⎛ n  ⎞ = tan −1 ⎜⎜ 2 ⎟⎟ ⎝ n1  ⎠

or, for n1 = 1, θ  p = tan −1 (n2 ) For silicate flint glass, n2 ≈ 1.62 and:

 p,silicate flint

For borate flint glass, n2 ≈ 1.57 and:

 p, borate flint

For quartz glass, n2 ≈ 1.54 and:

 p, quartz

For silicate crown glass, n2 ≈ 1.51 and:

= tan −1 (1.62) = 58.3° = tan −1 (1.57) = 57.5°

= tan −1 (1.54) = 57.0°

 p,silicate crown

= tan −1 (1.51) = 56.5°

A light ray passes through a prism with an apex angle of α , as shown 78 •• 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 

(

)⎤⎦ = n sin ( α ).

refraction of the prism material by sin ⎡⎣ 12 α  + δ 

1 2

Properties of Light

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 θ r  in terms of  α  and δ 1 in terms of  θ r  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 90° −θ r 

θ  i

δ 

1

θ 

n1

n2

θ r  = 12 α 

Express the angle of deviation at the refracting surface:

δ 1

Referring to triangle ABC, we see that:

δ  = 2δ 1

= θ i − θ r  = θ i − 12 α 

= 2(θ 1 − 12 α ) = 2θ i − α 

α

A

n1

δ 

1

δ  δ 

1

n2

C

δ 

r 

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

B

2945

δ 

n1

= 12 (α  + δ )

Solving for θ I yields:

θ i

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

sin θ i

= n sin 12 α 

n1

2946

Chapter 31

Substitute for θ I to obtain:

sin

α  + δ 

2

= n sin

α 

(1)

2

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 θ  p = sin θ c. (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 θ  p = sin θ c.

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

n1 sin θ 1

= n2 sin θ r 

For θ 1 = θ c, n1 = n, and n2 = 1:

n sin θ c

= sin 90° = 1

For θ 1 = θ  p, n1 = n, and n2 = 1:

Because both expressions equal one: (b) For any value of θ  :

=

tan θ  p

tan θ  p

n2 n1

=

1 n

⇒ n tan θ  p = 1

= sin θ c

tan θ  > sin θ 

⇒

θ  p

> θ c

Light in air is incident on the surface of a transparent substance at an 80 •• 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).

= n sin θ 2

(a) Apply Snell’s law at the airmedium interface:

sin θ 1

Because the reflected and refracted rays are mutually perpendicular:

θ 1 + θ 2

= 90° ⇒ θ 2 = 90° − θ 1

Properties of Light Substitute for θ 2 to obtain:

sin θ 1

= n sin(90° − θ 1 ) = n cosθ 1

or  n = tan θ 1

= tan θ  p

Substitute for θ  p and evaluate n:

n = tan 58° = 1.6

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

n2 sin θ c

Because n1 = 1:

Substitute for n and evaluate θ c:

2947

= n1 sin 90° = n1

θ c

⎛  1  ⎞ 1 ⎞ = sin −1 ⎜⎜ ⎟⎟ = sin −1 ⎛  ⎜ ⎟ ⎝ n ⎠ ⎝ n2  ⎠

θ c

1  ⎞ = sin −1 ⎛  ⎜ ⎟ = 39° ⎝ 1.6 ⎠

A light ray in dense flint glass that has an index of refraction of 1.655 81 •• 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: Substitute numerical values and evaluate nliquid : (b) With the liquid removed:

nliquid

sin θ c

=

nliquid

= (1.655)sin 53.7° = 1.33

θ c

nglass

⇒ nliquid = nglass sin θ c

⎛  1  ⎞ ⎟ = sin −1 ⎜⎜ ⎟ ⎝ nglass  ⎠

2948

Chapter 31

Substitute numerical values and evaluate θ c: (c) Apply Snell’s law at the glass−liquid interface: Solve for θ 2:

θ c

1  ⎞ = sin −1 ⎛  ⎜ ⎟ = 37.2° ⎝ 1.655 ⎠

nglass sin θ 1

θ 2

Letting θ 1 be the angle of incidence found in Part (b) yields:

θ 2

Substitute numerical values and evaluate θ 2:

θ 2

= nliquid sin θ 2

⎡n ⎤ = sin −1 ⎢ glass sin θ 1 ⎥ ⎥⎦ ⎣⎢ nliquid

⎡⎛  n  ⎞⎛  1  ⎞⎤ ⎟⎥ = sin −1 ⎢⎜⎜ glass ⎟⎟⎜⎜ n n ⎢⎣⎝  liquid  ⎠⎝  glass  ⎠⎟⎥⎦ ⎛  1  ⎞ ⎟ = sin −1 ⎜⎜ ⎟ n ⎝  liquid  ⎠ 1  ⎞ = sin −1 ⎛  ⎜ ⎟ = 48.6° ⎝ 1.333 ⎠

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  I T = I 0 ⎡ 4n ⎣

2

(n + 1) ⎤⎦ . (b) Use the Part (a) result to find 2

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 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.

