Chapter 31
Properties of Light
Conceptual Problems
1
•
[SSM] A ray of light reflects from a plane mirror. The angle
between the incoming ray and the reflected ray is 70°. What is the angle of
reflection? (a) 70°, (b) 140°, (c) 35°, (d) Not enough information is given to
determine the reflection angle.
Determine the Concept Because the angles of incidence and reflection are equal,
their sum is 70° and the angle of reflection is 35°. (c ) is correct.
2
•
A ray of light passes in air is incident on the surface of a piece of
glass. The angle between the normal to the surface and the incident ray is 40°, and
the angle between the normal and the refracted ray 28°. What is the angle between
the incident ray and the refracted ray? (a) 12°, (b) 28°, (c) 40° ,(d) 68°
Determine the Concept The angle between the incident ray and the refracted ray
is the difference between the angle of incidence and the angle of refraction.
(a ) is correct.
3
•
During a physics experiment, you are measuring refractive indices of
different transparent materials using a red helium-neon laser beam. For a given
angle of incidence, the beam has an angle of refraction equal to 28° in material A
and an angle of refraction equal to 26° in material B. Which material has the
larger index of refraction? (a) A, (b) B, (c) The indices of refraction are the same.
(d) You cannot determine the relative magnitudes of the indices of refraction from
the data given.
Determine the Concept The refractive index is a measure of the extent to which
a material refracts light that passes through it. Because the angles of incidence are
the same for materials A and B and the angle of refraction is smaller (more
bending of the light) for material B, its index of refraction is larger than that of A.
(b ) is correct.
4
•
A ray of light passes from air into water, striking the surface of the
water at an angle of incidence of 45º. Which, if any, of the following four
quantities change as the light enters the water: (a) wavelength, (b) frequency,
(c) speed of propagation, (d) direction of propagation, (e) none of the above?
2889
2890
Chapter 31
Determine the Concept When light passes from air into water its wavelength
changes ( λ water = λair n water ), 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.
5
•
Earth’s atmosphere decreases in density as the altitude increases. As a
consequence, the index of refraction of the atmosphere also decreases as altitude
increases. Explain how one can see the Sun when it is below the horizon. (The
horizon is the extension of a plane that is tangent to Earth’s surface.) Why does
the setting Sun appear flattened?
Determine the Concept The decrease in the index of refraction n of the
atmosphere with altitude results in refraction of the light from the Sun, bending it
toward the normal to the surface of Earth. Consequently, the Sun can be seen even
after it is just below the horizon.
Atmosphere
Earth
6
•
A physics student playing pocket billiards wants to strike her cue ball
so that it hits a cushion and then hits the eight ball squarely. She chooses several
points on the cushion and then measures the distances from each point to the cue
ball and to the eight ball. She aims at the point for which the sum of these
distances is least. (a) Will her cue ball hit the eight ball? (b) How is her method
related to Fermat’s principle? Neglect any effects due to ball rotation.
Determine the Concept (a) Yes. (b) Her procedure is based on Fermat’s
principle of least time. The ball presumably bounces off the cushion with an angle
of reflection equal to the angle of incidence, just as a light ray would do if the
cushion were a mirror. The least time would also be the shortest distance of travel
for the light ray.
7
•
[SSM] A swimmer at point S in Figure 31-53 develops a leg cramp
while swimming near the shore of a calm lake and calls for help. A lifeguard at
point L hears the call. The lifeguard can run 9.0 m/s and swim 3.0 m/s. She knows
physics and chooses a path that will take the least time to reach the swimmer.
Which of the paths shown in the figure does the lifeguard take?
Determine the Concept The path through point D is the path of least time. In
analogy to the refraction of light, the ratio of the sine of the angle of incidence to
the sine of the angle of refraction equals the ratio of the speeds of the lifeguard in
Properties of Light 2891
each medium. Careful measurements from the figure show that path LDS is the
path that best satisfies this criterion.
8
•
Material A has a higher index of refraction than material B. Which
material has the larger critical angle for total internal reflection when the material
is in air? (a) A, (b) B, (c) The angles are the same. (d) You cannot compare the
angles based on the data given.
Determine the Concept Because the product of the index of refraction on the
incident side of the interface and the sine of the critical angle is equal to one (the
index of refraction of air), the material with the smaller index of refraction will
have the larger critical angle. (b ) is correct.
9
•
[SSM] A human eye perceives color using a structure which is
called a cone that is is located on the retina. Three types of molecules compose
these cones and each type of molecule absorbs either red, green, or blue light by
resonance absorption. Use this fact to explain why the color of an object that
appears blue in air appears blue underwater, in spite of the fact that the
wavelength of the light is shortened in accordance with Equation 31-6.
Determine the Concept In resonance absorption, the molecules respond to the
frequency of the light through the Einstein photon relation E = hf. Neither the
wavelength nor the frequency of the light within the eyeball depend on the index
of refraction of the medium outside the eyeball. Thus, the color appears to be the
same in spite of the fact that the wavelength has changed.
10 •
Let θ be the angle between the transmission axes of two polarizing
sheets. Unpolarized light of intensity I is incident upon the first sheet. What is the
intensity of the light transmitted through both sheets? (a) I cos2 θ,
(b) (I cos2 θ)/2, (c) (I cos2 θ)/4, (d) I cos θ, (e) (I cos θ)/4, (f) None of the above
Picture the Problem The intensity of the light transmitted by the second
polarizer is given by 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
2892 Chapter 31
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
θP θP
θr
Smooth surface
12 •
Use the photon model of light to explain why, biologically, it is far
less dangerous to stand in front of an intense beam of red light than a very weak
beam of gamma radiation. HINT: Ionization of molecules in tissue can cause
biological damage. Molecules absorb light in what form, wave or particle?
Determine the Concept Molecules require that a certain minimum energy be
absorbed before they ionize. The red light photons contain considerably less
energy than the gamma photons so, even though there are likely to be fewer
photons in the gamma beam, each one is potentially dangerous.
13 •
Three energy states of an atom are A, B and C. State B is 2.0 eV above
state A and state C is 3.00 eV above state B. Which atomic transition results in the
emission of the shortest wavelength of light? (a) B → A, (b) C → B, (c) C → A,
(d) A→ C
Determine the Concept Because the wavelength of the light emitted in an atomic
transition is inversely proportional to the energy difference between the energy
levels, the highest energy difference produces the shortest wavelength light.
(c ) is correct.
14 •
In Problem 13, if the atom is initially in state A, which transition
results in the emission of the longest wavelength light? (a) A → B, (b) B → C, (c)
A → C, (d) B→ A
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 E2,Ne and the
state 1.96 eV below it (see Figure 31-51) is achieved by inelastic collisions
between neon atoms and helium atoms excited to the state E2,He.
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
18 •
Estimate the time required for light to make the round trip during
Galileo’s experiment to measure the speed of light. Compare the time of the round
trip to typical human response times. How accurate do you think this experiment
is?
Picture the Problem We can use the distance, rate, and time relationship to
estimate the time required to travel 6 km.
2894 Chapter 31
Express the distance d the light in
Galileo’s experiment traveled in
terms of its speed c and the
elapsed time Δt:
d = cΔt ⇒ Δt =
Substitute numerical values and
evaluate Δt:
6 km
= 2 × 10 −5 s
8
2.998 × 10 m/s
Δt reaction
0 .3 s
=
≈ 2 × 10 4
Δt
2 × 10 −5 s
or Δt reaction ≈ 2 × 10 4 Δt
Because human reaction is
approximately 0.3 s:
d
c
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.
19 •
Estimate the time delay in receiving a light on your retina when you
are wearing eyeglasses compared to when you are not wearing your eyeglasses.
Determine the Concept We’ll assume that the source of the photon is a distance
L from your retina and express the difference in the photon’s travel time when
you are wearing your glasses. Let the thickness of your glasses by 2 mm and the
index of refraction of the material from which they are constructed by 1.5.
When you are not wearing your
glasses, the time required for a light
photon, originating a distance L
away, to reach your retina is given
by:
Δt 0 =
L
c
L−d
d
+
c
cn
(n − 1)d
L + (n − 1)d
=
= Δt 0 +
c
c
If glass of thickness d and index of
refraction n is inserted in the path of
the photon, its travel time becomes:
Δt = t in air + t in glass =
The time delay is the difference
between Δt and Δt0:
t delay = Δt − Δt 0 =
Substitute numerical values and
evaluate tdelay:
t delay =
(n − 1)d
c
(1.5 − 1)(2 mm ) ≈
2.998 × 10 8 m/s
3 ps
20 ••
Estimate the number of photons that enter your eye if you look for a
tenth of a second at the Sun. What energy is absorbed by your eye during that
Properties of Light 2895
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:
The intensity of the radiation from
the Sun is given by:
Pincident
dN
on pupil
=
dt E per photon
I Sun =
Solving for Pincident yields:
Pincident
on pupil
Apupil
Pincident =
on pupil
on pupil
(1)
=
PSun
Asphere at Earth's
distance from the Sun
Apupil
Asphere at Earth's
PSun (2)
distance from the Sun
Substituting in equation (1) yields:
Apupil
dN
=
dt
Asphere at Earth's
PSun
distance from the Sun
E per photon
1
πd pupil
PSun
dN
= 4 2
dt 4πREarth-Sun Eper photon
2
Substitute for the two areas and
simplify to obtain:
⎛ d pupil
= ⎜⎜
⎝ 4 REarth-Sun
2
⎞
PSun
⎟⎟
⎠ Eper photon
Substitute numerical values and evaluate dN/dt:
dN ⎛
1 mm
= ⎜⎜
dt ⎝ 4 1.50 × 1011 m
2
⎞
4.2 × 10 26 W
⎟⎟
= 3.237 × 1015 s −1
−19
1.602
×
10
J
⎠ 2.25 eV ×
eV
or, separating variables and integrating this expression,
(
)
(
)
N = 3.237 ×1015 s −1 t
2896 Chapter 31
(
)
N (0.1 s ) = 3.237 ×1015 s −1 (0.1 s )
Evaluating N for t = 0.1 s yields:
≈ 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:
Apupil
PSun t
Asphere at Earth's
distance from the Sun
Substitute for the two areas and
simplify to obtain:
E=
1
4
2
πd pupil
2
4πREarth
-Sun
⎛ d pupil
= ⎜⎜
⎝ 4 REarth -Sun
PSun t
2
⎞
⎟⎟ PSun t
⎠
Substitute numerical values and evaluate E for t = 0.1 s:
⎛
1 mm
E (0.1 s ) = ⎜⎜
11
⎝ 4 1.50 × 10 m
(
2
⎞
⎟⎟ 4.2 × 10 26 W (0.1 s ) = 0.1167 mJ ≈ 0.1 mJ
⎠
) (
)
21 ••
Römer was observing the eclipses of Jupiter’s moon Io with the hope
that they would serve as a highly accurate clock that would be independent of
longitude. (Prior to GPS, such a clock was needed for accurate navigation.) Io
eclipses (enters the umbra of Jupiter’s shadow) every 42.5 h. Assuming an eclipse
of Io is observed on Earth on June 1 at midnight when Earth is at location A (as
shown in Figure 31-54), predict the expected time of observation of an eclipse
one-quarter of a year later when Earth is at location B, assuming (a) the speed of
light is infinite and (b) the speed of light is 2.998 × 108 m/s.
Picture the Problem We can use the period of Io’s motion and the position of the
earth at B to find the number of eclipses of Io during Earth’s movement and then
use this information to find the number of days before a night-time eclipse.
During the 42.5 h between eclipses of Jupiter’s moon, Earth moves from A to B,
increasing the distance from Jupiter by approximately the distance from Earth to
the Sun, making the path for the light longer and introducing a delay in the onset
of the eclipse.
Properties of Light 2897
Tearth 365.24 d 24 h
=
×
4
4
d
= 2191.4 h
(a) Find the time it takes Earth to
travel from point A to point B:
t A→ 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:
N=
t A→ B 2191.4 h
=
= 51.56
42.5 h
TIo
Hence, in one-fourth of a year, there will be 51.56 eclipses. Because we want to
find the next occurrence that happens in the evening hours, we’ll use 52 as the
number of eclipses. We’ll also assume that Jupiter is visible so that the eclipse of
Io can be observed at the time we determine.
