ch24 giancoli7e manual

8/15/2019 Ch24 Giancoli7e Manual

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24-1

Responses to Questions

1. Huygens’ principle applies to both sound waves and water waves. Huygens’ principle applies to all

waves that form a wave crest. Sound, water, and light waves all can be represented in this way. The

maximum or crest of a water wave is its highest point above the equilibrium level of the water, so a

wave front can be formed by connecting all of the local maximum points of a wave.

2. A ray shows the direction of propagation of a wave front. If this information is enough for the situation

under discussion, then light can be discussed as rays. This is true in particular for geometric optics.

Sometimes, however, the wave nature of light is essential to the discussion. For instance, the double-

slit interference pattern depends on the interference of the waves and could not be explained by

examining light as only rays.

3. The bending of waves around corners or obstacles is called diffraction. Diffraction is most prominent

when the size of the obstacle is on the order of the size of the wavelength. Sound waves have much

longer wavelengths than do light waves. As a result, the diffraction of sound waves around a corner of

a building or through a doorway is noticeable, and we can hear the sound in the “shadow region,” but

the diffraction of light waves around a corner is not noticeable because of the very short wavelength of

the light.

4. For destructive interference, the path lengths must differ by an odd number of half wavelengths, such

as λ/2, 3 λ/2, 5 λ/2, 7 λ/2, etc. In general, the path lengths must differ by λ(m + ½), where m is an integer.

Under these conditions, the wave crests from one ray match up with the wave troughs from the other

ray and cancellation occurs (destructive interference).

5. As red light is switched to blue light, the wavelength of the light is decreased. Thus, sind mθ λ = says

that θ is decreased for a constant m and d . This means that the bright spots on the screen are moreclosely packed together with blue light than with red light.

6. The wavelength of light in a medium such as water is decreased when compared to the wavelength in

air. Thus, sind mθ λ = says that θ is decreased for a particular m and d . This means that the bright

spots on the screen are more closely packed together in water than in air.

7. The reason you do not get an interference pattern from the two headlights of a distant car is that they

are not coherent light sources. The phase relationship between the two headlights is not constant—they

THE WAVE NATURE OF LIGHT 24


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24-2 Chapter 24



have randomly changing phases relative to each other. Thus, you cannot produce zones of destructive

and constructive interference where the crests and troughs match up or the crests and crests match up.

Also, the headlights are far enough apart that even if they were coherent, the interference pattern

would be so tightly packed that it would not be observable with the unaided eye.

8. For a very thin film, there are only a select few distinct visible wavelengths that meet the constructive

interference criteria, because the film is only a few wavelengths thick. But for a thick film, there might

be many different wavelengths that meet the constructive interference criteria for many different m

values. Accordingly, many different colors will constructively interfere, and the reflected light will be

white (containing all visible wavelengths).

9. As you move farther away from the center of the curved piece of glass on top, the path differences

change more rapidly due to the curvature. Thus, you get higher order interference patterns more

closely spaced together. An “air wedge,” as in Fig. 24–33, has equally spaced interference patterns

because as the observation point is moved farther from the contact point of the flat piece of glass on

top, the path differences change linearly.

10. These lenses probably are designed to eliminate reflected wavelengths at both the red and the blueends of the spectrum. The thickness of the coating is designed to cause destructive interference for

reflected red and blue light. The reflected light then appears yellow-green.

11. The index of refraction of the oil must be less than the index of refraction of the water. If the oil film

appears bright at the edge, then the interference between the light reflected from the top of the oil film

and that reflected from the bottom of the oil film at that point must be constructive. The light reflecting

from the top surface (the air/oil interface) undergoes a 180° phase shift since the index of refraction of

the oil is greater than that of air. The thickness of the oil film at the edge is negligible, so for there to

be constructive interference, the light reflecting from the bottom of the oil film (the oil/water interface)

must also undergo a 180° phase shift. This will occur only if the index of refraction of the oil is less

than that of the water: 1.00 < n < 1.33.

12. Radio waves have a much longer wavelength than visible light and will diffract around normal-sizedobjects (like hills). The wavelengths of visible light are very small and will not diffract around normal-

sized objects. Thus diffraction allows the radio waves to be “picked up” even if the broadcasting tower

is not visible on a line of sight.

13. You see a pattern of dark and bright lines parallel to your fingertips in the narrow opening between

your fingers, due to diffraction.

14. (a) When you increase the slit width in a single-slit diffraction experiment, the spacing of the fringes

decreases. The equation for the location of the minima, sin ,m

D

λ θ = indicates that θ is

decreased for a particular m and λ when the width D increases. This means that the bright spotson the screen are more closely packed together for a wider slit.

(b) When you increase the wavelength of light used in a single-slit diffraction experiment, the

spacing of the fringes increases. The equation for the location of the minima, sin ,m

D

λ θ =

indicates that θ is increased for a particular m and D when the wavelength increases. This

means that the bright spots on the screen are spread farther apart for a longer wavelength.

15. (a) A slit width of 60 nm would produce a central maximum so spread out that it would cover the

entire width of the screen. No minimum (and therefore no diffraction pattern) will be seen,


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The Wave Nature of Light 24-3



because the first minimum would have to satisfy sin 10, D

λ θ = ≈ which is not possible. The

different wavelengths will all overlap, so the light on the screen will be white. It will also be

dim, compared to the source, because it is spread out.

(b) For the 60,000-nm slit, the central maximum will be very narrow, about a degree in width for the blue end of the spectrum and about a degree and a half for the red. The diffraction pattern will

not be distinct, because most of the intensity will be in the small central maximum and the

fringes for the different wavelengths of white light will not coincide.

16. (a) If the apparatus is immersed in water, then the wavelength of the light will decrease .n

λ λ

⎛ ⎞′ =⎜ ⎟

⎝ ⎠

and the diffraction pattern will become more compact.

(b) If the apparatus is placed in a vacuum, then the wavelength of the light will increase slightly and

the diffraction pattern will spread out very slightly.

17. The interference pattern created by the diffraction grating with 410 lines/cm has bright maxima that

are more sharply defined and narrower than the interference pattern created by the two slits 410− cm

apart. The spacing of the bright maxima would be the same in both patterns, but for the grating, each

maximum would be essentially the same brightness, while for the two-slit pattern, slit width effects

would make the maxima for m > 1 much less bright than the central maximum.

18. (a) The advantage of having many slits in a diffraction grating is that this makes the bright maxima

in the interference pattern more sharply defined, brighter, and narrower.

(b) The advantage of having closely spaced slits in a diffraction grating is that this spreads out the

bright maxima in the interference pattern and makes them easier to resolve.

19. (a) Violet light will be at the top of the rainbow created by the diffraction grating. Principal maxima

for a diffraction grating are at positions given by sin .m

d

λ θ = Violet light has a shorter

wavelength than red light so will appear at a smaller angle away from the direction of the

horizontal incident beam.

(b) Red light will appear at the top of the rainbow created by the prism. The index of refraction for violet

light in a given medium is slightly greater than for red light in the same medium, so the violet light

will bend more and will appear farther from the direction of the horizontal incident beam.

20. Polarization demonstrates the transverse wave nature of light and cannot be explained if light is

considered as a longitudinal wave or as classical particles.

21. Polarized sunglasses completely block horizontally polarized glare at certain reflected angles and also

block unpolarized light by 50%. Regular tinted sunglasses reduce the intensity of all incoming light but

don’t preferentially reduce the “glare” from smooth reflective surfaces.

22. Take the sunglasses outside and look up at the sky through them. Rotate the sunglasses (about an axis

perpendicular to the lens) through at least 180°. If the sky seems to lighten and darken as you rotate the

sunglasses, then they are polarizing. You could also look at a liquid crystal display or reflections from

a tile floor (with a lot of “glare”) while rotating the glasses and again look for the light to be lighter or

darker depending on the rotation angle. Finally, you could put one pair of glasses on top of the other as

in Fig. 24–44 and rotate them relative to each other. If the intensity of light that you see through the

glasses changes as you rotate them, then the glasses are polarizing.