2949

Glass

Air 

 I glass

 I 0  I R,1

 I T

 I R,2

n

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

 I glass = I 0 − I R,1

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

⎛ 1 − n ⎞  I   I R,1 = ⎜ ⎟ 0 ⎝ 1 + n ⎠

Substitute and simplify to obtain:

⎛ 1 − n ⎞  I   I glass = I 0 − ⎜ ⎟ 0 ⎝ 1 + n ⎠ ⎡ ⎛ 1 − n ⎞ 2 ⎤ = I 0 ⎢1 − ⎜ ⎟ ⎥ ⎢⎣ ⎝ 1 + n ⎠ ⎥⎦ ⎡ 4n ⎤ = I 0 ⎢ 2⎥ ⎣ (1 + n ) ⎦

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

 I T

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

⎛ 1 − n ⎞  I   I R,2 = ⎜ ⎟ glass ⎝ 1 + n ⎠ 2 1 − n ⎞ ⎡ 4n ⎤ ⎛  =⎜  I  ⎟ ⎢ 2⎥ 0 + 1 n 1 ( ) + n ⎝   ⎠ ⎣ ⎦

where I R,1 is the intensity of the light reflected at the air-glass interface. 2

2

= I glass − I R,2

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

2950

Chapter 31

Substitute and simplify to obtain:

 I T

⎡ 4n ⎤ ⎛ 1 − n ⎞ 2 ⎡ 4n ⎤ = I 0 ⎢ −⎜  I  ⎟ ⎢ 2⎥ 2⎥ 0 + n 1 ⎣ (1 + n ) ⎦ ⎝   ⎠ ⎣ (1 + n ) ⎦ ⎡ ⎛ 1 − n ⎞ 2 ⎤ ⎡ 4n ⎤ = I 0 ⎢1 − ⎜ ⎟ ⎥⎢ 2⎥ + n 1 ⎝   ⎠ ⎣⎢ ⎦⎥ ⎣ (1 + n ) ⎦ ⎡ 4n ⎤ ⎡ 4n ⎤ = I 0 ⎢ 2 ⎥⎢ 2⎥ ⎣ (1 + n ) ⎦ ⎣ (1 + n ) ⎦ ⎡ 4n ⎤ =  I 0 ⎢ 2⎥ ⎣ (1 + n ) ⎦

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

For  N slabs:

⎡ 4n ⎤ ⎢ 2 ⎥ ⎣ (n + 1) ⎦  I t

2

2

⎡ 4n ⎤ =  I 0 ⎢ 2 ⎥ ⎣ (n + 1) ⎦

2 N 

and  I t  I 0

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

⎡ 4n ⎤ = ⎢ 2⎥ ⎣ (n + 1) ⎦

2 N 

(1)

2 N 

⎡ 4n ⎤ ⎛  I   ⎞ log⎜⎜ t ⎟⎟ = log ⎢ 2⎥ ⎝  I 0  ⎠ ⎣ (n + 1) ⎦ ⎡ 4n ⎤ = 2 N log ⎢ 2⎥ ⎣ (n + 1) ⎦

Solve for  N :  N  =

⎛  I   ⎞ log⎜⎜ t ⎟⎟ ⎝  I 0  ⎠ ⎡ 4n ⎤ 2⎥ ⎣ (n + 1) ⎦

2 log ⎢

Substitute numerical values and evaluate N :

 N  =

log(0.1)

⎡ 4(1.5) ⎤ 2 log ⎢ 2 ⎥ ⎣ (1.5 + 1) ⎦

= 28.2 ≈ 28

Properties of Light

2951

83 ••• Equation 31-14 gives the relation between the angle of deviation φ d of  a light ray incident on a spherical drop of water in terms of the incident angle θ 1 and the index of refraction of water. (a) Assume that nair  = 1, and derive an expression for d φd  /d θ1  . 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 θ 1m is given by

cos θ 1m

=

1 3

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

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:

For nair  = 1and nwater  = n:

Use the hint to differentiate φ d with respect to θ 1:

