PHY 1161C

Homework

Chapter 23: Reflection and Refraction of Light

Ch 23: 8, 10, 14, 20, 26, 28, 33, 38, 43, 45, 52

**23.8** What is the speed of
light in water (n = 1.33)?

n = c/vv = c/n

v = (3.00 x 10

^{8}^{m}/_{s})/(1.33)v = 2.26 x 10

^{8}^{m}/_{s}

**23.10** What is the wavelength
in water of **blue**
light, with a wavelength of 440 nm in air?

**[Visible light has
wavelengths from about ****700
nm for red**

v = f for all wavesc = f

_{o}for light in airv = f ’ for light in water

Divide one of these by the other to obtain

’ =

_{o}/(c/v)’ =

_{o}/n’ =

^{440 nm}/_{1.33}’ = 330 nm

**23.14** Two light pulses are
emitted simultaneously from a source. Both pulses travel to a
detector over the same physical distance but one passes through
**6.2
cm** of ice. What is the
difference in the pulses' times of arrival at the
detector?

**The index of refraction for ice
is 1.305**

The time for the light to travel a distance of 6.2 cm = 0.062 m through theiceist = d/v

v = c/n

v = (3.00 x 10

^{8}m/s)/1.305v = 2.299 x 10

^{8}m/st

_{ice}= 0.062 m/(2.299 x 10^{8}m/s)t

_{ice}= 0.0270 x 10^{ - 8}sThe time for the other light beam to travel a distance of 6.2 cm = 0.062 m through the

airist = d/v

v = 3.00 x 10

^{8}m/st

_{air}= 0.062 m/(3.00 x 10^{8}m/s)t

_{air}= 0.0207 x 10^{ - 8}sThe

differencein these two times ist = t

_{ice}- t_{air}= 0.0063 x 10^{ - 8}st = 0.063 x 10

^{ - 9}s = 0.063 ns

**23.20** A
SCUBA diver, underwater, shines a light up toward the smooth surface
of the water with an angle of incidence of 35°. At what angle
does the light leave the water?

We begin with Snell’s law,n

_{1}sin_{}_{1}= n_{2}sin_{}_{2}(1.33)(sin 35°) = (1.00) sin

_{}_{2}(1.33)(0.5736) = (1.00) sin

_{}_{2}sin

_{}_{2}= 0.7269

_{}_{2}= 49.7°

**23.26** An aquarium with glass
walls (n = 1.50) is filled with water (n = 1.33). A ray of light,
passing through the air strikes the glass side with an angle of
incidence of 37°. What is the angle of refraction inside the
water?

We can apply Snell’s Lawtwice,n

_{1}sin_{1}= n_{2}sin_{}_{2}n

_{2}sin_{}_{2}= n_{3}sin_{}_{3}n

_{1}sin_{}_{1}= n_{3}sin_{}_{3}(1.00) sin 37° = (1.33) sin

_{}_{3}(1.00) (0.60) = (1.33) sin

_{}_{3}sin

_{}_{3}= 0.60/1.33 = 0.452

_{}_{3}=26.9°

**23.28** A SCUBA diver,
underwater, shines a light up toward the smooth surface of the water
with an angle of incidence of 53°. Explain what happens to the
light.

The critical angle for total internal reflection at the water-air interface issin

_{}_{c}= n_{2}/n_{1}=^{1.00}/_{1.33}= 0.75

_{}_{c}= 48.6°

Since this light comes in to the water-air interface at an angle of incidencegreaterthan the critical angle,allof it is reflected. That is, the light undergoestotal internal reflectionso all of it is reflected with an angle of reflection of 53° andnoneof it is refracted into the air.

**23.33** What is the critical
angle for diamond in air? What is the critical angle for diamond in
water?

The critical angle is given by

_{}_{c}= sin^{ - 1}[n_{2}/ n_{1}]The index of refraction of diamond is n

_{1}= 2.4.For a diamond in

air(n_{2}= 1.00), then, the critical angle is

_{}_{c}= sin^{ - 1}[1.00/2.4] = sin^{ - 1}[0.417]

_{}_{c}= 24.6°For a diamond in

water(n_{2}= 1.33), then, the critical angle is

_{}_{c}= sin^{ - 1}[1.33/2.4] = sin^{ - 1}[0.554]

_{}_{c}= 33.6°

**23.38** A ray of light,
containing only wavelengths of 490 and 660 nm, is incident on a piece
of heavy flint glass at 42°. What is the angular separation of
the refracted rays inside the glass?

Table 23.3, on page 876, lists the index of refraction of heavy flint glass for various wavelengths; n(490 nm) = 1.664, n(660 nm) = 1.644. That means the angle of refraction qr will be different for the two wavelengths. We apply Snell’s Law for both values of the index of refraction.n

_{1}sin_{}_{1}= n_{2}sin_{}_{2}For = 490 nm and n

_{2}= 1.664 this becomes(1.00) sin 42° = (1.664) sin

_{}_{2}(1.00) (0.6691) = (1.664) sin

_{}_{2}sin

_{}_{2}= 0.6691/1.664 = 0.4021

_{}_{2}= 23.71°For = 660 nm and n

_{2}= 1.644 Snell’s Law gives us(1.00) sin 42° = (1.644) sin

_{}_{2}(1.00) (0.6691) = (1.644) sin

_{}_{2}sin

_{}_{2}= 0.6691/1.644 = 0.4070

_{}_{2}= 24.02°The

differencein these two angles is 0.32°. The two colors of light are dispersed by this angle

_{}= 0.32°

**23.41** A ray of white light
shines on an equilateral prism made of zinc crown glass so that the
incoming angle of incidence equals the outgoing angle of refraction
(for 590 nm light). What is the angular spread of the
spectrum?