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24-4 Chapter 24



23. If Earth had no atmosphere, then the color of the sky would be black (and dotted with stars and

planets) at all times. This is the condition of the sky that the astronauts found on the Moon, which has

no atmosphere. If there were no air molecules to scatter the light from the Sun, then the only light we

would see would be from the stars, the planets, the Moon, and direct sunlight. The rest of the sky

would be black.

24. If the atmosphere were 50% more dense, then sunlight (after passing through the atmosphere) would

be much redder than it is now. As the atmosphere increased in density, more and more of the blue light

would be scattered away in all directions, making the light that reaches the ground very red. Think of

the color of a deep red sunset, but this might be the color even when the sun was at high elevations.

Responses to MisConceptual Questions

1. (a) The width of the fringes is proportional to the wavelength and inversely proportional to the slit

spacing. Therefore, since red light has the longer wavelength, red light with small slit spacing

will have the largest fringe width.

2. (a) At point A the green laser light is at a maximum. The next maximum (B) occurs when the pathdifference between the point and each of the two slits is equal to one full wavelength, or 530 nm.

The minimum (halfway between points A and B) is the point where the difference in distances is

equal to a half wavelength (or 265 nm).

3. (c) The index of refraction in the water droplets varies slightly with wavelength. Therefore, as

sunlight passes through the droplets, the light of different frequencies (colors) refracts at slightly

different angles.

4. (a) The cause of single-slit diffraction can be difficult to understand. The slit can be divided into a

number of individual regions. If the path difference between each pair of regions and a point on

the screen is a half wavelength, then destructive interference occurs at that point.

5. (b) It might be common to think that the lines are due to the inability of the eye to focus on the

point. However, the lines are due to diffraction of the light as it passes through the very small

opening between your fingers.

6. (b) The spreading out of the light is due to diffraction, with diffraction being significant when the

slit size is about the same size as the wavelength of light. For visible light, the wavelength is

about –7

5 10 m,× which is on the order of the slit width. Therefore, the light will spread out

wider than the slit. The height opening, which is 4 orders of magnitude larger than the

wavelength, will have almost no discernible diffraction, so the light will remain about the same

height.

7. (d ) In Eq. 24–2a the angles of constructive interference are proportional to the wavelength and

inversely proportional to the slit width. Therefore, if the slit width and wavelength are changed

by the same factor, then the diffraction pattern will not change.

8. (b) A common misconception is that the center of the shadow will be the darkest. However, since

the center of the disk is equidistant from all points on the edge of the disk, the diffracting light

constructively interferes at the center, making a bright point. This bright point is independent of

the wavelength of the light.


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9. (c) It can be difficult to recognize the different scales of length involved with the diffraction of

sound versus the diffraction of light. The sound waves of a person’s voice have wavelengths

about the same size as a doorway and therefore diffract easily around the corner. The wavelength

of light is much smaller, approximately 10 –7 m, so light does not diffract around the corner.

10. (c) The CD is made of thinly spaced grooves. As light reflects off of these grooves, the light

interferes in the same way as when it passes through a diffraction grating. Note that the

diffraction pattern is the same regardless of the data and/or security on the CD. The bit pattern

does not determine the diffraction—just the presence of the thinly spaced grooves.

11. (e) If the film has regions of lower index of refraction on both sides of the film (or higher on both

sides), then a phase shift will occur at one surface but not the other. In this case constructive

interference will occur when the thickness is ¼ of a wavelength and destructive interference will

occur when the thickness is ½ of a wavelength. If the film has a lower index on one side and a

higher index on the other, then either no phase shift will occur at either surface or phase shifts

will occur at both surfaces. In this case destructive interference occurs when the film has a

thickness of ¼ wavelength and constructive interference occurs when the thickness is ½ of a

wavelength. Since the indices of refraction are not specified for the surfaces of the film, none ofthe answers are always true.

12. (b) If two successive polarizers are perpendicular to each other, then light cannot get through. In case

1, the first two polarizers are perpendicular, and in case 2, the last two polarizers are perpendicular.

In case 2, no successive polarizers are perpendicular, so some light can pass through.

Solutions to Problems

1. For constructive interference, the path difference is a multiple of the wavelength, as given by Eq. 24–2a.

Apply this to the fifth order.

57sin (1.8 10 m)sin 8.6

sin 5.4 10 m5

d

d m m

θ

θ λ λ

−−× °

= → = = = ×


Apply this to the third order (m = 3).

963(610 10 m)sin 3.6 10 m

sin sin 31

md m d

λ θ λ

θ

−−×

= → = = = ×°


The location on the screen is given by tan , x θ = as seen in Fig. 24–7(c). For small angles, we have

sin tan / . xθ θ ≈ ≈ Adjacent fringes will have 1.m∆ =

11 2 2 1

57 7

814

7

10

sin

( 1) ( 1);

(4.8 10 m)(0.085 m)6.277 m 6.3 10 m

6.50 m

3.00 10 m/s4.8 10 Hz

6.277 10 m

x m

d m d m x d

m m m m x x x x x

d d d d d

d x

c f

λ

θ λ λ

λ λ λ λ λ

λ

λ

−− −

−

×

= → = → =

+ += = → ∆ = − = − =

∆ ×= = = ≈ ×

×= = = ×

×


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24-6 Chapter 24





sin tan / . xθ θ ≈ ≈ Second order means m = 2.

1 2

1 2

942 1

2 1 4

sin ; ;

( ) [(720 660) 10 ](2)(1.0 m)1.935 10 m 0.2 mm

(6.2 10 m)

m m x m

d m d m x x xd d d

m m x x x

d

λ λ λ

θ λ λ

λ λ − −−

= → = → = = = →

− − ×∆ = − = = = × ≈

×

This justifies using the small angle approximation, since . x

5. For destructive interference, the path difference is as follows.

( ) ( ) ( )

( )1 12 21 1

2 2

(4.5 cm)sin sin (0.60), 0,1, 2, 3, .

(7.5 cm)

m md m m m

d

λ θ λ θ

+ += + → = = = + = …

The angles for the first three regions of complete destructive interference are calculated.

( )

( )

( )

1 110 2

1 111 2

1 112 2

sin 0 (0.60) sin 0.30 17.46 17

sin 1 (0.60) sin 0.90 64.16 64

sin 2 (0.60) sin 1.50 impossible

θ

θ

θ

− −

− −

− −

⎡ ⎤= + = = ° ≈ °⎣ ⎦

⎡ ⎤= + = = ° ≈ °⎣ ⎦

⎡ ⎤= + = =⎣ ⎦

There are only two regions of destructive interference, at 17° and 64 .°

6. The slit spacing and the distance from the slits to the screen are the same in both cases. The distance

between bright fringes can be taken as the position of the first bright fringe (m = 1) relative to the central

fringe. We indicate the lab laser with subscript 1 and the laser pointer with subscript 2. For constructive

interference, the path difference is a multiple of the wavelength, as given by Eq. 24–2a. The location on the

screen is given by tan , x θ = as seen in Fig. 24–7(c). For small angles, we have sin tan / . xθ θ ≈ ≈

1 21 2

12 2 2

1

sin ; ;

5.14 mm(632.8 nm) 650.52 nm 651 nm

5.00 mm

x md m d m x x x

d d d

d x x

x

λ λ λ θ λ λ

λ λ

= → = → = = = →

= = = = ≈



sin tan / . xθ θ ≈ ≈ 9

4

3

(680 10 m)(3)(2.8 m)sin 1.5 10 m

38 10 m

x md m d m d

x

λ θ λ λ

−−

−

×= → = → = = = ×

×

8. Using a ruler on Fig. 24–9a, the width of 18 fringes is found to be about 25 mm. For constructiveinterference, the path difference is a multiple of the wavelength, as given by Eq. 24–2a. The location

on the screen is given by tan , x θ = as seen in Fig. 24–7(c). For small angles, we have

sin tan / . xθ θ ≈ ≈

47(1.7 10 m)(0.025 m)sin 6.4 10 m

(18)(0.37 m)

x dx dxd m d m

m mθ λ λ λ

−−×= → = → = = = = ×


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sin tan / . xθ θ ≈ ≈ For adjacent fringes, 1.m∆ =