Relate the time t(N) at which the
Nth eclipse occurs to N and the
period TIo of Io:
Evaluate t(52) to obtain:
Subtract the number of whole days to
find the clock time t:
t ( N ) = NTIo
⎛
1d ⎞
⎟
t (52) = (52)⎜⎜ 42.5 h ×
24 h ⎟⎠
⎝
= 92.083 d
t = t (52 ) − 92 d = 92.083 d − 92 d
24 h
= 0.083 d ×
= 1.992 h
d
≈ 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 =
rearth -sun
c
1.5 × 1011 m
1 min
Δt =
= 500 s ×
8
60 s
2.998 × 10 m/s
= 8.34 min
Hence, the eclipse will actually occur at 2:08 a.m., September 1
2898 Chapter 31
22 ••
If the angle of incidence is small enough, the small angle
approximation sin θ ≈ θ may be used to simplify Snell’s law of refraction.
Determine the maximum value of the angle that would make the value for the
angle differ by no more than one percent from the value for the sine of the angle.
(This approximation will be used in connection with image formation by spherical
surfaces in Chapter 32.)
Picture the Problem We can express the relative error in using the small angle
approximation and then either use 1) trial-and-error methods, 2) a spreadsheet
program, or 3) the Solver capability of a scientific calculator to solve the
transcendental equation that results from setting the error function equal to 0.01.
Express the relative error δ in using
the small angle approximation:
δ (θ ) =
θ − sin θ
θ
=
−1
sin θ
sin θ
A spreadsheet program was used to plot the following graph of δ (θ ).
0.016
0.014
delta(theta)
0.012
0.010
0.008
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 θ = 0.244 radians.
The Speed of Light
23 •
Mission Control sends a brief wake-up call to astronauts in a spaceship
that is far from Earth. 5.0 s after the call is sent, Mission Control can hear the
groans of the astronauts. How far from Earth is the spaceship? (a) 7.5 × 108 m,
(b) 15 × 108 m, (c) 30 × 108 m, (d) 45 × 108 m, (e) The spaceship is on the moon.
Properties of Light 2899
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 8 m/s (2.5 s )
= 7.5 × 10 m
8
and (a) is correct.
24 •
The distance from a point on the surface of Earth to a point on the
surface of the moon is measured by aiming a laser light beam at a reflector on the
surface of the moon and measuring the time required for the light to make a round
trip. The uncertainty in the measured distance Δx is related to the uncertainty in
the measured time Δt by Δx = 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 Δx is related to the uncertainty in the time Δt by Δx = cΔt to
evaluate Δx.
(a) The uncertainty in the distance is:
Δx = 12 cΔt
Substitute numerical values and
evaluate Δx:
Δx =
(b)The percent uncertainty in the
distance to the Moon is:
ΔxEarth to Moon
15.0 cm
=
xEarth to Moon
3.84 × 108 m
1
2
(2.998 ×10
8
)
m/s (± 1.00 ns )
= ± 15.0 cm
≈ 10 −8 %
25 ••
[SSM]
Ole Römer discovered the finiteness of the speed of light by
observing Jupiter’s moons. Approximately how sensitive would the timing
apparatus need to be in order to detect a shift in the predicted time of the moon’s
eclipses that occur when the moon happens to be at perigee ( 3.63 × 105 km ) and
those that occur when the moon is at apogee ( 4.06 × 105 km )? Assume that an
2900 Chapter 31
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.
Δt =
d moon
at apogee
Sensitivity =
Δt =
at perigee
c
Substituting for Δt yields:
Substitute numerical values and
evaluate the required sensitivity:
− d moon
d moon
at apogee
− d moon
at perigee
10c
4.06 × 105 km − 3.63 × 105 km
(10) 2.998 ×108 m/s
(
)
= 14 ms
Remarks: Instruments with this sensitivity did not exist in the 17th 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.
Express the intensity I of the light
reflected from an air-water interface
at normal incidence in terms of the
indices of refraction and the intensity
I0 of the incident light:
⎛ n − nwater
I = ⎜⎜ air
⎝ nair + nwater
Solve for the ratio I/I0:
I ⎛ nair − nwater
=⎜
I 0 ⎜⎝ nair + nwater
2
⎞
⎟⎟ I 0
⎠
⎞
⎟⎟
⎠
2
Properties of Light 2901
Substitute numerical values and
evaluate I/I0:
2
I ⎛ 1.00 − 1.33 ⎞
=⎜
⎟ = 2.0%
I 0 ⎝ 1.00 + 1.33 ⎠
27 •
A ray of light is incident on one of a pair of mirrors set at right angles
to each A ray of light is incident on one of two mirrors that are set at right angles
to each other. The plane of incidence is perpendicular to both mirrors. Show that
after reflecting from each mirror, the ray will emerge traveling in the direction
opposite to the incident direction, regardless of the angle of incidence.
Picture the Problem The diagram
shows ray 1 incident on the vertical
surface at an angle θ1, reflected as ray
2, and incident on the horizontal
surface at an angle of incidence θ3.
We’ll prove that rays 1 and 3 are
parallel by showing that θ1 = θ4, i.e.,
by showing that they make equal
angles with the horizontal. Note that
the law of reflection has been used in
identifying equal angles of incidence
and reflection.
1
θ1
θ1
3
2
θ2
θ3 θ3
θ4
θ 2 + 90° + (90° − θ1 ) = 180°
We know that the angles of the right
triangle formed by ray 2 and the two
mirror surfaces add up to 180°:
or
The sum of θ2 and θ3 is 90°:
θ 3 = 90° − θ 2
Because θ1 = θ 2 :
θ 3 = 90° − θ1
The sum of θ4 and θ3 is 90°:
θ 3 + θ 4 = 90°
Substitute for θ3 to obtain:
(90° − θ1 ) + θ 4 = 90° ⇒ θ1 =
θ1 = θ 2
θ4
28 ••
(a) A ray of light in air is incident on an air–water interface. Using a
spreadsheet or graphing program, plot the angle of refraction as a function of the
angle of incidence from 0º to 90º. (b) Repeat Part (a), but for a ray of light in
water that is incident on a water–air interface. [For Part (b), there is no reflected
ray for angles of incidence that are greater than the critical angle.]
2902 Chapter 31
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.
(b)
(a)
θ1
n1
Air
Water
θ2
Air
Water
θ1
n2
n2
n1
θ2
(a) Apply Snell’s law to the airwater interface to obtain:
Solving for θ2 yields:
n1 sin θ1 = n2 sin θ 2
where the angles of incidence and
refraction are θ1 and θ2, respectively.
⎞
⎛ n1
sin θ1 ⎟⎟
⎠
⎝ n2
θ 2 = sin −1 ⎜⎜
A spreadsheet program to graph θ2 as a function of θ1 is shown below. The
formulas used to calculate the quantities in the columns are as follows:
Cell
B1
B2
A6
A7
B6
Content/Formula
1
1.33333
0
A6 + 5
A6*PI()/180
C6
ASIN(($B$1/$B$2)*SIN(B6))
D6
C6*180/PI()
1
2
A
B
n1= 1
n2= 1.33333
Algebraic Form
n1
n2
θ1 (deg)
θ1 + Δθ
π
θ1 ×
180
⎛n
⎞
sin −1 ⎜⎜ 1 sin θ1 ⎟⎟
⎝ n2
⎠
θ2 ×
C
180
π
D
Properties of Light 2903
3
4
5
6
7
8
9
θ1
θ1
θ2
θ2
(deg)
0
1
2
3
(rad)
0.00
0.02
0.03
0.05
(rad)
0.000
0.013
0.026
0.039
(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
Angle of refraction, deg
A graph of θ2 as a function of θ1 follows:
50
45
40
35
30
25
20
15
10
5
0
0
10
20
30
40
50
60
70
80
90
Angle of incidence, deg
Angle of refraction, 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
80
70
60
50
40
30
20
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
29 ••
The red light from a helium-neon laser has a wavelength of 632.8 nm
in air. Find the (a) speed, (b) wavelength, and (c) frequency of helium-neon laser
light in air, water, and glass. (The glass has an index of refraction equal to 1.50.)
Picture the Problem We can use the definition of the index of refraction to find
the speed of light in the three media. The wavelength of the light in each medium
is its wavelength in air divided by the index of refraction of the medium. The
frequency of the helium-neon laser light is the same in all media and is equal to its
value in air. The wavelength of helium-neon laser light in air is 632.8 nm.
The speed of light in a medium
whose index of refraction is n is
given by:
v=
The wavelength of light in a medium
whose index of refraction is n is
given by:
λn =
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:
c
n
f =
(1)
λ air
(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 λwater:
λ water =
632.8 nm
= 476 nm
1.33
The other speeds and wavelengths are found similarly and are summarized in the
following table:
(a) speed
(m/s)
(b) wavelength (c) frequency
(nm)
(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
30 ••
The index of refraction for silicate flint glass is 1.66 for violet light
that has a wavelength in air equal to 400 nm and 1.61 for red light that has a
wavelength in air equal to 700 nm. A ray of 700-nm–wavelength red light and a
ray of 400-nm-wavelength violet light both have angles of refraction equal to 30º
upon entering the glass from air. (a) Which is greater, the angle of incidence of
the ray of red light or the angle of incidence of the ray of violet light? Explain
your answer. (b) What is the difference between the angles of incidence of the
two rays?
Picture the Problem Let the subscript 1 refer to the air and the subscript 2 to the
silicate glass and apply Snell’s law to the air-glass interface.
(a) Because the index of refraction for violet light is larger than that of red light,
for a given incident angle violet light would refract more than red light. Thus to
exhibit the same refraction angle, violet light would require an angle of incidence
larger than that of red light.
(b) Express the difference in their
angles of incidence:
Δθ = θ 1, violet − θ 1, red
Apply Snell’s law to the air-glass
interface to obtain:
n1 sin θ1 = n2 sin θ 2
Solving for θ1 yields:
(1)
⎛ n2
⎞
sin θ 2 ⎟⎟
⎝ n1
⎠
θ 1 = sin −1 ⎜⎜
Substitute for θ 1, violet and θ 1, red in equation (1) to obtain:
⎞
⎛n
⎞
⎛n
Δθ = sin −1 ⎜⎜ violet sin θ 2, violet ⎟⎟ − sin −1 ⎜⎜ red sin θ 2, red ⎟⎟
⎠
⎝ nair
⎠
⎝ nair
For θ 2, violet = θ 2, red = 30° :
⎛n
⎞
⎛n
⎞
⎛n
Δθ = sin −1 ⎜⎜ violet sin 30° ⎟⎟ − sin −1 ⎜⎜ red sin 30° ⎟⎟ = sin −1 ⎜⎜ violet
⎝ 2n air
⎠
⎝ n air
⎠
⎝ n air
⎛n
⎞
⎟⎟ − sin −1 ⎜⎜ red
⎝ 2n air
⎠
Substitute numerical values and evaluate Δθ:
⎛ 1.66 ⎞
⎛ 1.61 ⎞
⎟⎟ − sin −1 ⎜⎜
⎟⎟ = 56.10° − 53.61° = 2.49°
Δθ = sin −1 ⎜⎜
⎝ 2(1.00) ⎠
⎝ 2(1.00) ⎠
⎞
⎟⎟
⎠
2906 Chapter 31
Remarks: Note that Δθ is positive. This means that the angle for violet light is
greater than that for red light and confirms our answer in Part (a).
31 ••
[SSM] A slab of glass that has an index of refraction of 1.50 is
submerged in water that has an index of refraction of 1.33. Light in the water is
incident on the glass. Find the angle of refraction if the angle of incidence is
(a) 60º, (b) 45º, and (c) 30º.
Picture the Problem Let the subscript 1 refer to the water and the subscript 2 to
the glass and apply Snell’s law to the water-glass interface.
Apply Snell’s law to the waterglass interface to obtain:
Solving for θ2 yields:
(a) Evaluate θ2 for θ1 = 60°:
(b) Evaluate θ2 for θ1 = 45°:
(c) Evaluate θ2 for θ1 = 30°:
n1 sin θ1 = n2 sin θ 2
⎛ n1
⎞
sin θ1 ⎟⎟
⎝ n2
⎠
θ 2 = sin −1 ⎜⎜
⎛ 1.33
⎞
sin 60° ⎟ = 50°
⎝ 1.50
⎠
θ 2 = sin −1 ⎜
⎛ 1.33
⎞
sin 45° ⎟ = 39°
⎝ 1.50
⎠
θ 2 = sin −1 ⎜
⎛ 1.33
⎞
sin 30° ⎟ = 26°
⎠
⎝ 1.50
θ 2 = sin −1 ⎜
32 ••
Repeat Problem 31 for a beam of light initially in the glass that is
incident on the glass–water interface at the same angles.