φ d

⎛ n sin θ 1 ⎞ ⎟⎟ = π  + 2θ 1 − 4 sin −1 ⎜⎜ air  n ⎝  water   ⎠

φ d

sin θ 1 ⎞ = π  + 2θ 1 − 4 sin −1 ⎛  ⎜ ⎟ ⎝  n  ⎠

d φ d d θ 1

⎡ −1 ⎛ sin θ 1 ⎞ ⎤ 2 4 sin π  + θ  − ⎜ ⎟ 1 ⎢ d θ 1 ⎣ ⎝  n  ⎠⎥⎦ d 

=

4 cos θ 1

= 2−

n

(b) Set d φd  /d θ1  = 0:

− sin 2 θ 1

2

4 cosθ 1

2−

n

2

− sin 2 θ 1

= 0 for extrema

Simplify to obtain:

16 cos 2 θ 1

= 4 n 2 − sin 2 θ 1

Replace sin2θ 1 with 1 − cos2θ 1 and simplify to obtain:

12 cos2 θ 1

= 4n 2 − 4

Solve for cosθ 1 = cosθ 1m: cos θ 1m

=

n

2

−1 3

2952

Chapter 31

(c) Express the angular separation Δφ  of blue and red: From Equation 31-18, with nair  = 1 and nwater  = n: From Part (b):

Δφ  = φ d, blue − φ d,red

φ d

(1)

sin θ 1 ⎞ = π  + 2θ 1 − 4 sin −1 ⎛  ⎜ ⎟ ⎝  n  ⎠

θ 1m

⎡ = cos ⎢ ⎣⎢ −1

−1⎤ ⎥ 3 ⎦⎥

n2

Substitute to obtain:

φ d

⎡ = π  + 2 cos −1 ⎢ ⎢⎣

⎛  ⎧⎪ ⎜ sin cos −1 ⎡⎢ ⎜ ⎨⎪ 2 ⎤ ⎢⎣ n −1 ⎥ − 4 sin −1 ⎜ ⎩ ⎜ 3 ⎥⎦ n ⎜ ⎜ ⎝ 

n

2

− 1 ⎤ ⎫⎪ ⎞⎟ ⎥⎬ 3 ⎥⎦ ⎪ ⎟ ⎭⎟ ⎟ ⎟ ⎟  ⎠

Evaluate φ d for blue light in water:

φ d, blue

⎡ = π  + 2 cos −1 ⎢ ⎢⎣

2 ⎛  ⎧⎪  ⎞ ⎜ sin cos −1 ⎡⎢ (1.3435) − 1 ⎤⎥ ⎫⎪ ⎟ ⎬⎟ ⎜ ⎨ 2 3 ⎤ ⎢ ⎥ ⎪ (1.3435) − 1 ⎣ ⎦ ⎪⎭ ⎟ ⎥ − 4 sin −1 ⎜⎜ ⎩ ⎟ 3 1.3435 ⎥⎦ ⎜ ⎟ ⎜ ⎟ ⎝   ⎠

= 139.42° Evaluate φ d for red light in water:

φ d,red

⎡ = π  + 2 cos ⎢ ⎢⎣ −1

= 137.75°

2 ⎛  ⎧⎪  ⎞ ⎜ sin cos −1 ⎡⎢ (1.3318) − 1 ⎤⎥ ⎫⎪ ⎟ ⎬⎟ ⎜ ⎨ 3 ⎢ ⎥ ⎪ (1.3318)2 − 1 ⎤ ⎣ ⎦ ⎪⎭ ⎟ ⎥ − 4 sin −1 ⎜⎜ ⎩ ⎟ 3 1.3318 ⎥⎦ ⎜ ⎟ ⎜ ⎟ ⎝   ⎠

Properties of Light

2953

Δφ  = 139.42 − 137.75° = 1.67°

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

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° − θ 2, and 90° − θ 3. Consequently, θ 2 + θ 3 = α . We can apply Snell’s law at both

interfaces to express the angle of deviation δ  as a function of  θ 3 and then set the derivative of this function equal to zero to find the conditions on θ 3 and θ 2 that result in δ  being a minimum. a α

d δ 

θ 1 θ  u22

θ 3 θ 4

u3

n 

Express the angle of deviation:

δ  = θ 1 + θ 2

Apply Snell’s law to relate θ 1 to θ 2 and θ 3 to θ 4:

sin θ 1

Solve equation (2) for θ 1 and equation (3) for θ 4:

− α 

(1)

= n sin θ 2

(2)

and n sin θ 3 θ 1

= sin θ 4

(3)

= sin −1 (n sin θ 2 )

and θ 4 = sin −1 (n sin θ 3 )

Substitute in equation (1) to obtain: δ  = sin −1 (n sin θ 2 ) + sin −1 (n sin θ 3 ) − α  = sin −1 [n sin (θ 3

− α )] + sin −1 (n sin θ 3 ) − α