9

5

sin

(633 10 m)(3.3 m)(1) 0.0307 m 3.1 cm

(6.8 10 m)

x md m d m x d

x md

λ θ λ λ

λ −

−

= → = → = →

×∆ = ∆ = = =

×



sin tan / . xθ θ ≈ ≈

96(633 10 m)(1)(5.0 m)sin 9.0 10 m

(0.35 m)

x md m d m d

x

λ θ λ λ

−−×= → = → = = = ×

11. The 180° phase shift produced by the glass is equivalent to a path length of 12

.λ For constructive

interference on the screen, the total path difference is a multiple of the wavelength:

( )1 1max max2 2sin , 0, 1, 2, sin , 1, 2,d m m d m mλ θ λ θ λ + = = → = − =

We could express the result as ( )1max 2sin , 0, 1, 2, .d m mθ λ = + =

For destructive interference on the screen, the total path difference is

( )1 1min min2 2sin , 0, 1, 2, sin , 0, 1, 2,d m m d m mλ θ λ θ λ + = + = → = =

Thus the pattern is just the reverse of the

usual double-slit pattern. There will be a darkcentral line instead of a bright central

maximum. Every place there was a bright

fringe will now have a dark line, and vice

versa.

Figure 24–10 is reproduced here, and

immediately to the right is the pattern that

would appear from the situation described by

this problem. Notice that maxima have

changed to minima, and vice versa.

12. We equate the expression from Eq. 24–2a for the second-order blue light to Eq. 24–2b, since the slit

separation and angle must be the same for the two conditions to be met at the same location.

( )

( )

1 b 2

12

sin (2)(480 nm) 960 nm; sin , 0, 1, 2,

960 nm 1920 nm; 1 640 nm

2 384 nm

d m d m m

m m m

m

θ λ θ λ

λ λ λ

λ

′ ′= = = = + =

′ ′ ′+ = = 0 → = = → =

′ = → =

The only one visible is 640 nm . 384 nm is near the low-wavelength limit for visible light.

Original ShiftedPattern Pattern


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24-8 Chapter 24





sin tan / . xθ θ ≈ ≈ For adjacent fringes, 1.m∆ =

93

3

sin

(544 10 m)(4.0 m)(1) 2.2 10 m

(1.0 10 m)

x m

d m d m x d

x md

λ

θ λ λ

λ − −−

= → = → = →

×∆ = ∆ = = ×

×

14. We have the same setup as in Example 24–3, so the two slits are 0.50 mm apart, and the screen is

2.5 m away. We use Eq. 24–21 with the small-angle approximation as described in that example, and

solve for the wavelength.

4 37(5.0 10 m)(2.9 10 m)

sin 5.8 10 m(1)(2.5 m)

x dxd d m

mθ λ λ

− −−× ×

= = → = = = ×

From Fig. 24–12, that wavelength is yellow.

15. An expression is derived for the slit separation from the data for the 480-nm light. That expression is

then used to find the location of the maxima for the 650-nm light. For constructive interference, the

path difference is a multiple of the wavelength, as given by Eq. 24–2a. The location on the screen is

given by tan , x θ = as seen in Fig. 24–7(c). For small angles, we have sin tan / . xθ θ ≈ ≈

1 1

1

2 2 2 22 1

1 1 1 1 1

sin

(650 nm)(2)(16 mm) 14.44 mm 14 mm

/ (480 nm)(3)

m x m md m d m d x

x x d

m m x x

m x m

λ λ λ θ λ λ

λ λ

λ λ

= → = → = = → = →

= = = = ≈

16. The presence of the water changes the wavelength according to Eq. 24–1, so we must change λ to

/ .n nλ λ = For constructive interference, the path difference is a multiple of the wavelength, as given by Eq. 24–2a. The location on the screen is given by tan , x θ = as seen in Fig. 24–7(c). For small

angles, we have sin tan / . xθ θ ≈ ≈ Adjacent fringes will have 1.m∆ =

11 2

9

2 1 5

3 310

( 1)sin ; ;

( 1) (470 10 m)(0.400 m)

(1.33)(6.00 10 m)

2.356 m 2.4 10 m

nn n

n n n

m m x md m d m x x x

d d d

m m x x x

d d d nd

λ λ λ θ λ λ

λ λ λ λ −

−

− −×

+= → = → = = = →

+ ×∆ = − = − = = =

×

= ≈ ×

17. To change the center point from constructive to destructive interference, the phase shift produced by

inserting the plastic must be equivalent to half a wavelength. The wavelength of the light is shorter inthe plastic than in the air, so the number of wavelengths in the plastic must be ½ greater than the

number in the same thickness of air. The number of wavelengths in the distance equal to the thickness

of the plate is the thickness of the plate divided by the appropriate wavelength.

plastic 1 plastic air plastic 2

plastic

plastic

( 1)

680 nm570 nm

2( 1) 2(1.60 1)

tnt t t t N N n

t n

λ λ λ λ λ

λ

− = − = − = − = →

= = =− −


9/33




18. We find the speed of light from the index of refraction, / .c nυ = We calculate the % difference using

the red as the standard. Note that answers will vary due to differences in reading the graph.

red bluered blue blue red

red blue

red

( ) ( ) (1.615 1.640) 0.015 1.5% less(1.640)

c c

n n n nnc

n

υ υ υ

⎡ ⎤⎛ ⎞ ⎛ ⎞−⎢ ⎥⎜ ⎟ ⎜ ⎟

− − −⎝ ⎠ ⎝ ⎠⎣ ⎦= = = = − =⎛ ⎞⎜ ⎟⎝ ⎠

19. We find the angles of refraction in the glass from Snell’s law, 1 1 2 2sin sin .n nθ θ =

2,450 2,450

2,700 2,700

2,700 2,450

(1.00)sin 65.00 (1.4831)sin 37.67

(1.00)sin 65.00 (1.4754)sin 37.90

37.90 37.67 0.23

θ θ

θ θ

θ θ

° = → = °

° = → = °

− = ° − ° = °

20. We use Snell’s law for the refraction at the first surface,entering the prism. The lower wavelength has the higher

index of refraction. The indices of refraction are found

from Fig. 24–14.

air

,455 ,455

,642 ,642

sin sin ;

(1.00)sin 45 (1.64)sin 25.54

(1.00)sin 45 (1.62)sin 25.88

a b

b b

b b

n nθ θ

θ θ

θ θ

=

° = → = °

° = → = °

We find the angle of incidence at the second surface, leaving the

prism, from the second diagram.

,455 ,455

,642 ,642

(90 ) (90 ) 180

60.00 25.54 34.46

60.00 25.88 34.12

b c

c b

c b

A

A

A

θ θ

θ θ

θ θ

° − + ° − + = °

= − = ° − ° = °

= − = ° − ° = °

Use Snell’s law again for the refraction at the second surface.

air

,455 ,455 1

,642 ,642 2

sin sin

(1.64)sin 34.46 (1.00)sin 68.1

(1.62)sin 34.12 (1.00)sin 65.3

c d

d d

d d

n nθ θ

θ θ θ

θ θ θ

=

° = → = = °

° = → = = °

The values may vary slightly from differences in reading the graph.

21. We use Eq. 24–3a to calculate the angular distance from the middle of the central peak to the firstminimum. The width of the central peak is twice this angular distance.

91 1

1 1 3

1

680 10 msin sin sin 0.9168

0.0425 10 m

2 2(0.9168) 1.834 1.8

D D

λ λ θ θ

θ θ

−− −

−

⎛ ⎞×⎛ ⎞= → = = = °⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ×⎝ ⎠

∆ = = = ° ≈ °


10/33


11/33




(b) We repeat the process from part (a) using a wavelength of 1.0 cm.

1 11 2

1 1.0 cm 2 1.0 cmsin 38.7 sin no real solution

1.6 cm 1.6 cmθ θ − −

⎛ ⎞ ⎛ ⎞× ×= = ° = =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

The only diffraction minimum is at 39°.