Picture the Problem Let the subscript 1 refer to the glass and the subscript 2 to
the water and apply Snell’s law to the glass-water interface.
Apply Snell’s law to the waterglass interface to obtain:
Solving for θ2 yields:
(a) Evaluate θ2 for θ1 = 60°:
n1 sin θ1 = n2 sin θ 2
⎛ n1
⎞
sin θ1 ⎟⎟
⎝ n2
⎠
θ 2 = sin −1 ⎜⎜
⎛ 1.50
⎞
sin 60° ⎟ = 78°
⎝ 1.33
⎠
θ 2 = sin −1 ⎜
Properties of Light 2907
(b) Evaluate θ2 for θ1 = 45°:
(c) Evaluate θ2 for θ1 = 30°:
⎛ 1.50
⎞
sin 45° ⎟ = 53°
⎝ 1.33
⎠
θ 2 = sin −1 ⎜
⎛ 1.50
⎞
sin 30° ⎟ = 34°
⎝ 1.33
⎠
θ 2 = sin −1 ⎜
33 ••
A beam of light in air strikes a glass slab at normal incidence. The
glass slab has an index of refraction of 1.50. (a) Approximately what percentage
of the incident light intensity is transmitted through the slab (in one side and out
the other)? (b) Repeat Part (a) if the glass slab is immersed in water.
Picture the Problem Let the subscript
1 refer to the medium to the left (air) of
the first interface, the subscript 2 to
glass, and the subscript 3 to the
medium (air) to the right of the second
interface. Apply the equation relating
the intensity of reflected light at normal
incidence to the intensity of the
incident light and the indices of
refraction of the media on either side of
the interface to both interfaces. We’ll
neglect multiple reflections at glass-air
interfaces.
(a) Express the intensity of the
transmitted light in the second
medium:
Express the intensity of the
transmitted light in the third
medium:
Substitute for I2 to obtain:
n1 = 1.00
I r,1
I1
I 2 = I 1 − I r,1
n2 = 1.50
n3 = 1.00
I r, 2
I2
I3
⎛ n − n2
= I 1 − ⎜⎜ 1
⎝ n1 + n 2
⎞
⎟⎟ I 1
⎠
⎡ ⎛n −n
2
= I 1 ⎢1 − ⎜⎜ 1
+
n
n
⎢⎣ ⎝ 1
2
⎞
⎟⎟
⎠
2
2
⎤
⎥
⎥⎦
2
I 3 = I 2 − I r,2
⎛ n − n3 ⎞
⎟⎟ I 2
= I 2 − ⎜⎜ 2
⎝ n 2 + n3 ⎠
⎡ ⎛ n − n ⎞2 ⎤
3
⎟⎟ ⎥
= I 2 ⎢1 − ⎜⎜ 2
⎢⎣ ⎝ n 2 + n3 ⎠ ⎥⎦
⎡ ⎛n −n
2
I 3 = I 1 ⎢1 − ⎜⎜ 1
⎢⎣ ⎝ n1 + n2
⎞
⎟⎟
⎠
2
⎤⎡ ⎛ n − n ⎞2 ⎤
3
⎟⎟ ⎥
⎥ ⎢1 − ⎜⎜ 2
⎥⎦ ⎢⎣ ⎝ n 2 + n3 ⎠ ⎥⎦
2908 Chapter 31
Solve for the ratio I3/I1 to obtain:
I 3 ⎡ ⎛ n1 − n 2
= ⎢1 − ⎜
I 1 ⎢ ⎜⎝ n1 + n 2
⎣
⎞
⎟⎟
⎠
2
⎤⎡ ⎛ n − n ⎞2 ⎤
3
⎟⎟ ⎥
⎥ ⎢1 − ⎜⎜ 2
⎥⎦ ⎢⎣ ⎝ n 2 + n3 ⎠ ⎥⎦
Substitute numerical values and evaluate I3/I1:
2
2
I 3 ⎡ ⎛ 1.00 − 1.50 ⎞ ⎤ ⎡ ⎛ 1.50 − 1.00 ⎞ ⎤
= ⎢1 − ⎜
⎟ ⎥ = 92%
⎟ ⎥ ⎢1 − ⎜
I1 ⎢⎣ ⎝ 1.00 + 1.50 ⎠ ⎥⎦ ⎢⎣ ⎝ 1.50 + 1.00 ⎠ ⎥⎦
(b) With n1 = n3 = 1.33:
2
2
I 3 ⎡ ⎛ 1.33 − 1.50 ⎞ ⎤ ⎡ ⎛ 1.50 − 1.33 ⎞ ⎤
= ⎢1 − ⎜
⎟ ⎥ = 99%
⎟ ⎥ ⎢1 − ⎜
I1 ⎢⎣ ⎝ 1.33 + 1.50 ⎠ ⎥⎦ ⎢⎣ ⎝ 1.50 + 1.33 ⎠ ⎥⎦
34 ••
This problem is a refraction analogy. A band is marching down a
football field with a constant speed v1. About midfield, the band comes to a
section of muddy ground that has a sharp boundary making an angle of 30º with
the 50-yd line, as shown in Figure 31-55. In the mud, each marcher moves at a
speed equal to 12 v1 in a direction perpendicular to the row of markers they are in.
(a) Diagram how each line of marchers is bent as it encounters the muddy section
of the field so that the band is eventually marching in a different direction.
Indicate the original direction by a ray and the final direction by a second ray.
(b) Find the angles between these rays and the line normal to the boundary. Is
their direction of motion ″bent″ toward the normal or away from it? Explain your
answer in terms of refraction.
Picture the Problem We can apply Snell’s law to find the angle of refraction of
the line of marchers as they enter the muddy section of the field
(a)
30°
θ1
θ2
(b) Apply Snell’s law at the interface
to obtain:
n1 sin θ1 = n2 sin θ 2
Properties of Light 2909
Solving for θ2 yields:
⎤
⎡ n1
sin θ1 ⎥
⎦
⎣ n2
θ 2 = sin −1 ⎢
The ratio of the indices of refraction
is the reciprocal of the ratio of the
speeds of the marchers in the two
media:
v
n1 v1 v2 12 v1
=
= =
=
v v1
v1
n2
v2
Because the left and right sides of
the 30° angle and θ1 are mutually
perpendicular, θ1 = 30°. Substitute
numerical values and evaluate θ2:
θ 2 = sin −1 [ 12 sin 30°] = 14°
1
2
As the line enters the muddy field, its speed is reduced by half and the direction of
the forward motion of the line is changed. In this case, the forward motion in the
muddy field makes an angle θ2 with respect to the normal to the boundary line.
Note that the separation between successive lines in the muddy field is half that in
the dry field.
35 ••
[SSM] In Figure 31-56, light is initially in a medium that has an
index of refraction n1. It is incident at angle θ1 on the surface of a liquid that has
an index of refraction n2. The light passes through the layer of liquid and enters
glass that has an index of refraction n3. If θ3 is the angle of refraction in the glass,
show that n1 sin θ1 = n3 sin θ3. That is, show that the second medium can be
neglected when finding the angle of refraction in the third medium.
Picture the Problem We can apply Snell’s law consecutively, first to the n1-n2
interface and then to the n2-n3 interface.
Apply Snell’s law to the n1-n2
interface:
n1 sin θ1 = 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 θ1 = n3 sin θ 3
36 ••
On a safari, you are spear fishing while wading in a river. You observe
a fish gliding by you. If your line of sight to the fish is 64.0o degrees below the
horizontal in air, and assuming the spear follows a straight-line path through the
air and water after it is released, determine the angle below the horizontal that you
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.
α
θ1
n1
Air
Water
n2
d = 1.20 m
h = 1.50 m
64.0°
θ2
L
l
⎡h + d ⎤
⎥
⎣L+l⎦
Use the pictorial representation to
express the aiming angle α:
α = tan −1 ⎢
Referring to the diagram, note that:
L = h tan θ1 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 θ2 yields:
⎤
h+d
⎥
⎣ h tan θ 1 + d tan θ 2 ⎦
⎡
α = tan −1 ⎢
n1 sin θ 1 = n 2 sin θ 2
⎤
⎡ n1
sin θ 1 ⎥
⎦
⎣ n2
θ 2 = sin −1 ⎢
Properties of Light 2911
Substitute for θ2 to obtain:
⎡
⎤
⎢
⎥
h+d
⎥
−1 ⎢
α = tan ⎢
⎛
⎤⎞⎥
⎡
⎢ h tan θ1 + d tan⎜ sin −1 ⎢ n1 sin θ1 ⎥ ⎟ ⎥
⎜
⎟
⎢⎣
⎦ ⎠ ⎥⎦
⎣ n2
⎝
Noting that θ1 is the complement of 64.0°, substitute numerical values and
evaluate α:
⎡
⎤
⎢
⎥
1.50 m + 1.20 m
⎥ = 66.9°
α = tan −1 ⎢
⎢
⎛ −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
hstanding
α
θ1
Air
Water
θ2
L
n2
α'
θ1'
Air
Water
n1
n2
θ2'
h
h
l
d
hsitting
n1
L'
'
l
d
2912 Chapter 31
Use the fact that angles α and θ1 are
complementary, as are α′ and θ1′ to
determine θ1 and θ1′:
Express the distances L and L′ in
terms of θ1 and θ1′:
Assuming that your eyes are 1.7 m
above the level of the water when
you are standing and 0.7 m above the
water when you are sitting, evaluate
L and L′:
Referring to the pictorial
representations, note that:
θ 1 = 90° − α = 90° − 28° = 62°
and
θ 1 ' = 90° − α ' = 90° − 14° = 76°
L = hstanding tan θ 1
and
L' = hsitting tan θ 1 '
L = (1.7 m ) tan 62° = 3.197 m
and
L' = (0.7 m) tan 76° = 2.808 m
tan θ 2 =
and
tan θ 2 ' =
l d −L
=
h
h
(1)
l' d − L'
=
h
h
Divide the first of these equations by
the second to obtain:
tan θ 2 d − L
=
tan θ 2 ' d − L'
Solving for d yields:
d=
Apply Snell’s law to the air-water
interface when you are standing:
θ 2 = sin −1 ⎢
Substitute numerical values and
evaluate θ 2 :
θ 2 = sin −1 ⎢
Apply Snell’s law to the air-water
interface when you are sitting:
θ 2 ' = sin −1 ⎢
Substitute numerical values and
evaluate θ 2 ' :
θ 2 ' = sin −1 ⎢
L' tan θ 2 − L tan θ 2 '
tan θ 2 − tan θ 2 '
(2)
⎤
⎡ n1
sin θ1 ⎥
⎦
⎣ n2
⎡1.00
⎤
sin 62°⎥ = 41.60°
⎦
⎣1.33
⎤
⎡ n1
sin θ1 '⎥
⎦
⎣ n2
⎤
⎡1.00
sin 76°⎥ = 46.85°
⎦
⎣1.33
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° = 5.130 m =
tan 41.60° − tan 46.85°
5.1 m wide
Solving equation (1) for h yields:
h=
d −L
tan θ 2
Substitute numerical values and
evaluate h:
h=
5.130 m − 3.197 m
= 2.2 m deep
tan 41.60°
38 ••• Figure 31-57 shows a beam of light incident on a glass plate of
thickness d and index of refraction n. (a) Find the angle of incidence so that the
separation b between the ray reflected from the top surface and the ray reflected
from the bottom surface and exiting the top surface is a maximum. (b) What is
this angle of incidence if the index of refraction of the glass is 1.60? (c) What is
the separation of the two beams if the thickness of the glass plate is 4.0 cm?