First, we must find the angle of incidence,_{}_{1}. We start with Snell’s Law,n

_{1}sin_{}_{1}= n_{2}sin_{}_{2}(1.000) sin

_{}_{1}= (1.517) sin_{}_{2}From the symmetry of having the incoming incident angle

_{}_{1}equal to the final angle of reflection_{}_{4}, the two angles inside the glass must also be equal;_{}_{2}=_{}_{3}. This requires that the light inside the glass be parallel to the base of the prism or that

_{}_{2}= 30°(1.000) sin

_{}_{1}= (1.517) sin 30°(1.000) sin

_{}_{1}= (1.517) (0.500)sin

_{}_{1}= 0.7585

_{}_{1}= 49.33°Now we must go back and calculate the final angle of refraction,

_{}_{4}, for the extremes.For = 430 nm (violet light), n

_{2}= n_{3}= 1.528;for = 770 nm (red light), n

_{2}= n_{3}= 1.511.We may as well start with violet light,

= 430 nm, n

_{2}= n_{3}= 1.528n

_{1}sin_{}_{1}= n_{2}sin_{}_{2}(1.000) sin 49.33° = (1.528) sin

_{}_{2}(1.000)(0.7585) = (1.528) sin

_{}_{2}sin

_{}_{2}=^{0.7585}/_{1.528}= 0.496

_{}_{2}= 29.76°We can use the geometry of a triangle -- that the sum of a triangle’s three angles is 180° -- to determine

_{}_{3},60° + (90° -

_{}_{2}) + (90° -_{}_{3}) = 180°

_{}_{3}= 60° -_{}_{2}

_{}_{3}= 60° - 29.76°

_{}_{3}= 30.24°Again, for this glass to air interface, apply Snell’s Law,

n

_{4}sin_{}_{4}= n_{3}sin_{}_{3}(1.00) sin

_{}_{4}= (1.528) sin 30.24°sin

_{}_{4}= (1.528) (0.5036) = 0.769

_{}_{4}= 50.31°, forlight.violetNow we go through the same calculations but, this time,

for

redlight, = 770 nm, n_{2}= n_{3}= 1.511n

_{1}sin_{}_{1}= n_{2}sin_{}_{2}(1.000) sin 49.33° = (1.511) sin

_{}_{2}(1.000)(0.7585) = (1.511) sin

_{}_{2}sin

_{}_{2}=^{0.7585}/_{1.511}= 0.502

_{}_{2}= 30.13°We can use the geometry of a triangle -- that the sum of a triangle’s three angles is 180° -- to determine

_{}_{3},60° + (90° -

_{}_{2}) + (90° -_{}_{3}) = 180°

_{}_{3}= 60° -_{}_{2}

_{}_{3}= 60° - 30.13°

_{}_{3}= 29.87°Again, for this glass to air interface, apply Snell’s Law,

n

_{4}sin_{}_{4}= n_{3}sin_{}_{3}(1.00) sin

_{}_{4}= (1.511) sin 29.87°sin

_{}_{4}= (1.511) (0.4980) = 0.7525

_{}_{4}= 48.81°, forlight.redThe entire spectrum of colors, then, will be spread out between the

violetandredextremes,

_{}= 50.31° - 48.81°

_{}= 1.50°

**23.43**
Through what angle should an analyzer be rotated
from the incoming plane of polarization to reduce the intensity to
one-fourth?

The intensity is given byS = S

_{o}cos^{2}_{}

[This is containedinthe text, on page 885, before Example 23.8; the equation given in Example 23.8, at the top of page 886, is NOT correct].S = S

_{o}cos^{2}_{}S = S

_{o}cos^{2}_{}= So/4cos

^{2}_{}= 0.25cos

_{}= 0.50

_{}= 60°That is, cos (60°) = 0.5

**23.45 **[Oops, this is
essentially a repeat of the previous problem; sorry about that].
Light coming from a polarizer passes through another polarizer, used
as an analyzer, that has been rotated 30°. What is the intensity
of the exiting light (compared to the intensity arriving at the
analyzer)?

The intensity is given byS = S

_{o}cos^{2}_{}S = S

_{o}cos^{2}30° = S_{o}(0.866)^{2}= S_{o}(0.75)S = 0.75 S

_{o}

**23.52** What should be the
initial angle of incidence onto an equilateral prism so that it
equals the final angle of refraction? The index of refraction of the
prism is 1.45.

This is essentially the same as problem 23.41 with a different index of refraction.

We must find the angle of incidence,

_{}_{1}. We start with Snell’s Law,n

_{1}sin_{}_{1}= n_{2}sin_{}_{2}(1.000) sin

_{}_{1}= (1.45) sin_{}_{2}From the symmetry of having the incoming incident angle

_{}1 equal to the final angle of reflection_{}_{3}, the two angles inside the glass must also be equal;_{}_{2}=_{}_{3}. This requires that the light inside the glass be parallel to the base of the prism or that

_{}_{2}= 30°(1.000) sin

_{}_{1}= (1.45) sin 30°(1.000) sin

_{}_{1}= (1.45) (0.500)sin

_{}_{1}= 0.725

_{}_{1}= 46.47°(c) Doug Davis, 2003; all rights reserved

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