(c) We repeat the process from part (a) using a wavelength of 3.0 cm.

11

1 3.0 cmsin no real solution

1.6 cmθ −

⎛ ⎞×= =⎜ ⎟

⎝ ⎠

There are no diffraction minima.

27. (a) There will be no diffraction minima if the angle for the first minimum is greater than 90°. We

set the angle in Eq. 24–3a equal to 90° and solve for the slit width.

sin

sin90

D

D

λ λ θ λ = → = =

°

(b) For no visible light to exhibit a diffraction minimum, the slit width must be equal to the shortest

visible wavelength.

min 400 nm D λ = =

28. The first bright region occurs at3

sin .2 D

λ θ ≈ We then find the distance on the screen from the central

maximum by multiplying the distance to the screen by the tangent of the angle.

91 1

max6

1

3 3(620 10 m)sin sin 14.17

2 2(3.80 10 m)

tan (10.0 m) tan (14.17 ) 2.52 m

D

x

λ θ

θ

−− −

−

⎡ ⎤×⎛ ⎞= = = °⎢ ⎥⎜ ⎟

⎝ ⎠ ×⎢ ⎥⎣ ⎦= = ° =

29. The angle from the central maximum to the first bright maximum is half the angle between the first

bright maxima on either side of the central maximum. The angle to the first maximum is about halfway

between the angles to the first and second minima. We use Eq. 24–3b, setting 3/2,m = to calculate the

slit width, D.

1 11 2 2

m1

(32 ) 16

(3/2)(633 nm)sin 3445 nm 3.4 m

sin sin16

m D m D

θ θ

λ θ λ µ

θ

= ∆ = ° = °

= → = = = ≈°

30. (a) For vertical diffraction we use the height of the slit (1.5 µ m) as the slit width in Eq. 24–3a tocalculate the angle between the central maximum to the first minimum. The angular separation

of the first minima is equal to twice this angle.

91 1

1 1 6

1

780 10 msin sin sin 31.3

1 5 10 m

2 2(31.3 ) 63

D D

λ λ θ θ

θ θ

−− − ×

= → = = = °. ×

∆ = = ° ≈ °

2


12/33

24-12 Chapter 24



(b) To find the horizontal diffraction we use the width of the slit (3.0 µ m) in Eq. 24–3a.

91 1

1 1 6

1

780 10 msin sin sin 15.07

3.0 10 m

2 2(15.07 ) 30

D D

λ λ θ θ

θ θ

−− −

−

×= → = = = °

×

∆ = = ° ≈ °

Note that the 30° has 2 significant figures.

31. The path difference between the top and bottom of the slit for the

incident wave is isin . D θ The path difference between the top

and bottom of the slit for the diffracted wave is sin . D θ

Therefore, the net path difference is isin sin . D Dθ θ − When

i ,θ θ = the net path difference is 0, and there will be

constructive interference. Thus there will be a central maximum

at 23.0 .θ = ° When the net path difference is equal to a multiple

of the wavelength, there will be an even number of segments of

the wave having a path difference of λ /2. We set the pathdifference equal to m (an integer) times the wavelength and solve

for the angle of the diffraction minimum.

i isin sin sin sin , 1, 2m

D D m m D

λ θ θ λ θ θ − = → = − = ± ± , …

From this equation we see that when i 28.0 ,θ = ° the minima will be symmetrically distributed around

a central maximum at 28.0°.

32. We use Eq. 24–4 to calculate the angle for the second-order maximum.

91 1

52(510 10 m)sin sin sin 4.31.35 10 m

md md λ θ λ θ −− −

−⎛ ⎞×⎛ ⎞= → = = = °⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ×⎝ ⎠

33. The slit separation, in cm, is the reciprocal of the number of slits per cm. We use Eq. 24–4 to find the

wavelength.

5

1cm sin22.0

sin 3800sin 3.29 10 cm 330 nm

3

d d m

m

θ θ λ λ −

⎛ ⎞°⎜ ⎟

⎝ ⎠= → = = = × ≈

34. Because the angle increases with wavelength, to have a complete order we use the longest wavelength.

We set the maximum angle to 90° to determine the largest integer m in Eq. 24–4. The slit separation, in

cm, is the reciprocal of the number of lines per cm.

9

1 1 mcm sin 90

7400 100 cmsinsin 1.93

(700 10 m)

d d m m

N

θ θ λ

λ −

⎛ ⎞⎛ ⎞°⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠= → = = =

×

Thus only one full order (all the way out to red) can be seen on each side of the central white line,

although the second order is almost fully visible.

θ i

θ

θ i

Slit, width a D


13/33




We can also evaluate for the shortest wavelength.

9

1 1 mcm sin 90

7400 100 cmsinsin 3.38

(400 10 m)

d d m m

N

θ θ λ

λ −

⎛ ⎞⎛ ⎞°⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠= → = = =

×

Thus part of the third order will be visible—at least the violet part.

35. Since the same diffraction grating is being used for both wavelengths of light, the slit separation will

be the same. Solve Eq. 24–4 for the slit separation for both wavelengths and set the two equations

equal. The resulting equation is then solved for the unknown wavelength.

1 1 2 2 1 22 1

1 2 2 1

sin 2 sin 20.6sin (632.8 nm) 556 nm

sin sin sin 1 sin 53.2

m m md m d

m

λ λ θ θ λ λ λ

θ θ θ

°= ⇒ = = ⇒ = = =

°

36. The number of lines per cm is the reciprocal of the slit width in cm. Use Eq. 24–4.

9

1 sin sin15.0sin 1392 lines/cm 1400 lines/cm

100 cm(3)(620 10 m)

m

d md m

θ θ λ

λ −

°= → = = = ≈

⎛ ⎞× ⎜ ⎟

⎝ ⎠

37. We use Eq. 24–4 to calculate the wavelengths from the given angles. The slit separation, d , is the

inverse of the number of slits per cm.

sinsin

d d m

m

θ θ λ λ = → =

51

52

1cm sin 28.8 4.92 10 cm 492 nm 490 nm

9800

1cm sin 36.7 6.10 10 cm 610 nm

9800

λ

λ

−

−

⎛ ⎞= ° = × = ≈⎜ ⎟

⎝ ⎠

⎛ ⎞= ° = × ≈⎜ ⎟⎝ ⎠

53

54

1cm sin 38.6 6.37 10 cm 640 nm

9800

1cm sin 41.2 6.72 10 cm 670 nm

9800

λ

λ

−

−

⎛ ⎞= ° = × ≈⎜ ⎟

⎝ ⎠

⎛ ⎞= ° = × ≈⎜ ⎟

⎝ ⎠

38. We find the first-order angles for the maximum and minimum wavelengths using Eq. 24–4, where the

number of lines per centimeter is the reciprocal of the slit separation distance, in cm. Then we set the

distance from the central maximum of the maximum and minimum wavelength equal to the distance to

the screen multiplied by the tangent of the first-order angle. The width of the spectrum is the difference

in these distances.

1

1 71

1 72

2 1 2 1

sin sin

sin [(410 10 cm)(7800 lines/cm)] 18.65

sin [(750 10 cm)(7800 lines/cm)] 35.80

( tan tan ) (3.40 m)( tan 35.80 tan18.65 ) 1.3 m

md m

d

x x x

λ θ λ θ

θ

θ

θ θ

−

− −

− −

⎛ ⎞= → = ⎜ ⎟

⎝ ⎠

= × = °

= × = °

∆ = − = − = ° − ° =


14/33

24-14 Chapter 24



39. We find the second-order angles for the maximum and minimum wavelengths using Eq. 24–4, where

the slit separation distance is the inverse of the number of lines per cm. Subtracting these two angles

gives the angular width.

1

1 7 51

1 7 52

2 1

sin sin

sin [2(4.5 10 m)(6 5 10 /m)] 35.80

sin [2(7.0 10 m)(6.5 10 /m)] 65.50

65.50 35.80 29.7 30

md m

d

λ θ λ θ

θ

θ

θ θ θ

−

− −

− −

⎛ ⎞= → = ⎜ ⎟

⎝ ⎠= × . × = °

= × × = °

∆ = − = ° − ° = ° ≈ °

Note that the answer has 2 significant figures.