Picture the Problem Let x be the
perpendicular separation between the
two rays and let l be the separation
between the points of emergence of the
two rays on the glass surface. We can
use the geometry of the refracted and
reflected rays to express x as a function
of l, d, θr,
and θi. Setting the
derivative of the resulting equation
equal to zero will yield the value of θi
that maximizes x.
l
θi θi
x
θi
Air
θr
θr
d
Glass
Air
(a) Express l in terms of d and the
angle of refraction θr:
l = 2d tan θ r
Express x as a function of l, d,
θr, and θi:
x = 2d tan θ r cosθ i
Differentiate x with respect toθi:
⎛
⎞
d
dx
(tan θ r cos θ i ) = 2d ⎜⎜ − tan θ r sin θ i + sec 2 θ r cos θ i dθ r ⎟⎟
= 2d
dθ i ⎠
dθ i
dθ i
⎝
(1)
2914 Chapter 31
Apply Snell’s law to the air-glass
interface:
n1 sin θ i = n2 sin θ r
or, because n1 = 1 and n2 = n,
sin θ i = n sin θ r
Differentiate implicitly with
respect toθI to obtain:
cosθ i dθ i = n cosθ r dθ r
or
dθ r 1 cos θ i
=
dθ i n cos θ r
(2)
Substitute in equation (1) to obtain:
⎛ 1 cos 2 θ i sin θ r sin θ i
⎛ sin θ r
dx
1 cos θ i cos θ i ⎞
⎟
⎜
sin θ i +
= 2d ⎜ −
2
⎟ = 2d ⎜⎜ n cos 3 θ − cos θ
dθ i
θ
n
θ
cos
cos
θ
cos
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:
n
⎛ 1 − sin 2 θ i sin 2 θ i
dx
= 2d ⎜⎜
−
3
dθ i
⎝ n cos θ r n cos θ r
⎞
⎟⎟
⎠
⎞
⎟⎟
⎠
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
dx
= 2d ⎜⎜
−
3
dθ i
n cos 3 θ r
⎝ n cos θ r
⎞
2d
⎟⎟ =
1 − sin 2 θ i − sin 2 θ i cos 2 θ r
3
⎠ n cos θ r
(
Substitute 1 − sin 2 θ r for cos 2 θ r :
[
(
dx
2d
=
1 − sin 2 θ i − sin 2 θ i 1 − sin 2 θ r
3
dθ i n cos θ r
Substitute
)]
1
sin θ i for sin θ r to obtain:
n
2d
dx
=
dθ i n cos 3 θ r
⎡
1
⎛
⎞⎤
2
2
2
⎢1 − sin θ i − sin θ i ⎜1 − n 2 sin θ i ⎟⎥
⎝
⎠⎦
⎣
Factor out 1/n2, simplify, and set equal to zero to obtain:
[
]
dx
2d
= 3
sin 4 θ i − 2n 2 sin 2 θ i + n 2 = 0 for extrema
dθ i n cos 3 θ r
)
Properties of Light 2915
If dx/dθ1 = 0, then it must be true
that:
Solve this quartic equation for θi
to obtain:
(b) Evaluate θI for n = 1.60:
sin 4 θ i − 2n 2 sin 2 θ i + n 2 = 0
⎛
θ i = sin − 1 ⎜ n 1 − 1 −
⎜
⎝
1 ⎞⎟
n 2 ⎟⎠
⎛
θ i = sin −1 ⎜1.6 1 − 1 −
⎜
⎝
1
(1.60)2
⎞
⎟
⎟
⎠
= 48.5°
(c) In (a) we showed that:
Solve equation (2) for θr:
Substitute numerical values and
evaluate θr:
Substitute numerical values and
evaluate x:
x = 2d tan θ r cosθ i
⎛ n1
⎞
sin θ i ⎟⎟
⎝ n2
⎠
θ r = sin −1 ⎜⎜
⎛ 1
⎞
sin 48.5° ⎟ = 27.9°
⎝ 1.60
⎠
θ r = sin −1 ⎜
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 θ1 = n2 sin θ 2
When there is total internal
reflection:
θ1 = θ c and θ 2 = 90°
Substitute to obtain:
n1 sin θ c = n2 sin 90° = n2
Solving for θc yields:
⎛ n2 ⎞
⎟⎟
⎝ n1 ⎠
θ c = sin −1 ⎜⎜
2916 Chapter 31
Substitute numerical values and
evaluate θc:
⎛ 1.00 ⎞
⎟ = 48.8°
⎝ 1.33 ⎠
θ c = sin −1 ⎜
40 •
A glass surface (n = 1.50) has a layer of water (n = 1.33) on it. Light in
the glass is incident on the glass–water interface. Find the critical angle for total
internal reflection.
Picture the Problem Let the index of
refraction of glass be represented by n1
and the index of refraction of water by
n2 and apply Snell’s law to the glasswater interface under total internal
reflection conditions.
n2 = 1.33
θ2
Water
Glass
n1 = 1.50
θ1
Apply Snell’s law to the glasswater interface:
n1 sin θ1 = n2 sin θ 2
At the critical angle, θ1 = θc and
θ2 = 90°:
n1 sin θ c = n2 sin 90°
Solve for θc:
Substitute numerical values and
evaluate θc:
⎡ n2
⎤
sin 90°⎥
⎣ n1
⎦
θ c = sin −1 ⎢
⎡1.33
⎤
sin 90°⎥ = 62.5°
⎣1.50
⎦
θ c = sin −1 ⎢
41 •
A point source of light is located 5.0 m below the surface of a large
pool of water. Find the area of the largest circle on the pool’s surface through
which light coming directly from the source can emerge.
Air
Picture the Problem We can apply
Snell’s law to the water-air interface
to express the critical angle θc in
terms of the indices of refraction of
water (n1) and air (n2) and then relate
the radius of the circle to the depth d
of the point source and θc.
Express the area of the circle whose
radius is r:
r
n2 = 1.00
Water
n1 = 1.33
d = 5.0 m
θc
A = π r2
90º
θc
Properties of Light 2917
Relate the radius of the circle to the
depth d of the point source and the
critical angle θc:
r = d tan θ c
Apply Snell’s law to the water-air
interface to obtain:
n1 sin θ c = n2 sin 90° = n2
Solving for θc yields:
Substitute for r and θc to obtain:
⎛ n2 ⎞
⎟⎟
⎝ n1 ⎠
θ c = sin −1 ⎜⎜
A = π [d tan θ c ]
2
⎡
⎛
⎧ n ⎫ ⎞⎤
= π ⎢d tan ⎜⎜ sin −1 ⎨ 2 ⎬ ⎟⎟⎥
⎩ n1 ⎭ ⎠⎦⎥
⎝
⎣⎢
Substitute numerical values and
evaluate A:
2
⎡
⎛
⎧ 1 ⎫ ⎞⎤
A = π ⎢(5.0 m ) tan⎜⎜ sin −1 ⎨
⎬ ⎟⎟⎥
⎩1.33 ⎭ ⎠⎦
⎝
⎣
2
= 1.0 ×10 2 m 2
42 ••
Light traveling in air strikes the largest face of an isosceles-righttriangle 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 θc = 45°.
Express the speed of light v in the
prism in terms of its index of
refraction n1:
45º
n1
θc
45º
n2 = 1.00
v=
c
n1
2918 Chapter 31
Apply Snell’s law to the prism-air
interface to obtain:
n1 sin θ c = n2 sin 90° = 1
Solving for n1 yields:
n1 =
Substitute for n1 and simplify to
obtain:
v = c sin θ c
Substitute numerical values and
evaluate v:
v = 2.998 ×10 8 m/s sin 45°
1
sin θ c
(
)
= 2.1×10 8 m/s
43 ••
A point source of light is located at the bottom of a steel tank, and an
opaque circular card of radius 6.00 cm is placed horizontally over it. A
transparent fluid is gently added to the tank so that the card floats on the fluid
surface with its center directly above the light source. No light is seen by an
observer above the surface until the fluid is 5.00 cm deep. What is the index of
refraction of the fluid?
Picture the Problem The observer above the surface of the fluid will not see any
light until the angle of incidence of the light at the fluid-air interface is less than
or equal to the critical angle for the two media. We can use Snell’s law to express
the index of refraction of the fluid in terms of the critical angle and use the
geometry of card and light source to express the critical angle.
r
n2
n1
θ2
θc
d
θc
Apply Snell’s law to the fluid-air
interface to obtain:
n1 sin θ1 = n2 sin θ 2
Light is seen by the observer
when θ1 = θc and θ2 = 90°:
n1 sin θ c = n2 sin 90° = n2
Properties of Light 2919
1
sin θ c
Because the medium above the
interface is air, n2 = 1. Solve for
n1 to obtain:
n1 =
From the geometry of the
diagram:
tan θ c =
Substitute for θ c to obtain:
Substitute numerical values and
evaluate n1:
n1 =
n1 =
r
⎛r⎞
⇒ θ c = tan −1 ⎜ ⎟
d
⎝d ⎠
1
⎡
⎛ r ⎞⎤
sin ⎢ tan −1 ⎜ ⎟⎥
⎝ d ⎠⎦
⎣
1
⎡
⎛ 6.00 cm ⎞⎤
⎟⎟⎥
sin ⎢ tan −1 ⎜⎜
5
.
00
cm
⎠⎦
⎝
⎣
= 1.30
44 ••
An optical fiber allows rays of light to propagate long distances by
using total internal reflection. Optical fibers are used extensively in medicine and
in digital communications. As shown in Figure 31-58 the fiber consists of a core
material that has an index of refraction n2 and radius b surrounded by a cladding
material that has an index of refraction n3 < n2. The numerical aperture of the
fiber is defined as sinθ1, where θ1 is the angle of incidence of a ray of light that
impinges on the center of the end of the fiber and then reflects off the corecladding interface just at the critical angle. Using the figure as a guide, show that
the numerical aperture is given by sin θ1 = 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 θ1 = n22 − n32 .
Referring to the figure, note that:
Apply the Pythagorean theorem
to the right triangle to obtain:
Solving for
b
yields:
c
sin θ c =
n3 a
b
= and sin θ 2 =
n2 c
c
a 2 + b 2 = c 2 or
b
a2
= 1− 2
c
c
a 2 b2
+
=1
c2 c2
2920 Chapter 31
Substitute for
b
a
and to obtain:
c
c
Use the law of refraction to relate
θ1 and θ2:
Substitute for sinθ2, let n1 = 1
(air), and simplify to obtain:
sin θ 2 = 1 −
n32
n22
n1 sin θ1 = n2 sin θ 2
sin θ1 = n2 1 −
n32
=
n22
n22 − n32
45 ••
[SSM] Find the maximum angle of incidence θ1 of a ray that would
propagate through an optical fiber that has a core index of refraction of 1.492, a
core radius of 50.00 μm, and a cladding index of 1.489. See Problem 44.
Picture the Problem We can use the result of Problem 44 to find the maximum
angle of incidence under the given conditions.
From Problem 44:
sin θ1 = n22 − n32
Solve for θ1 to obtain:
θ1 = sin −1 n22 − n32
Substitute numerical values and
evaluate θ1:
θ1 = sin −1 ⎛⎜ (1.492 )2 − (1.489 )2 ⎞⎟
(
⎝
)
⎠
= 5°
46 ••
Calculate the difference in time needed for two pulses of light to travel
down 15.0 km of the fiber that is described in Problem 44. Assume that one pulse
enters the fiber at normal incidence, and the second pulse enters the fiber at the
maximum angle of incidence calculated in Problem 45. In fiber optics, this effect
is known as modal dispersion.
Picture the Problem We can derive an expression for the difference in the travel
times by expressing the travel time for a pulse that enters the fiber at the
maximum angle of incidence and a pulse that enters the fiber at normal incidence.
Examination of Figure 31-58 reveals that, if the length of the tube is L, the
distance traveled by the pulse that enters at an angle θ1 is the ratio of c to a
multiplied by L.
The difference in the travel times
Δt is given by:
Δt = tθ1 − t normal
incidence
(1)
Properties of Light 2921
The travel time for the pulse that
enters the fiber at the maximum
angle of incidence is:
The travel time for the pulse that
enters the fiber normally is:
Substitute for tθ1 and t normal
in
incidence
equation (1) to obtain:
c
distance traveled
tθ1 =
= a
c
speed in the medium
n2
L
=
t normal
incidence
L
c
n2
c
L
Δt = a −
c
c
n2
n2
L
n2 L ⎛ c ⎞
⎜ − 1⎟
c ⎝a ⎠
Simplify to obtain:
Δt =
Referring to the figure, note that:
sin θ c =
a
c
From Snell’s law, the sine of the
critical angle is also given by:
sin θ c =
n3
a n
⇒ = 3
n2
c n2
Substitute for c/a in equation (2)
to obtain:
Δt =
n2 L ⎛ n2 ⎞
⎜ − 1⎟
c ⎜⎝ n3 ⎟⎠
Substitute numerical values and
evaluate Δt:
Δt =
(1.492)(15 km ) ⎛ 1.492 − 1⎞
⎜
2.998 × 108 m/s ⎝ 1.489
(2)
⎟
⎠
= 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 θc for glass-to-air refraction and for which the light rays will leave the glass,
travel through the water and then pass into the air?
Picture the Problem Let the index of refraction of glass be represented by n1, the
index of refraction of water by n2, and the index of refraction of air by n3. We can
apply Snell’s law to the glass-water interface under total internal reflection
conditions to find the critical angle for total internal reflection. The application of
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 θc 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
θ3
θ2
Water
n2 = 1.33
Glass
n1 = 1.50
θ2
θ1
(a) Apply Snell’s law to the glasswater interface:
n1 sin θ1 = n2 sin θ 2
At the critical angle, θ1 = θc and
θ2 = 90°:
n1 sin θ c = n2 sin 90°
Solving for θc yields:
⎤
⎡ n2
sin 90°⎥
⎦
⎣ n1
θ c = sin −1 ⎢
⎡1.33
⎤
sin 90°⎥ = 62.5°
⎣1.50
⎦
Substitute numerical values and
evaluate θc:
θ c = sin −1 ⎢
(b) Apply Snell’s law to a water-air
interface at the critical angle for a
water-air interface:
n2 sin θ c = n3 sin 90°
Solving for θ c yields:
⎛ n3 ⎞
⎟⎟
⎝ n2 ⎠
θ c = sin −1 ⎜⎜
⎛ 1.00 ⎞
⎟ = 48.8°
⎝ 1.33 ⎠
Substitute numerical values and
evaluate θ c :
θ c = sin −1 ⎜
Apply Snell’s law to a ray incident at
the glass-water interface:
n1 sin θ1 = n2 sin θ 2
and
⎞
⎛n
θ1 = sin −1 ⎜⎜ 2 sin θ 2 ⎟⎟
⎠
⎝ n1
Properties of Light 2923
For θ 2 = θ c :
Substitute numerical values and
evaluate θ1:
⎡⎛ n2 ⎞ ⎛ n3 ⎞⎤ −1 ⎛ n3 ⎞
⎟⎟ ⎜⎜ ⎟⎟⎥ sin ⎜⎜ ⎟⎟
⎝ n1 ⎠
⎣⎝ n1 ⎠ ⎝ n2 ⎠⎦
θ1 = sin −1 ⎢⎜⎜
⎛ 1.00 ⎞
⎟ = 41.8°
⎝ 1.53 ⎠
θ1 = sin −1 ⎜
Yes, if θ ≥ 41.8°, where is the angle of incidence for the rays in glass that are
incident on the glass-water boundary, the rays will leave the glass through the
water and pass into the air.
48 ••• A laser beam is incident on a plate of glass that is 3.0-cm thick (Figure
31-57). The glass has an index of refraction of 1.5 and the angle of incidence is
40º. The top and bottom surfaces of the glass are parallel. What is the distance b
between the beam formed by reflection off the top surface of the glass and the
beam reflected off the bottom surface of the glass.
Picture the Problem The situation
is shown in the adjacent figure. We
can use the geometry of the diagram
and trigonometric relationships to
derive an expression for d in terms of
the angles of incidence and
refraction. Applying Snell’s law will
yield θr.
b
θ i θi
θi
x
Air
Glass
n = 1.5
θr
t
θr
Express the distance x in terms of t
and θr:
x = 2t tanθ r
The separation of the reflected rays
is:
b = x cosθ i
Substitute for x to obtain:
b = 2t tanθ r cosθ i
Apply Snell’s law at the air-glass
interface to obtain:
⎛ sin θ i ⎞
sin θi = n sin θ r ⇒ θ r = sin −1 ⎜
⎟
⎝ n ⎠
Substitute for θ r in equation (1) to
obtain:
⎡
⎛ sin θ i
b = 2t tan ⎢sin −1 ⎜
⎝ n
⎣
(1)
⎞⎤
⎟⎥ cosθ i
⎠⎦
2924
Chapter 31
Substitute numerical values and
evaluate b:
⎡
⎛ sin 40° ⎞⎤
b = 2(3.0 cm ) tan ⎢sin −1 ⎜
⎟⎥ cos 40°
⎝ 1.5 ⎠⎦
⎣
= 2.2 cm
Dispersion
49 •
A beam of light strikes the plane surface of silicate flint glass at an
angle of incidence of 45º. The index of refraction of the glass varies with
wavelength (see Figure 31-59). How much smaller is the angle of refraction for
violet light of wavelength 400 nm than the angle of refraction for red light of
wavelength 700 nm?
Picture the Problem We can apply Snell’s law of refraction to express the angles
of refraction for red and violet light in silicate flint glass.
Express the difference between the
angle of refraction for violet light
and for red light:
Δθ = θ r,red − θ r,violet
Apply Snell’s law of refraction to
the interface to obtain:
⎛ 1 ⎞
sin 45° = n sin θ r ⇒ θ r = sin −1 ⎜
⎟
⎝ 2n ⎠
Substituting for θ r in equation (1)
yields:
⎞
⎛
⎛ 1 ⎞
1
⎟
⎟ − sin −1 ⎜
Δθ = sin −1 ⎜⎜
⎟
⎟
⎜ 2n
2
n
red ⎠
violet ⎠
⎝
⎝
Substitute numerical values and
evaluate Δθ :
⎞
⎛
⎞
⎛
1
1
⎟
⎟ − sin −1 ⎜
Δθ = sin −1 ⎜⎜
⎜ 2 (1.66 ) ⎟
⎟
⎠
⎝
⎝ 2 (1.60 ) ⎠
(1)
= 26.2° − 25.2° = 1.0°
50 ••
In many transparent materials, dispersion causes different colors
(wavelengths) of light to travel at different speeds. This can cause problems in
fiber-optic communications systems where pulses of light must travel very long
distances in glass. Assuming a fiber is made of silicate crown glass (see Figure
31-19), calculate the difference in travel times that two short pulses of light take
to travel 15.0 km in this fiber if the first pulse has a wavelength of 700 nm and the
second pulse has a wavelength of 500 nm.
Picture the Problem The transit times will be different because the speed with
which light of various wavelengths propagates in silicate crown glass is
dependent on the index of refraction. We can use Figure 31-19 to estimate the
indices of refraction for pulses of wavelengths 500 and 700 nm.
Properties of Light 2925
Express the difference in time
needed for two short pulses of
light to travel a distance L in the
fiber:
Δt =
L
L
−
v500 v700
Substitute for L, v500, and v700 and
simplify to obtain:
Δt =
n500 L n700 L L
−
= (n500 − n700 )
c
c
c
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 =
15.0 km
(1.55 − 1.50)
2.998 ×10 8 m/s
≈ 3 μs
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:
⎛ n2 ⎞
⎟⎟
⎝ n1 ⎠
θ p = tan −1 ⎜⎜
⎛ 1.33 ⎞
⎟ = 53.1°
⎝ 1.00 ⎠
(a) For n1 = 1 and n2 = 1.33:
θ p = tan −1 ⎜
(b) For n1 = 1 and n2 = 1.50:
θ p = tan −1 ⎜
⎛ 1.50 ⎞
⎟ = 56.3°
⎝ 1.00 ⎠
52 •
Light that is horizontally polarized is incident on a polarizing sheet. It
is observed that only 15 percent of the intensity of the incident light is transmitted
through the sheet. What angle does the transmission axis of the sheet make with
the horizontal?
2926
Chapter 31
Picture the Problem The intensity of the transmitted light I is related to the
intensity of the incident light I0 and the angle the transmission axis makes with the
horizontal θ according to 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:
⎛ I ⎞
⎟
I = I 0 cos 2 θ ⇒ θ = cos −1 ⎜⎜
⎟
I
⎝ 0⎠
Substitute numerical values and
evaluate θ :
θ = cos −1 0.15 = 67°
(
)
53 •
Two polarizing sheets have their transmission axes crossed so that no
light gets through. A third sheet is inserted between the first two so that its
transmission axis makes an angle θ with the transmission axis of the first sheet.
Unpolarized light of intensity I0 is incident on the first sheet. Find the intensity of
the light transmitted through all three sheets if (a) θ = 45º and (b) θ = 30º.
Picture the Problem Let In be the intensity after the nth polarizing sheet and use
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:
I1 = 12 I 0
The intensity of the light between the
second and third sheets is:
I 2 = I1 cos 2 θ1, 2 = 12 I 0 cos 2 45° = 14 I 0
The intensity of the light that has
passed through the third sheet is:
I 3 = I 2 cos 2 θ 2,3 = 14 I 0 cos 2 45° =
(b) The intensity of the light between
the first and second sheets is:
I1 = 12 I 0
The intensity of the light between the
second and third sheets is:
I 2 = I1 cos 2 θ1, 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 cos 2 θ 2,3 = 83 I 0 cos 2 60° =
1
8
I0
3
32
I0
Properties of Light 2927
54 •
A horizontal 5.0 mW laser beam that is vertically polarized is incident
on a sheet that is oriented with its transmission axis vertical. Behind the first sheet
is a second sheet that is oriented so that its transmission axis makes an angle of
27º with respect to the vertical. What is the power of the beam transmitted
through the second sheet?
Picture the Problem Because the light is polarized in the vertical direction and
the first polarizer is also vertically polarized, no loss of intensity results from the
first transmission. We can use Malus’s law to find the intensity of the light after it
has passed through the second polarizer.
The intensity of the beam is the ratio
of its power to cross-sectional area:
I=
Express the intensity of the light
between the first and second
polarizers:
I1 = I 0 and P1 = P0
Express the law of Malus in terms of
the power of the beam:
P P0
= cos 2 θ ⇒ P = P0 cos 2 θ
A A
Express the power of the beam after
the second transmission:
P2 = P1 cos 2 θ1, 2 = P0 cos 2 θ12
Substitute numerical values and
evaluate P2:
P2 = (5.0 mW ) cos 2 27° = 4.0 mW
P
A
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 θp 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 θp and evaluate n2:
n 2 = tan 60° = 1.7
56 ••
Two polarizing sheets have their transmission axes crossed so that no
light is transmitted. A third sheet is inserted so that its transmission axis makes an
angle θ with the transmission axis of the first sheet. (a) Derive an expression for
the intensity of the transmitted light as a function of θ. (b) Show that the intensity
transmitted through all three sheets is maximum when θ = 45º.
Picture the Problem Let In be the intensity after the nth polarizing sheet and use
I = I 0 cos 2 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:
I1 = 12 I 0
The intensity of the light between
the second and third sheets is:
I 2 = I1 cos 2 θ1, 2 = 12 I 0 cos 2 θ
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
= 12 I 0 cos 2 θ cos 2 (90° − θ )
= 12 I 0 cos 2 θ sin 2 θ
= 18 I 0 (2 cos θ sin θ )
2
=
1
8
I 0 sin 2 2θ
(b) Because the sine function is a maximum when its argument is 90°, the
maximum value of I3 occurs when θ = 45°.
57 ••
If the middle polarizing sheet in Problem 56 is rotating at an angular
speed ω about an axis parallel with the light beam, find an expression for the
intensity transmitted through all three sheets as a function of time. Assume that
θ = 0 at time t = 0.
Picture the Problem Let In be the intensity after the nth polarizing sheet, use
I = I 0 cos 2 θ to find the intensity of the light transmitted through each sheet, and
replace θ with ωt.