40. We set the diffraction angles as one-half the difference between the angles on opposite sides of the center.

Then we solve Eq. 24–4 for the wavelength, with d equal to the inverse of the number of lines per cm.

1

51

2 22

52

26 38 ( 26 18 )26 28 26 28/60 26.47

2 2

1sin cm sin 26.47 4.618 10 cm 462 nm9650

41 02 ( 40 27 )40 44.5 40 44.5/60 40.74

2 2

1cm sin 40.74 6.763 10 cm 676 nm

9650

r

r

d

θ θ θ

λ θ

θ θ θ

λ

−

−

′ ′− ° − − °′= = = ° = + = °

⎛ ⎞= = ° = × =⎜ ⎟⎝ ⎠

′ ′− ° − − °′= = = ° = + = °

⎛ ⎞= ° = × =⎜ ⎟

⎝ ⎠

41. The maximum angle is 90°. The slit separation is the reciprocal of the line spacing.

7

1cm sin90

sin 6500sin 2.43

633 10 cm

d d m m

θ θ λ

λ −

⎛ ⎞°⎜ ⎟

⎝ ⎠= → = = =

×

Thus the second order is the highest order that can be seen.

42. We solve Eq. 24–4 for the slit separation width, d, using the given information. Then, setting m = 3, we

solve for the angle of the third-order maximum.

1 13

1(589 nm)sin 2352 nm 2.35 m

sin sin14.5

3 589 nmsin sin 48.7

2352 nm

m md

d

m

d

λ λ θ µ

θ

λ θ − −

= → = = = =°

⎛ ⎞×⎛ ⎞= = = °⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

43. Because the angle of the diffracted light increases with wavelength, to have a full order use the largest

wavelength, 700 nm. See Fig. 24–26 for an illustration. The maximum angle of diffraction is 90°. Use

that angle with the largest wavelength to find the minimum slit separation. The reciprocal of the slit

separation gives the number of slits per cm.

76 4

4

2(7.00 10 m)sin 1.40 10 m 1.40 10 cm

sin sin 90

1 17140 slits/cm

1.40 10 cm

md m d

d

λ θ λ

θ

−− −

−

×= → = = = × = ×

°

= =×

We have assumed that the original wavelength is known to 3 significant figures.


15/33




44. From Example 24–11, we see that the thickness is related to the

interference maximum wavelength by /4 .t nλ =

/4 4 4(1.32)(120 nm) 634 nm 630 nmt n nt λ λ = → = = = ≈

45. Between the 25 dark lines there are 24 intervals. When we add the half-interval at the wire end, we

have 24.5 intervals over the length of the plates.

21.5 cm0.878 cm

24.5 intervals=

46. (a) An incident wave that reflects from the outer surface of

the bubble has a phase change of 1 .φ π = An incident

wave that reflects from the inner surface of the bubble

has a phase change due to the additional path length, so

2film

22 .

t φ π

λ

⎛ ⎞= ⎜ ⎟

⎝ ⎠ For destructive interference with a

minimum nonzero thickness of bubble, the net phase change must be .π

1net 2 1 film2

film

2 480 nm2 180 nm

2 2(1.33)

t t

n

λ φ φ φ π π π λ

λ

⎡ ⎤⎛ ⎞= − = − = → = = = =⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

(b) For the next two larger thicknesses, the net phase change would be 3π and 5 .π

net 2 1 filmfilm

3net 2 1 film 2

film

2 480 nm2 3 360 nm(1.33)

2 480 nm2 5 540 nm

(1.33)

t t n

t t

n

λ φ φ φ π π π λ λ


λ

⎡ ⎤⎛ ⎞= − = − = → = = = =⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞= − = − = → = = = =⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

(c) If the thickness were much less than one wavelength, then there would be very little phase

change introduced by additional path length, so the two reflected waves would have a phase

difference of about 1 .φ π = This would produce destructive interference.

47. An incident wave that reflects from the top surface of the

coating has a phase change of 1 .φ π = An incident wave that

reflects from the glass ( 1.52)n = at the bottom surface of

the coating has a phase change due to both the additional

path length and a phase change of π on reflection, so

2film

22 .

t φ π π

λ

⎛ ⎞= +⎜ ⎟

⎝ ⎠ For constructive interference with a

minimum nonzero thickness of coating, the net phase change must be 2 .π

1 1net 2 1 film2 2

film film

22 2

t t

n

λ φ φ φ π π π π λ

λ

⎡ ⎤⎛ ⎞ ⎛ ⎞= − = + − = → = =⎢ ⎥⎜ ⎟ ⎜ ⎟

⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

1φ π =

( )2 film2 2 0t φ λ π = +

1.32n =

1φ π =

( )2 film2 2t φ λ π π = +

1.52n =

1.25n =


16/33

24-16 Chapter 24



The lens reflects the most for 570 nmλ = . The minimum nonzero thickness occurs for 1:m =

minfilm

(570 nm)230 nm

2 2(1.25)t

n

λ = = =

Since the middle of the spectrum is being selectively reflected, the transmitted light will be stronger inthe red and blue portions of the visible spectrum.

48. (a) When illuminated from above at point A, a light ray reflected from the air–oil interface

undergoes a phase shift of 1 .φ π = A ray reflected at the oil–water interface undergoes no phase

shift. If the oil thickness at point A is negligible compared to the wavelength of the light, then

there is no significant shift in phase due to a path length traveled by a ray in the oil. Thus the

light reflected from the two surfaces will destructively interfere for all visible wavelengths, and

the oil will appear black when viewed from above.

(b) From the discussion in part (a), the ray reflected from the air–oil interface undergoes a phase

shift of 1 .φ π = A ray that reflects from the oil–water interface has no phase change due to

reflection, but has a phase change due to the additional path length of 2oil

22 .

t

φ π λ

⎛ ⎞= ⎜ ⎟⎝ ⎠

For

constructive interference, the net phase change must be a multiple of 2 .π

( ) ( )1 1 1 1net 2 1 oil2 2 2 2oil o

22 (2 )

t m t m m

n


λ

⎡ ⎤⎛ ⎞= − = − = → = + = +⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

From the diagram, we see that point B is the second thickness that yields constructive

interference for 580 nm, so we use m = 1. (The first location that yields constructive interference

would be for m = 0.)

( ) ( )1 1 1 12 2 2 2o

580 nm1 290 nm

1.50t m

n

λ = + = + =

49. According to Section 24–8, at each surface approximately 4% of the light is reflected. So passing each

surface of a lens results in only 96% transmission either into the lens (at the front surface) or on to the next

piece of optics (at the second surface). So the light exiting a lens would experience two of these reductions

and so would pass only 96% of 96% of the light incident on the lens. This is a factor of2

(0.96) for each

lens. Reducing to 50% would involve that factor for each lens. Let n be the number of lenses.

2 2 ln (0.50)0.50 [(0.96) ] (0.96) ln (0.50) 2 ln (0.96) 8.492ln(0.96)

n n n n= = → = → = =

It would take nine lenses for the light to be reduced to 50% or less.

50. An incident wave that reflects from the convex surface of

the lens has no phase change, so 1 0.φ = An incident wavethat reflects from the glass underneath the lens has a phase

change due to both the additional path length and a phase

change of π on reflection, so 22

2 .t

φ π π λ

⎛ ⎞= +⎜ ⎟

⎝ ⎠ For

destructive interference (dark rings), the net phase change

must be an odd-integer multiple of ,π so

net 2 1 (2 1) , 0, 1, 2, .m mφ φ φ π = − = + =

10φ =

( )2 2 2t φ λ π π = +


17/33




Because 0m = corresponds to the dark center, m represents the number of the ring.

net 2 1

1 1air 2 2

22 0 (2 1) , 0, 1, 2,

(35)(560 nm) 9800 nm 9.8 m

t m m

t m

φ φ φ π π π λ

λ µ

⎡ ⎤⎛ ⎞= − = + − = + = →⎜ ⎟⎢ ⎥

⎝ ⎠⎣ ⎦

= = = =

The thickness of the lens is the thickness of the air at the edge of the lens.