Properties of Light 2929
The intensity of the light between the
first and second sheets is:
I1 = 12 I 0
The intensity of the light between the
second and third sheets is:
I 2 = I1 cos 2 θ1, 2 = 12 I 0 cos 2 ωt
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
= 12 I 0 cos 2 ωt cos 2 (90° − ωt )
= 12 I 0 cos 2 ωt sin 2 ωt
= 18 I 0 (2 cos ωt sin ωt )
2
=
1
8
I 0 sin 2 2ωt
58 ••
A stack of N + 1 ideal polarizing sheets is arranged so that each sheet
is rotated by an angle of π/(2N) rad with respect to the preceding sheet. A linearly
polarized light wave of intensity I0 is incident normally on the stack. The incident
light is polarized along the transmission axis of the first sheet and is therefore
perpendicular to the transmission axis of the last sheet in the stack.
(a) Show that the intensity of the light transmitted through the entire stack is
given by I0 cos 2N ⎡⎣π (2N )⎤⎦ . (b) Using a spreadsheet or graphing program, plot
the transmitted intensity as a function of N for values of N from 2 to 100. (c) What
is the direction of polarization of the transmitted beam in each case?
Picture the Problem Let In be the intensity after the nth polarizing sheet and use
I = I 0 cos 2 θ to find the ratio of In+1 to In.
(a) Find the ratio of In+1 to In:
I n+1
π
= cos 2
In
2N
Because there are N such
reductions of intensity:
I N +1 I N +1
⎛ π ⎞
=
= cos 2 N ⎜
⎟
I1
I0
⎝ 2N ⎠
and
⎛ π ⎞
I N +1 = I 0 cos 2 N ⎜
⎟
⎝ 2N ⎠
(b) A spreadsheet program to graph IN+1/I0 as a function of N is shown below. The
formulas used to calculate the quantities in the columns are as follows:
Cell
A2
A3
Content/Formula
2
A2 + 1
Algebraic Form
N
N+1
2930 Chapter 31
B2
⎛ π ⎞
cos 2 N ⎜
⎟
⎝ 2N ⎠
(cos(PI()/(2*A2))^(2*A2)
1
2
3
4
5
A
N
2
3
4
5
B
I/I0
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/I0 as a function of N follows.
1.0
0.9
0.8
I /I0
0.7
0.6
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 In be the intensity after the nth polarizing sheet and use
I = I 0 cos 2 θ to find the ratio of In+1 to In. Because each sheet introduces a 2%
loss of intensity, the net transmission after N sheets (0.98)N.
Find the ratio of In+1 to In:
I n+1
π
= (0.98)cos 2
In
2N
Because there are N such
reductions of intensity:
I N +1
⎛ π ⎞
N
= (0.98) cos 2 N ⎜
⎟
I0
⎝ 2N ⎠
A spreadsheet program to graph IN+1/I0 for an ideal polarizer as a function of N,
the percent transmission, and IN+1/I0 for a real polarizer as a function of N is
shown below. The formulas used to calculate the quantities in the columns are as
follows:
Cell
A3
B2
Content/Formula
1
(cos(PI()/(2*A2))^(2*A2)
C3
(0.98)^A3
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
Algebraic Form
N
⎛ π ⎞
cos 2 N ⎜
⎟
⎝ 2N ⎠
(0.98)N
(0.98)N cos 2 N ⎛⎜
π ⎞
⎟
⎝ 2N ⎠
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/I0 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.
60 ••
Show mathematically that a linearly polarized wave can be thought of
as a superposition of a right and a left circularly polarized wave.
Picture the Problem A circularly polarized wave is said to be right circularly
polarized if the electric and magnetic fields rotate clockwise when viewed along
the direction of propagation and left circularly polarized if the fields rotate
counterclockwise.
For a circularly polarized wave, the x
and y components of the electric
field are given by:
E x = E0 cos ωt
and
E y = E0 sin ωt or E y = − E0 sin ωt
for left and right circular polarization,
respectively.
For a wave polarized along the x
axis:
r
r
E right + Eleft = E0 cos ωt iˆ + E0 cos ωt iˆ
= 2 E0 cos ωt iˆ
61 ••
Suppose that the middle sheet in Problem 53 is replaced by two
polarizing sheets. If the angle between the transmission axes in the second, third,
and fourth sheets in the stack make angles of 30º, 60º and 90º, respectively, with
the transmission axis of the first sheet, (a) what is the intensity of the transmitted
Properties of Light 2933
light? (b) How does this intensity compare with the intensity obtained in Part (a)
of Problem 53?
Picture the Problem Let In be the intensity after the nth polarizing sheet and use
I = I 0 cos 2 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:
I1 = 12 I 0
The intensity of the light between the
second and third sheets is:
I 2 = I1 cos 2 θ1, 2 = 12 I 0 cos 2 30° = 83 I 0
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° =
The intensity of the light to the right
of the fourth sheet is:
I 4 = I 3 cos 2 θ 3, 4 =
9
32
9
32
I0
27
I 0 cos 2 30° = 128
I0
= 0.211I 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.
62 ••
Show that the electric field of a circularly polarized wave propagating
parallel with the x axis can be expressed by
r
E = E0 sin (kx − ωt )iˆ + E0 cos(kx − ωt )kˆ .
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 x2 + E y2
Substitute for Ex and Ey to obtain:
E=
[E0 sin (kx − ωt )]2 + [E0 cos(kx − ωt )] 2
= E0
[
]
= E02 sin 2 (kx − ωt ) + cos 2 (kx − ωt )
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
E = E0 sin (kx − ωt ) ˆj + E0 cos(kx − ωt )kˆ
The corresponding magnetic field is:
r
B = B0 sin (kx − ωt )kˆ − B0 cos(kx − ωt ) ˆj
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
E = E0 sin (kx + ωt ) ˆj − E0 cos(kx + ωt )kˆ
Sources of Light
64 •
A helium–neon laser emits light that has a wavelength equal to 632.8
nm and has a power output of 4.00 mW. How many photons are emitted per
second by this laser?
Picture the Problem We can express the number of photons emitted per second
as the ratio of the power output of the laser and energy of a single photon.
Relate the number of photons per
second n to the power output of the
pulse and the energy of a single
photon Ephoton :
n=
The energy of a photon is given by:
E photon =
P
Ephoton
hc
λ
Properties of Light 2935
Substitute for Ephoton to obtain:
n=
λP
hc
Substitute numerical values and evaluate n:
n=
(632.8 nm )(4.00 mW )
1240 eV ⋅ nm
1eV
= 1.27 × 1016 photons/s
−19
1.602 × 10 J
65 •
The first excited state of an atom of a gas is 2.85 eV above the ground
state. (a) What is the maximum wavelength of radiation for resonance absorption
by atoms of the gas that are in the ground state? (b) If the gas is irradiated with
monochromatic light that has a wavelength of 320 nm, what is the wavelength of
the Raman scattered light?
Picture the Problem We can use the Einstein equation for photon energy to find
the wavelength of the radiation for resonance absorption. We can use the same
relationship, with ERaman = Einc − ΔE where ΔE is the energy for resonance
absorption, to find the wavelength of the Raman scattered light.
(a) Use the Einstein equation for
photon energy to relate the
wavelength of the radiation to
energy of the first excited state:
λ=
hc
E
Substitute numerical values and
evaluate λ:
λ=
1240 eV ⋅ nm
= 435 nm
2.85 eV
(b) The wavelength of the Raman
scattered light is given by:
λ Raman =
Relate the energy of the Raman
scattered light ERaman to the
energy of the incident light Einc:
E Raman = E inc − ΔE
Substitute numerical values and
evaluate λRaman:
λRaman =
1240 eV ⋅ nm
E Raman
1240 eV ⋅ nm
− 2.85 eV
320 nm
= 1.025 eV
=
1240 eV ⋅ nm
= 1210 nm
1.025 eV
66 ••
A gas is irradiated with monochromatic ultraviolet light that has a
wavelength of 368 nm. Scattered light that has a wavelength equal to 368 nm is
observed, and scattered light that has a wavelength of 658-nm is also observed.
Assuming that the gas atoms were in their ground state prior to irradiation, find
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 ΔE = hf =
λ
The energy difference between the
ground state and the atomic state
excited by the irradiation is given by:
Substitute 368 nm for λ and evaluate
ΔE:
ΔE = hf =
ΔE =
hc
λ
=
1240 eV ⋅ nm
λ
1240 eV ⋅ nm
= 3.37 eV
368 nm
67 ••
[SSM] Sodium has excited states 2.11 eV, 3.20 eV, and 4.35 eV
above the ground state. Assume that the atoms of the gas are all in the ground
state prior to irradiation. (a) What is the maximum wavelength of radiation that
will result in resonance fluorescence? What is the wavelength of the fluorescent
radiation? (b) What wavelength will result in excitation of the state 4.35 eV above
the ground state? If that state is excited, what are the possible wavelengths of
resonance fluorescence that might be observed?
Picture the Problem The ground state
and the three excited energy levels are
shown in the diagram to the right.
Because the wavelength is related to the
energy of a photon by λ = hc/ΔE,
longer wavelengths correspond to
smaller energy differences.
(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:
λmax =
3
4.35 eV
2
3.20 eV
1
2.11 eV
0
0
1240 eV ⋅ nm
= 388 nm
3.20 eV
Properties of Light 2937
The fluorescence wavelengths are:
λ2→1 =
and
(b) For excitation:
1240 eV ⋅ nm
= 1140 nm
3.20 eV − 2.11eV
λ1→0 =
1240 eV ⋅ nm
= 588 nm
2.11eV − 0
λ 0 →3 =
1240 eV ⋅ nm
= 285 nm
4.35 eV
(not in visible the spectrum)
The fluorescence wavelengths
corresponding to the possible
transitions are:
λ3→2 =
1240 eV ⋅ nm
= 1080 nm
4.35 eV − 3.20 eV
(not in visible spectrum)
λ2→1 =
1240 eV ⋅ nm
= 1140 nm
3.20 eV − 2.11eV
(not in visible spectrum)
1240 eV ⋅ nm
= 588 nm
2.11eV − 0
1240 eV ⋅ nm
λ3→1 =
= 554 nm
4.35 eV − 2.11eV
λ1→0 =
and
λ2→0 =
1240 eV ⋅ nm
= 388 nm
3.20 eV − 0
(not in visible spectrum)
68 ••
Singly ionized helium is a hydrogen-like atom that has a nuclear
charge of +2e. Its energy levels are given by En = –4E0/n2, where n = 1, 2, … and
E0 = 13.6 eV. If a beam of visible white light is sent through a gas of singly
ionized helium, at what wavelengths will dark lines be found in the spectrum of
the transmitted radiation? (Assume that the ions of the gas are all in the same state
with energy E1 prior to irradiation.)
Determine the Concept The energy difference between the ground state and the
first excited state is 3E0 = 40.8 eV, corresponding to a wavelength of 30.4 nm.
This is in the far ultraviolet, well outside the visible range of wavelengths. There
will be no dark lines in the transmitted radiation.
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 Ephoton :
N=
The energy of a photon is given by:
E photon =
Substitute for Ephoton to obtain:
N=
E
E photon
hc
λ
λE
hc
Substitute numerical values (the wavelength of light emitted by a ruby laser is
694.3 nm) and evaluate N:
N=
(694.3 nm )(15 mJ )
1240 eV ⋅ nm
1eV
= 5.2 × 1016
1.602 × 10 −19 J
General Problems
70 •
A beam of red light that has a wavelength of 700 nm in air travels in
water. (a) What is the wavelength in water? (b) Does a swimmer underwater
observe the same color or a different color for this light?
Picture the Problem We can use v = fλ and the definition of the index of
refraction to relate the wavelength of light in a medium whose index of refraction
is n to the wavelength of light in air.
Properties of Light 2939
(a) The wavelength λn of light in a
medium whose index of refraction is
n is given by:
Substitute numerical values and
evaluate λwater:
λn =
λ
v
c
=
= 0
f nf
n
λwater =
700 nm 700 nm
=
= 526 nm
nwater
1.33
(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
(1)
n1 sin θ p = n2 sin (90° − θ p ) = n2 cosθ p
Apply Snell’s law, under
polarization conditions, to the
interface:
or
Solve equation (1) for the ratio of
n2 to n1:
n2
= sin θ c
n1
Substitute for n2/n1 in equation (2) to
obtain:
θ p = tan −1 (sin θ c )
Substitute numerical values and
evaluate θp:
θ p = tan −1 (sin 48°) = 37°
tan θ p =
⎛n ⎞
n2
⇒ θ p = tan −1 ⎜⎜ 2 ⎟⎟ (2)
n1
⎝ n1 ⎠
72 ••
Show that when a flat mirror is rotated through an angle θ about an
axis in the plane of the mirror, a reflected beam of light (from a fixed incident
beam) that is perpendicular to the rotation axis is rotated through 2θ.
Picture the Problem Let φ be the initial angle of incidence. Since the angle of
reflection with the normal to the mirror is alsoφ, the angle between incident and
reflected rays is 2φ. If the mirror is now rotated by a further angle θ, the angle of
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 θc in terms of the index of refraction of the glass and use Figure 31-59 to
find the index of refraction of the glass for the given wavelengths of light.
Apply Snell’s law at the glass-air
interface:
n1 sin θ1 = n2 sin θ 2
If θ1 = θc and n2 = 1:
n1 sin θ c = sin 90° = 1
and
⎛1⎞
⎟⎟
⎝ n1 ⎠
θ c = sin −1 ⎜⎜
⎛ 1 ⎞
⎟ = 36.8°
⎝ 1.67 ⎠
(a) For violet light of wavelength
400 nm, n1 = 1.67:
θ c = sin −1 ⎜
(b) For red light of wavelength
700 nm, n1 = 1.60:
θ c = sin −1 ⎜
⎛ 1 ⎞
⎟ = 38.7°
⎝ 1.60 ⎠
74 ••
Light is incident on a slab of transparent material at an angle θ1, as
shown in Figure 31-60. The slab has a thickness t and an index of refraction n.
Show that n = sin (θ1 ) sin ⎡⎣ tan -1 (d/ t )⎤⎦ , where d is the distance shown in the
figure.
Properties of Light 2941
Picture the Problem We can apply
Snell’s law at the air-slab interface to
express the index of refraction n in
terms of θ1 and θ2 and then use the
geometry of the figure to relate θ2 to
t and d.
θ1
n
θ2
t
d
sin θ1
sin θ 2
Applying Snell’s law to the first
interface yields:
sin θ1 = n sin θ 2 ⇒ n =
From the diagram:
⎛d ⎞
d = t tan θ 2 ⇒ θ 2 = tan −1 ⎜ ⎟
⎝t⎠
Substitute for θ2 to obtain:
n=
sin θ1
⎡
⎛ d ⎞⎤
sin ⎢ tan −1 ⎜ ⎟⎥
⎝ t ⎠⎦
⎣
75 ••
A ray of light begins at the point (–2.00 m, 2.00 m, 0.00 m), strikes a
mirror in the y = 0 plane at some point (x, 0, 0), and reflects through the point
(2.00 m, 6.00 m, 0.00 m). (a) Find the value of x that makes the total distance
traveled by the ray a minimum. (b) What is the angle of incidence on the
reflecting plane? (c) What is the angle of reflection?
y, m
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:
6
4
d2
2
d1
−2
x, m
0
xmin
2
D = d1 + d 2
=
(x + 2)2 + 4 + (2 − x )2 + 36
2942 Chapter 31
Differentiate D with respect to x:
dD d ⎡
=
dx dx ⎢⎣
=
1
2
(x + 2)2 + 4 + (2 − x )2 + 36 ⎤⎥
[(x + 2)
2
+4
]
− 12
[
⎦
2( x + 2) + 12 (2 − x ) + 36
2
]
− 12
2(2 − x )(− 1)
Setting this expression equal to zero for extreme values yields:
x min + 2
( x min + 2)2 + 4
Solve for xmin to obtain:
−
2 − x min
(2 − x min )2 + 36
=0
xmin = − 1.00 m
(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:
tan θ i =
⎛ x − x1
x − x1
⇒ θ i = tan −1 ⎜⎜
y − y1
⎝ y − y1
⎞
⎟
⎟
⎠
where (x, y) = (−1.00 m, 0).
Substitute numerical values and evaluate θi:
⎡ − 1.00 m − (− 2.00 m ) ⎤
⎥ = 26.6°
0 − 2.00 m
⎣
⎦
θ i = tan −1 ⎢
(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:
tan θ r =
⎛ x −x⎞
x2 − x
⎟⎟
⇒ θ r = tan −1 ⎜⎜ 2
y2 − y
⎝ y2 − y ⎠
where (x, y) = (−1.00 m, 0).
Substitute numerical values and evaluate θr:
⎡ 2.00 m − (− 1.00 m ) ⎤
⎥ = 26.6°
6.00 m − 0
⎣
⎦
θ r = tan −1 ⎢
76 ••
To produce a polarized laser beam a plate of transparent material,
(Figure 31-61) is placed in the laser cavity and oriented so the light strikes it at the
polarizing angle. Such a plate is called a Brewster window. Show that if θP1 is the
Properties of Light 2943
polarizing angle for the n1 to n2 interface, then θP2 is the polarizing angle for the
n2 to n1 interface.
Picture the Problem Let the angle of refraction at the first interface by θ1 and the
angle of refraction at the second interface be θ2. We can apply Snell’s law at each
interface and eliminate θ1 and n2 to show that θ2 = θP2.
Apply Brewster’s law at the n1-n2
interface:
tan θ P1 =
n2
n1
Draw a reference triangle consistent
with Brewster’s law:
2
2
n1
2
+n
θP1
n1
Apply Snell’s law at the n1-n2
interface:
Solve for θ1 to obtain:
Referring to the reference triangle
we note that:
n1 sin θ P1 = n2 sin θ1
⎞
⎛ n1
sin θ P1 ⎟⎟
⎠
⎝ n2
θ1 = sin −1 ⎜⎜
⎛n
θ1 = sin −1 ⎜
1
⎜ n2
⎝
⎞
⎟
2
2 ⎟
n1 + n2 ⎠
n2
⎛
⎞
n1
⎟
= sin −1 ⎜
⎜ n2 + n2 ⎟
2 ⎠
⎝ 1
i.e., θ1 is the complement of θp1.
Apply Snell’s law at the n2-n1
interface:
Solve for θ2 to obtain:
n2 sin θ1 = n1 sin θ 2
⎞
⎛ n2
sin θ1 ⎟⎟
⎠
⎝ n1
θ 2 = sin −1 ⎜⎜
n2
2944 Chapter 31
⎞
⎟
⎜ n1 n 2 + n 2 ⎟
1
2 ⎠
⎝
⎞
⎛
n2
⎟= θ
= sin −1 ⎜
P2
⎜ n2 + n2 ⎟
2 ⎠
⎝ 1
⎛n
Refer to the reference triangle again
to obtain:
θ 2 = sin −1 ⎜
Equate these expressions for n2 sin θ1
to obtain:
n1 sin θ P = n1 sin θ 2 ⇒ θ 2 = θ P
2
n1
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:
⎛ n2 ⎞
⎟⎟
⎝ n1 ⎠
θ p = tan −1 ⎜⎜
or, for n1 = 1,
θ p = tan −1 (n2 )
For silicate flint glass, n2 ≈ 1.62
and:
θ p,silicate flint = tan −1 (1.62) = 58.3°
For borate flint glass, n2 ≈ 1.57 and:
θ p,borate flint = tan −1 (1.57 ) = 57.5°
For quartz glass, n2 ≈ 1.54 and:
θ p, quartz = tan −1 (1.54) = 57.0°
For silicate crown glass, n2 ≈ 1.51
and:
θ p,silicate crown = tan −1 (1.51) = 56.5°
78 ••
A light ray passes through a prism with an apex angle of α, as shown
in Figure 31-63. The ray and the bisector of the apex angle bisect at right angles.
Show that the angle of deviation δ is related to the apex angle and the index of
refraction of the prism material by sin ⎡⎣ 12 (α + δ )⎤⎦ = n sin (12 α ).
Properties of Light 2945
Picture the Problem The diagram to
the right shows the angles of
incidence, refraction, and deviation
at the first interface. We can use the
geometry of this symmetric passage
of the light to express θr in terms of
α and δ1 in terms of θr and α. We
can then use a symmetry argument to
express the deviation at the second
interface and the total deviation δ.
Finally, we can apply Snell’s law at
the first interface to complete the
derivation of the given expression.
α
90° −θ r
θi
n1
θr
δ1
n2
With respect to the normal to the left
face of the prism, let the angle of
incidence be θi and the angle of
refraction be θr. From the geometry
of the figure, it is evident that:
θ r = 12 α
Express the angle of deviation at the
refracting surface:
δ1 = θ i − θ r = θ i − 12 α
Referring to triangle ABC, we see
that:
δ = 2δ1 = 2(θ1 − 12 α ) = 2θ i − α
α
A
n1
B δ
δ1 δ 1 C δ
n2
n1
(α + δ )
Solving for θI yields:
θi =
Apply Snell’s law, with n1 = 1 and
n2 = n, to the first interface:
sin θ i = n sin 12 α
1
2
δ
n1
2946 Chapter 31
Substitute for θI to obtain:
sin
α +δ
2
= n sin
α
2
(1)
79 ••
[SSM] (a) For light rays inside a transparent medium that is
surrounded by a vacuum, show that the polarizing angle and the critical angle for
total internal reflection satisfy tan θp = sin θc. (b) Which angle is larger, the
polarizing angle or the critical angle for total internal reflection?
Picture the Problem We can apply Snell’s law at the critical angle and the
polarizing angle to show that tan θp = sin θc.
(a) Apply Snell’s law at the
medium-vacuum interface:
n1 sin θ1 = n2 sin θ r
For θ1 = θc, n1 = n, and n2 = 1:
n sin θ c = sin 90° = 1
For θ1 = θp, n1 = n, and n2 = 1:
Because both expressions equal
one:
(b) For any value of θ :
tan θ p =
n2 1
= ⇒ n tan θ p = 1
n1 n
tan θ p = sin θ c
tan θ > sin θ ⇒ θ p > θ c
80 ••
Light in air is incident on the surface of a transparent substance at an
angle of 58º with the normal. The reflected and refracted rays are observed to be
mutually perpendicular. (a) What is the index of refraction of the transparent
substance? (b) What is the critical angle for total internal reflection in this
substance?
Picture the Problem Let the numeral 1 refer to the side of the interface from
which the light is incident and the numeral 2 to the refraction side of the interface.
We can apply Snell’s law, under the conditions described in the problem
statement, at the interface to derive an expression for n as a function of the angle
of incidence (also the polarizing angle).
(a) Apply Snell’s law at the airmedium interface:
sin θ1 = n sin θ 2
Because the reflected and refracted
rays are mutually perpendicular:
θ1 + θ 2 = 90° ⇒ θ 2 = 90° − θ1
Properties of Light 2947
Substitute for θ2 to obtain:
sin θ1 = n sin(90° − θ1 ) = n cosθ1
or
n = tan θ1 = tan θ p
Substitute for θp 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 = n1 sin 90° = n1
Because n1 = 1:
Substitute for n and evaluate θc:
⎛1⎞
⎛1⎞
⎟⎟ = sin −1 ⎜ ⎟
⎝n⎠
⎝ n2 ⎠
θ c = sin −1 ⎜⎜
⎛ 1 ⎞
⎟ = 39°
⎝ 1 .6 ⎠
θ c = sin −1 ⎜
81 ••
A light ray in dense flint glass that has an index of refraction of 1.655
is incident on the glass surface. An unknown liquid condenses on the surface of
the glass. Total internal reflection on the glass–liquid interface occurs for a
minimum angle of incidence on the glass–liquid interface of 53.7º. (a) What is the
refractive index of the unknown liquid? (b) If the liquid is removed, what is the
minimum angle of incidence for total internal reflection? (c) For the angle of
incidence found in Part (b), what is the angle of refraction of the ray into the
liquid film? Does a ray emerge from the liquid film into the air above? Assume
the glass and liquid have parallel planar surfaces.
Picture the Problem We can apply Snell’s law at the glass–liquid and liquid–air
interfaces to find the refractive index of the unknown liquid, the minimum angle
of incidence (glass-air interface) for total internal reflection, and the angle of
refraction of a ray into the liquid film.