51. Since the wedge is now filled with water, we just use the wavelength in the water, as instructed in the

problem statement. Since the index of refraction of the water is less than that of the glass, there are no

new phase shifts introduced by reflections, so we can use the relationships developed in that example.

We “count” the number of wavelengths at the position of the wire.

6

7

2 2(7.35 10 m)32.6 wavelengths

6.00 10 m/1.33n

t

λ

−

−

×= =

×

Thus there will be 33 dark bands across the plates, including the one at the point of contact.

52. An incident wave that reflects from the second

surface of the upper piece of glass has no phase

change, so 1 0.φ = An incident wave that reflects

from the first surface of the second piece of glass

has a phase change due to both the additional path

length and a phase change of π on reflection, so

2

22 .

t φ π π

λ

⎛ ⎞= +⎜ ⎟

⎝ ⎠ For destructive interference

(dark lines), the net phase change must be an odd-integer multiple of ,π so

net 2 1

(2 1) , 0, 1, 2, .m mφ φ φ π = − = + = Because 0m = corresponds to the left edge of the

diagram, the 24th dark line corresponds to m = 23. The 24th dark line also has a gap thickness of d .

1net 2 1 2

12

22 0 (2 1)

(23)(670 nm) 7705 nm 7.7 m

t m t m

d

φ φ φ π π π λ λ

µ

⎡ ⎤⎛ ⎞= − = + − = + → = →⎜ ⎟⎢ ⎥

⎝ ⎠⎣ ⎦

= = ≈

53. With respect to the incident wave, the wave that reflects from

the air at the top surface of the air layer has a phase change of

1 0.φ = With respect to the incident wave, the wave that

reflects from the glass at the bottom surface of the air layer

has a phase change due to both the additional path length andreflection, so 2

22 .

t φ π π

λ

⎛ ⎞= +⎜ ⎟

⎝ ⎠ For constructive interference,

the net phase change must be an even nonzero integer multiple of .π

( )1 1net 2 1 2 22

2 0 2 , 1, 2, .t

m t m mφ φ φ π π π λ λ

⎡ ⎤⎛ ⎞= − = + − = → = − =⎜ ⎟⎢ ⎥

⎝ ⎠⎣ ⎦

glass

air

glass

10φ =

( )2 2 2t φ λ π π = +


18/33

24-18 Chapter 24



The minimum thickness is with 1.m =

( )1 1min 2 2(450 nm) 1 113 nm 110 nmt = − = ≈

For destructive interference, the net phase change must be an odd-integer multiple of .π

1net 2 1 2

22 0 (2 1) , 0, 1, 2,

t m t m mφ φ φ π π π λ

λ

⎡ ⎤⎛ ⎞= − = + − = + → = =⎜ ⎟⎢ ⎥

⎝ ⎠⎣ ⎦

The minimum nonzero thickness is 1min 2 (450 nm)(1) 225 nm 230 nmt = = ≈

54. When illuminated from above, the light ray reflected from the air–oil interface undergoes a phase shift

of 1 .φ π = A ray reflected at the oil–water interface undergoes no phase shift due to reflection, but has

a phase change due to the additional path length of 2oil

22 .

t φ π

λ

⎛ ⎞= ⎜ ⎟

⎝ ⎠ For constructive interference to

occur, the net phase change must be a multiple of 2π .

( ) ( )1 1 1 1net 2 1 oil2 2 2 2oil o

22 (2 )

t m t m m

n


λ

⎡ ⎤⎛ ⎞= − = − = → = + = +⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

For 650 nm,λ = the possible thicknesses are as follows:

( )1 1650 2 2650 nm

108 nm, 325 nm, 542 nm,1.50

t m= + =

For 390 nm,λ = the possible thicknesses are as follows:

( )1 1390 2 2390 nm

65 nm,195 nm, 325 nm, 455 nm,1.50

t m= + =

The minimum thickness of the oil slick must be 325 nm .

55. With respect to the incident wave, the wave that reflects

from the top surface of the alcohol has a phase change of

1 .φ π = With respect to the incident wave, the wave that

reflects from the glass at the bottom surface of the

alcohol has a phase change due to both the additional

path length and a phase change of π on reflection, so

2film

22 .

t φ π π λ

⎛ ⎞= +⎜ ⎟⎝ ⎠

For constructive interference, the

net phase change must be an even nonzero integer multiple of .π

11 1net 2 1 1 1film 1 1 12 2

1film film

22 2 , 1, 2, 3, .

t m t m m m

n


λ

⎡ ⎤⎛ ⎞⎢ ⎥= − = + − = → = = =⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦


1φ π =

( )2 film2 2t φ λ π π = +


19/33




21net 2 1 2 2 24

2film film

22 (2 1) (2 1), 0, 1, 2, .

t m t m m

n

λ φ φ φ π π π π

λ

⎡ ⎤⎛ ⎞= − = + − = + → = + =⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

Set the two expressions for the thickness equal to each other.

1 2 2 11 11 22 4

film film 1 2

2 1 (655 nm) 5(2 1) 1.2476 1.25

2 (525 nm) 4

mm m

n n m

λ λ λ

λ

+= + → = = = ≈ =

Thus we see that 1 2 2,m m= = and the thickness of the film is

11 112 2

film

655 nm(2) 481.6 nm

1.36t m

n

λ ⎛ ⎞= = =⎜ ⎟

⎝ ⎠ or 21 124 4

film

525 nm(2 1) (5) 482.5 nm

1.36t m

n

λ ⎛ ⎞= + = =⎜ ⎟

⎝ ⎠

The average thickness, with 3 significant figures, is 482 nm.

56. From the discussion in Section 24–9, we see that the path length change is twice the distance that the

mirror moves. One fringe shift corresponds to a change in path length of ,λ so corresponds to a mirror

motion of

1

2 .λ Let N be the number of fringe shifts produced by a mirror movement of . x∆

9 41 12 21

2

(680)(589 10 m) 2.00 10 m x

N x N λ λ

− −∆= → ∆ = = × = ×


mirror moves. One fringe shift corresponds to a change in path length of ,λ so it corresponds to a

mirror motion of 12

.λ Let N be the number of fringe shifts produced by a mirror movement of . x∆

47

12

2 2(1.25 10 m)6.91 10 m 691 nm

362

x x N

N λ

λ

−−∆ ∆ ×= → = = = × =


mirror moves. One fringe shift corresponds to a change in path length of ,λ so it corresponds to a

mirror motion of 12

.λ Let N be the number of fringe shifts produced by a mirror movement of . x∆ The

thickness of the foil is the distance that the mirror moves during the 296 fringe shifts.

9 51 12 21

2

(296)(589 10 m) 8.72 10 m x

N x N λ λ

− −∆= → ∆ = = × = ×

59. One fringe shift corresponds to an effective change in path length of .λ The actual distance has not

changed, but the number of wavelengths in the depth of the cavity has. If the cavity has a length d , then

the number of wavelengths in vacuum is ,d

λ

and the (greater) number with the gas present is

gas

gas

.n d d

λ λ = Because the light passes through the cavity twice, the number of fringe shifts is twice

the difference in the number of wavelengths in the two media.

9gas

gas gas 2

(158)(632.8 10 m)2 2 ( 1) 1 1 1.004328

2 2(1.155 10 m)

n d d d N N n n

d

λ

λ λ λ

−

−

⎛ ⎞ ×= − = − → = + = + =⎜ ⎟

×⎝ ⎠

The answer is quoted to more significant figures than is justified.


20/33

24-20 Chapter 24



60. Use Eq. 24–5. Since the initial light is unpolarized, the intensity after the first polarizer will be half the

initial intensity. Let the initial intensity be 0. I

22 2 21 1

1 0 2 1 02 2

0

cos 72; cos cos 0.048 ,

2

I I I I I I

I

θ θ °

= = = → = = or 4.8%

61. We assume that the light is coming from air to glass and use Eq. 24–6b.

1 p glass ptan 1.56 tan 1.56 57.3nθ θ

−= = → = = °

62. Let the initial intensity of the unpolarized light be 0 . I The intensity after passing through the first

Polaroid will be 11 02 . I I = Then use Eq. 24–5.