(a) Apply Snell’s law, under
critical-angle conditions, at the
glass–liquid interface:
Substitute numerical values and
evaluate nliquid :
(b) With the liquid removed:
sin θ c =
nliquid
nglass
⇒ nliquid = nglass sin θ c
nliquid = (1.655) sin 53.7° = 1.33
⎛ 1 ⎞
⎟
⎟
n
⎝ glass ⎠
θ c = sin −1 ⎜⎜
2948 Chapter 31
⎛ 1 ⎞
⎟ = 37.2°
⎝ 1.655 ⎠
Substitute numerical values and
evaluate θc:
θ c = sin −1 ⎜
(c) Apply Snell’s law at the
glass−liquid interface:
nglass sin θ1 = nliquid sin θ 2
Solve for θ2:
⎡ nglass
θ 2 = sin −1 ⎢
⎢⎣ nliquid
Letting θ1 be the angle of incidence
found in Part (b) yields:
⎤
sin θ1 ⎥
⎥⎦
⎡⎛ nglass ⎞⎛ 1 ⎞⎤
⎟⎥
⎟⎜
⎢⎣⎝ nliquid ⎟⎠⎜⎝ nglass ⎟⎠⎥⎦
θ 2 = sin −1 ⎢⎜
⎜
⎛ 1 ⎞
⎟
= sin −1 ⎜
⎟
⎜n
liquid
⎠
⎝
Substitute numerical values and
evaluate θ2:
⎛ 1 ⎞
⎟ = 48.6°
⎝ 1.333 ⎠
θ 2 = sin −1 ⎜
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
2
2
approximately given by I T = I0 ⎡ 4n (n + 1) ⎤ . (b) Use the Part (a) result to find
⎣
⎦
the ratio of the transmitted intensity to the incident intensity through N parallel
slabs of glass for light of normal incidence. (c) How many slabs of a glass that has
an index of refraction of 1.5 are required to reduce the intensity to 10 percent of
the incident intensity?
Properties of Light 2949
Picture the Problem We’ll neglect
multiple reflections at the glass-air
interfaces. We can use the expression
(Equation 31-7) for the reflected
intensity at an interface to express the
intensity of the light in the glass slab as
the difference between the intensity of
the incident beam and the reflected
beam. Repeating this analysis at the
glass-air interface will lead to the
desired result.
(a) Express the intensity of the light
transmitted into the glass:
Glass
Air
I glass
I0
I R,1
IT
I R,2
n
I glass = I 0 − I R,1
where IR,1 is the intensity of the light
reflected at the air-glass interface.
2
The intensity of the light reflected at
the air-glass interface is given by
Equation 31-7:
⎛1− n ⎞
I R,1 = ⎜
⎟ I0
⎝1+ n ⎠
Substitute and simplify to obtain:
⎛1− n ⎞
I glass = I 0 − ⎜
⎟ I0
⎝1+ n ⎠
⎡ ⎛ 1 − n ⎞2 ⎤
= I 0 ⎢1 − ⎜
⎟ ⎥
⎢⎣ ⎝ 1 + n ⎠ ⎥⎦
⎡ 4n ⎤
= I0 ⎢
2⎥
⎣ (1 + n ) ⎦
Express the intensity of the light
transmitted at the glass-air
interface:
I T = I glass − I R,2
where IR,2 is the intensity of the light
reflected at the glass-air interface.
The intensity of the light reflected at
the glass-air interface is:
⎛1− n ⎞
I R,2 = ⎜
⎟ I glass
⎝1+ n ⎠
2
2
2
⎛ 1 − n ⎞ ⎡ 4n ⎤
I
=⎜
⎟ ⎢
2⎥ 0
⎝ 1 + n ⎠ ⎣ (1 + n ) ⎦
2950 Chapter 31
Substitute and simplify to obtain:
2
⎡ 4n ⎤ ⎛ 1 − n ⎞ ⎡ 4 n ⎤
IT = I0 ⎢
I
−⎜
⎟ ⎢
2⎥
2⎥ 0
⎣ (1 + n ) ⎦ ⎝ 1 + n ⎠ ⎣ (1 + n ) ⎦
⎡ ⎛ 1 − n ⎞ 2 ⎤ ⎡ 4n ⎤
= I 0 ⎢1 − ⎜
⎟ ⎥⎢
2⎥
⎢⎣ ⎝ 1 + n ⎠ ⎥⎦ ⎣ (1 + n ) ⎦
⎡ 4n ⎤ ⎡ 4n ⎤
= I0 ⎢
2 ⎥⎢
2⎥
⎣ (1 + n ) ⎦ ⎣ (1 + n ) ⎦
⎡ 4n ⎤
= I0 ⎢
2⎥
⎣ (1 + n ) ⎦
2
2
(b) From Part (a), each slab reduces
the intensity by the factor:
⎡ 4n ⎤
⎢
2⎥
⎣ (n + 1) ⎦
For N slabs:
⎡ 4n ⎤
I t = I0 ⎢
2⎥
⎣ (n + 1) ⎦
and
2N
⎡ 4n ⎤
It
= ⎢
2⎥
I0
⎣ (n + 1) ⎦
2N
(c) Begin the solution of equation (1)
for N by taking the logarithm
(arbitrarily to base 10) of both sides
of the equation:
⎡ 4n ⎤
⎛I ⎞
log⎜⎜ t ⎟⎟ = log ⎢
2⎥
⎝ I0 ⎠
⎣ (n + 1) ⎦
Solve for N:
⎛I ⎞
log⎜⎜ t ⎟⎟
⎝ I0 ⎠
N=
⎡ 4n ⎤
2 log ⎢
2⎥
⎣ (n + 1) ⎦
Substitute numerical values and
evaluate N:
N=
(1)
2N
⎡ 4n ⎤
= 2 N log ⎢
2⎥
⎣ (n + 1) ⎦
log(0.1)
= 28.2 ≈ 28
⎡ 4(1.5) ⎤
2 log ⎢
2⎥
⎣ (1.5 + 1) ⎦
Properties of Light 2951
83 ••• Equation 31-14 gives the relation between the angle of deviation φd of
a light ray incident on a spherical drop of water in terms of the incident angle θ1
and the index of refraction of water. (a) Assume that nair = 1, and derive an
expression for dφd/dθ1. Hint: If y = sin–1 x, then dy/dx = (1 – x2)–1/2. (b) Use this
result to show that the angle of incidence for minimum deviation θ1m is given by
cos θ1m =
1
3
(n
2
)
− 1 . (c) The index of refraction for a certain red light in water is
1.3318 and that the index of refraction for a certain blue light in water is 1.3435.
Use the result of Part (a) to find the angular separation of these colors in the
primary rainbow.
Picture the Problem (a) We can follow the directions given in the problem
statement and use the hint to establish the given result. (b) Treating the result of
Part (a) as an extreme-value problem will lead to the given result. (c) We can use
Equation 31-14 and the result of Part (b) to find the angular separation of these
colors in the primary rainbow.
(a) Equation 31-14 is:
⎛ nair sin θ1 ⎞
⎟⎟
⎝ nwater ⎠
φd = π + 2θ1 − 4 sin −1 ⎜⎜
⎛ sin θ1 ⎞
⎟
⎝ n ⎠
For nair = 1and nwater = n:
φd = π + 2θ1 − 4 sin −1 ⎜
Use the hint to differentiate φd with
respect to θ1:
dφ d
d ⎡
⎛ sin θ1 ⎞⎤
=
π + 2θ1 − 4 sin −1 ⎜
⎟⎥
⎢
d θ1 d θ1 ⎣
⎝ n ⎠⎦
= 2−
(b) Set dφd/dθ1 = 0:
2−
4 cos θ1
n 2 − sin 2 θ1
4 cos θ1
n 2 − sin 2 θ1
= 0 for extrema
(
Simplify to obtain:
16 cos 2 θ1 = 4 n 2 − sin 2 θ1
Replace sin2θ1 with 1 − cos2θ1 and
simplify to obtain:
12 cos 2 θ1 = 4n 2 − 4
Solve for cosθ1 = cosθ1m:
cos θ1m =
n2 − 1
3
)
2952 Chapter 31
(c) Express the angular separation Δφ
of blue and red:
Δφ = φd,blue − φd,red
From Equation 31-18, with nair = 1
and nwater = n:
φd = π + 2θ1 − 4 sin −1 ⎜
(1)
⎛ sin θ1 ⎞
⎟
⎝ n ⎠
⎡ n2 − 1 ⎤
⎥
3 ⎥⎦
⎢⎣
From Part (b):
θ1m = cos −1 ⎢
Substitute to obtain:
2
⎛ ⎧⎪
⎫⎞
⎜ sin cos −1 ⎡⎢ n − 1 ⎤⎥ ⎪ ⎟
⎨
⎬⎟
⎜ ⎪
2
3
⎡
⎤
⎢
⎥
n
1
−
⎣
⎦ ⎪⎭ ⎟
⎩
−1
φd = π + 2 cos −1 ⎢
⎥ − 4 sin ⎜
⎜
⎟
3 ⎥⎦
n
⎢⎣
⎜
⎟
⎜
⎟
⎝
⎠
Evaluate φd for blue light in water:
⎡
φd,blue = π + 2 cos −1 ⎢
⎢⎣
2
⎛ ⎧
⎫⎞
⎜ sin ⎪cos −1 ⎡⎢ (1.3435) − 1 ⎤⎥ ⎪ ⎟
⎬
⎜ ⎨
3
⎢⎣
⎥⎦ ⎪⎭ ⎟⎟
(1.3435)2 − 1 ⎤⎥ − 4 sin −1 ⎜ ⎪⎩
⎜
⎟
3
1.3435
⎥⎦
⎜
⎟
⎜
⎟
⎝
⎠
= 139.42°
Evaluate φd for red light in water:
⎡
φd,red = π + 2 cos −1 ⎢
⎢⎣
= 137.75°
2
⎛ ⎧
⎫⎞
⎜ sin ⎪cos −1 ⎡⎢ (1.3318) − 1 ⎤⎥ ⎪ ⎟
⎬⎟
⎜ ⎨
2
3
⎤
⎢
⎥
⎪
(1.3318) − 1 ⎥ − 4 sin −1 ⎜ ⎩
⎣
⎦ ⎪⎭ ⎟
⎜
⎟
3
1.3318
⎥⎦
⎜
⎟
⎜
⎟
⎝
⎠
Properties of Light 2953
Substitute numerical values in
equation (1) and evaluate Δφ:
Δφ = 139.42 − 137.75° = 1.67°
84 ••• Show that the angle of deviation δ is a minimum if the angle of
incidence is such that the ray and the bisector of the apex angle α (Figure 31-62)
intersect at right angles.
Picture the Problem The figure below shows the prism and the path of the ray
through it. The dashed lines are the normals to the prism faces. The triangle
formed by the interior ray and the prism faces has interior angles of α, 90° − θ2,
and 90° − θ3. Consequently, 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 θ3 and then set the
derivative of this function equal to zero to find the conditions on θ3 and θ2 that
result in δ being a minimum.
a
α
δd
θ1
θu22
θ3
u3
θ4
n
Express the angle of deviation:
δ = θ1 + θ 2 − α
(1)
Apply Snell’s law to relate θ1 to
θ2 and θ3 to θ4:
sin θ1 = n sin θ 2
and
n sin θ 3 = sin θ 4
(2)
Solve equation (2) for θ1 and
equation (3) for θ4:
(3)
θ1 = 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
2954 Chapter 31
Note that the only variable in this expression is θ3. To determine the condition
that minimizesδ, take the derivative of δ with respect to θ3 and set it equal to zero.
dδ
d
=
sin −1 [n sin (θ 3 − α )] + sin −1 (n sin θ 3 ) − α
dθ 3 d θ 3
{
=−
n cos(α − θ 3 )
1 − [n sin (α − θ 3 )]
2
+
n cos θ 3
1 − (n sin θ 3 )
2
}
= 0 for extrema
This equation is satisfied
provided:
α − θ 3 = θ 3 ⇒ θ 3 = 12 α
Because θ 2 = α − θ 3 :
θ 2 = α − 12 α = 12 α
Because θ 2 = θ 3 , we can conclude that the deviation angle is a minimum if the ray p
through the prism symmetrically.
Remarks: Setting dδ/dθ3 = 0 establishes the condition on θ3 that δ is either a
maximum or a minimum. To establish that δ is indeed a minimum when
θ3 = θ2 = 12 α, we can either show that d 2 δ dθ32 , evaluated at θ3 = 12 α , is
positive or, alternatively, plot a graph of δ (θ3) to show that it is concave
upward at θ3 = 12 α .
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Chapter 31 Properties of Light