2 2 1 212 1 02

0

2cos cos cos

I I I I

I θ θ θ −= = → =

(a) 1 12

0

2 2cos cos 35.3 35

3

I

I θ − −= = = ° ≈ °

(b)1 12

0

2 2cos cos 63.4 63

10

I

I θ − −= = = ° ≈ °

63. For the first transmission, the angle between the light and the polarizer is 21 0. °. For the second

transmission, the angle between the light and the polarizer is 42 0. °. Use Eq. 24–5 twice.

2 2 2 21 0 2 1 0 0cos 21.0 ; cos 42.0 ( cos 21.0 )(cos 42.0 ) 0.4813 I I I I I I = ° = ° = ° ° =

Thus the transmitted intensity is 48.1% of the incoming intensity.

64. For the first transmission, since the incoming light is unpolarized, the transmission is the same as if the

angle were 45 .° So 11 02 . I I = The second polarizer is at an angle of 30° relative to the first one, so

22 1 cos 30 . I I = ° And the third polarizer is at an angle of 60° relative to the second one, so

23 2 cos 60 . I I = ° Combine these to find the percent that is transmitted out through the third polarizer.

( )( )

2 2 2 2 213 2 1 02

2 23 31 1 12 2 4 4

1

cos 60 cos 30 cos 60 cos 30 cos 60

3cos 30 cos 60 0.09375 9.4%

32

I I I I

I

I

= ° = ° ° = ° ° →

= ° ° = = = ≈

65. The polarizing angle pθ is found using Eq. 24–6a,2

p1

tan .

n

nθ = For an oil–diamond interface,

p

2.42tan ,

1.43θ = which gives p 59.4 .θ = ° The material does NOT appear to be diamond.

66. If 0 I is the intensity passed by the first Polaroid, then the intensity passed by the second will be 0 I

when the two axes are parallel. To calculate a reduction to half intensity, we use Eq. 24–5.

2 21 10 02 2

cos cos 45 I I I θ θ θ = = → = → = °


21/33




67. The light is traveling from water to diamond. We use Eq. 24–6a.

1diamond p p

water

2.42tan 1.82 tan 1.82 61.2

1.33

n

nθ θ −= = = → = = °

68. The critical angle exists when light passes from a material with a higher index of refraction 1( )n into a

material with a lower index of refraction 2( ).n Use Eq. 23–6.

2C

1

sin sin 58n

nθ = = °

To find Brewster’s angle, use Eq. 24–6a. If light is passing from high index to low index, then we have

the following:

12 p p

1

tan sin 58 tan (sin 58 ) 40.3 40n

nθ θ −= = ° → = ° = ° ≈ °

If light is passing from low index to high index, then we have the following:

11 p p

2

1 1tan tan 49.7 50

sin 58 sin 58

n

nθ θ −

⎛ ⎞= = → = = ° ≈ °⎜ ⎟

° °⎝ ⎠

Note that both answers are given to 2 significant figures.

69. First case: the light is coming from water to air. Use Eq. 24–6a.

1 1air air p p

water water

1.00tan tan tan 36.9

1.33

n n

n nθ θ − −= → = = = °

Second case: for total internal reflection, the light must also be coming from water into air. UseEq. 23–6.

1 1air air C p

water water

1.00sin sin sin 48.8

1.33

n n

n nθ θ − −= → = = = °

Third case: the light is coming from air to water. Use Eq. 24–6b.

1 1 p water p water tan tan tan 1.33 53.1n nθ θ

− −= → = = = °

Note that the two Brewster’s angles add to give 90 0. °.

70. When plane-polarized light passes through a sheet oriented at an angle ,θ the intensity decreases

according to Eq. 24–5, 20 cos . I I θ = For the first sheet, with unpolarized light incident, we can treat

45 ,θ = ° 2 12

cos .θ = Then sheets two through six will each reduce the intensity by a factor of

2cos 35.0 .° Thus we have the following:

2 2 50 0( cos 45.0 )( cos 35.0) 0.068 I I I = ° =

The transmitted intensity is 6.8% of the incident intensity.


22/33

24-22 Chapter 24



71. Let 0 I be the initial intensity. Use Eq. 24–5 for both transmissions of the light.

2 2 2 21 0 1 2 1 2 0 1 2 0

1 11

2

cos ; cos cos cos 0.35

0.35 0.35cos cos 28

cos cos 48

I I I I I I θ θ θ θ

θ

θ

− −

= = = = →

⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟°⎝ ⎠ ⎝ ⎠

72. (a) We apply Eq. 24–5 through the successive polarizers. The initial light is unpolarized. Each

polarizer is then rotated 30 0. ° from the previous one.

2 2 2 2 21 1 11 0 2 1 2 0 2 3 2 3 0 2 32 2 2

2 2 2 2 2 2 21 14 3 4 0 2 3 4 0 02 2

; cos cos ; cos cos cos ;

cos cos cos cos cos 30.0 cos 30.0 cos 30.0 0.211

I I I I I I I I

I I I I I

θ θ θ θ θ

θ θ θ θ

= = = = =

= = = ° ° ° =

(b) If the second polarizer is removed, then the angle between polarizers 1 and 3 is now 60.0 .°

2 21 11 0 3 1 3 0 32 2

2 2 2 2 21 14 3 4 0 3 4 0 02 2

; cos cos ;

cos cos cos cos 60.0 cos 30.0 0.0938

I I I I I

I I I I I

θ θ

θ θ θ

= = =

= = = ° ° =

The same value would result by removing the third polarizer, because then the angle between

polarizers 2 and 4 would be 60 .° The intensity can be decreased by removing either the second

or third polarizer.

(c) If both the second and third polarizers are removed, then there are still two polarizers with their

axes perpendicular, so no light will be transmitted.

73. (a) For constructive interference, the path difference is a multiple of the wavelength, as given by

Eq. 24–2a. The location on the screen is given by tan , x θ = as seen in Fig. 24–7(c). For small

angles, we have sin tan / . xθ θ ≈ ≈ For adjacent fringes, 1.m∆ =

74 4

2

sin

(5.0 10 m)(5.0 m)(1)1.25 10 m 1.3 10 m

(2.0 10 m)

x md m d m x x m

d d

md

x

λ λ θ λ λ

λ − − −−

= → = → = → ∆ = ∆ →

∆ ×= = = × ≈ ×

∆ ×

(b) For minima, we use Eq. 24–2b. The fourth-order minimum corresponds to m = 3, and the fifth-

order minimum corresponds to m = 4. The slit separation, screen distance, and location on the

screen are the same for the two wavelengths.

( ) ( ) ( ) ( )

( )

( )

1 1 1 1A A B B2 2 2 2

1A 2 7 7

B A 1B 2

sin

3.5(5.0 10 m) 3.9 10 m

4.5

xd m d m m m

m

m

θ λ λ λ λ

λ λ − −

= + → = + → + = + →

+= = × = ×

+

74. The wavelength of the signal is8

6

(3.00 10 m/s)4.00 m.

(75 10 Hz) f

υ λ

×= = =

×

(a) There is a phase difference between the direct and reflected signals

from both the path difference, 2 ,h

π λ

⎛ ⎞⎜ ⎟⎝ ⎠

and the reflection, π .

The total phase difference is the sum of the two.


23/33




(122 m)2 2 62 31(2 )

(4.00 m)

hφ π π π π π π

λ

⎛ ⎞= + = + = =⎜ ⎟

⎝ ⎠

Since the phase difference is an integer multiple of 2 ,π the interference is constructive.

(b) When the plane is 22 m closer to the receiver, the phase difference is as follows.

( ) (122 m 22 m) 512 2 51 (2 )

(4.00 m) 2

h yφ π π π π π π

λ

⎡ ⎤− −⎡ ⎤= + = + = =⎢ ⎥⎢ ⎥

⎣ ⎦ ⎣ ⎦

Since the phase difference is an odd half-integer multiple of 2 ,π the interference is destructive.

75. We find the angles for the first order from Eq. 24–2a, sin .d mθ λ = We are considering first order, so

m = 1, and the slit separation (in meters) is the reciprocal of the lines per meter value.

7 1HH H H5

7 1 Ne Ne Ne Ne5

7 1Ar Ar Ar Ar 5

sinsin 6.56 10 m sin 0.2362 13.7

3.60 10 lines/m

sinsin 6.50 10 m sin 0.2340 13.53.60 10 lines /m

sinsin 6.97 10 m sin 0.2509 14.5

3.60 10 lines /m

d

d

d

θ λ θ θ

θ λ θ θ

θ λ θ θ

− −

− −

− −

= = = × → = = °×

= = = × → = = °×

= = = × → = = °×


from the top surface of the material has a phase change of

1 .φ π = If we assume that the film has an index less than that

of the glass, then the wave that reflects from the glass has a

phase change due to the additional path length and a phase

change on reflection, for a total phase change of2 film

2 22 2 .

/

t t

nφ π π π π

λ λ

⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟

⎜ ⎟⎝ ⎠⎝ ⎠


( )1 1net 2 1 2 22

2 (2 1) , 1, 2,/

t m n m m

n t

λ φ φ φ π π π π

λ

⎛ ⎞ ⎛ ⎞= − = + − = − → = − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ …

The minimum index of refraction is found from 1.m =

( )1 1min 2 2675 nm

1 1.35125 nm

n ⎛ ⎞= − =⎜ ⎟

⎝ ⎠

This index is smaller than the typical index of glass (about 1.5), so choose a material with n = 1.35.

If we now assume that the film has an index greater than glass, then the wave that reflects from the

glass has a phase change due to the additional path length and no phase change on reflection, for a total

phase change of 22

2 0./

t

nφ π

λ

⎛ ⎞= +⎜ ⎟

⎝ ⎠

Repeat the calculation from above for this new phase change.


24/33

24-24 Chapter 24



net 2 1

min

22 (2 1) , 1, 2,

/ 2

1 675 nm2.70

2 2 125 nm

t mm n m

n t

nt

λ φ φ φ π π π

λ

λ

⎛ ⎞= − = − = − → = =⎜ ⎟

⎝ ⎠

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

…

That is a very high index—higher than diamond. It probably is not realistic to look for this material.



sin tan / . xθ θ ≈ ≈ Second order means 2.m =

1 21 2

1 21 2

sin ; ;m m x m

d m d m x x xd d d

m m x x x

d d

λ λ λ θ λ λ

λ λ

= → = → = = = →

∆ = − = − →

4 39 7

2 1

(6.6 10 m)(1.23 10 m)650 10 m 4.81 10 m 480 nm

2(2.40 m)

d x

m

λ λ − −

− −∆ × ×= − = × − = × ≈

78. With respect to the incident wave, the wave that reflects at the

top surface of the film has a phase change of 1 .φ π = With

respect to the incident wave, the wave that reflects from the

bottom surface of the film has a phase change due to the

additional path length and no phase change due to reflection, so

2film

22 0.

t φ π

λ

⎛ ⎞= +⎜ ⎟

⎝ ⎠ For constructive interference, the net phase change must be an integer multiple of

2 .π

1 1 1 1net 2 1 film2 2 2 2film film2 2 2 ( ) ( ) , 0, 1, 2,t m t m m m

nλ φ φ φ π π π λ

λ

⎛ ⎞= − = − = → = + = + = …⎜ ⎟

⎝ ⎠

Evaluate the thickness for the two wavelengths.

1 21 1 1 11 22 2 2 2

film film

312 12 2 1 3 1

1 1 12 21 121 12 2

( ) ( )

( ) ( ) 688.0 nm 71.40 5( ) 7( ) 2

491.4 nm 5( ) ( )

t m mn n

m mm m m

m m

λ λ

λ

λ

= + = + →

+ += = = = = → + = + → =

+ +

Thus 2 3m = and 1 2.m = Evaluate the thickness with either value and the corresponding wavelength.

1 25 71 1 1 1 1 1

1 22 2 2 2 2 2 2 2film film

688.0 nm 491.4 nm

( ) ( ) 544 nm ; ( ) ( ) 544 nm1.58 1.58t m t mn n

λ λ

= + = = = + = =

79. Because the angle increases with wavelength, we compare the maximum angle for the second order

with the minimum angle for the third order, using Eq. 24–4, by calculating the ratio of the sines for

each angle. Since this ratio is greater than one, the maximum angle for the second order is larger than

the minimum angle for the first order and the spectra overlap.

2 2 2

3 3 3

sin 2 / 2 2(700 nm)sin sin ; 1.2

sin 3 / 3 3(400 nm)

d md m

d d

θ λ λ λ θ λ θ

θ λ λ

⎛ ⎞= → = = = = =⎜ ⎟

⎝ ⎠

1φ π =

( )2 film2 2t φ λ π =


25/33




To determine which wavelengths overlap, we set this ratio of sines equal to one and solve for the

second-order wavelength that overlaps with the shortest wavelength of the third order. We then repeat

this process to find the wavelength of the third order that overlaps with the longest wavelength of the

second order.

2 2 23 2, max

3 3 3

2 3,min

sin 2 / 2 2 21 (700 nm) 467 nmsin 3 / 3 3 3

3 3 (400 nm) 600 nm

2 2

d d

θ λ λ λ λ θ λ λ

λ λ

= = = → = = =

→ = = =

Therefore, the wavelengths 600 nm–700 nm of the second order overlap with the wavelengths

400 nm–467 nm of the third order. Note that these wavelengths are independent of the slit spacing.

80. Because the measurements are made far from the antennas, we can use the analysis for the double slit.

Use Eq. 24–2a for constructive interference, and Eq. 34–2b for destructive interference. The

wavelength of the signal is8

6

(3.00 10 m/s)3.322 m

(90.3 10 Hz) f

υ λ

×= = = .

×

For constructive interference, the path difference is a multiple of the wavelength.

1

1 11 2max max

13max

sin , 0, 1, 2, 3, ....; sin

(1)(3.322 m) (2)(3.322 m)sin 21.7 22 ; sin 47.6 48 ;

9.0 m 9.0 m

(3)(3.322 m)sin impossible

9.0 m

md m m

d

λ θ λ θ

θ θ

θ

−

− −

−

= = → =

= = ° ≈ ° = = ° ≈ °

= =

For destructive interference, the path difference is an odd multiple of half a wavelength.

11 21

2

311 12 2

0 1max max

5 71 12 2

2 3max max

( )

sin ( ) , 0,1, 2, 3, ...; sin

( )(3.322 m) ( )(3.39 m)sin 10.6 11 ; sin 34.4 34 ;

9.0 m 9.0 m

( )(3.39 m) ( )(3.39 m)sin 70.3 70 ; sin impossible

9.0 m 9.0 m

m

d m m d

λ

θ λ θ

θ θ

θ θ

−

− −

− −

+

= + = → =

= = ° ≈ ° = = ° ≈ °

= = ° ≈ ° = =

These angles are applicable both above and below the midline and both to the left and to the right of

the antennas.


from the top surface of the coating has a phase change of

1 .φ π = With respect to the incident wave, the wave thatreflects from the glass ( 1 5)n ≈ . at the bottom surface of

the coating has a phase change due to both the additional

path length and reflection, so 2film

22 .

t φ π π

λ

⎛ ⎞= +⎜ ⎟

⎝ ⎠ For

destructive interference, the net phase change must be an odd-integer multiple of π .

1φ π =

( )2 film2 2t φ λ π π = +


26/33

24-26 Chapter 24



net 2 1film

1 1film4 4

film

22 (2 1)

(2 1) (2 1) , 0, 1, 2,

t m

t m m mn

φ φ φ π π π π λ

λ λ

⎡ ⎤⎛ ⎞= − = + − = + →⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

= + = + = …

The minimum thickness has 0,m = so 1min 4film

.t n

λ =

(a) For the blue light, 1min 4(450 nm)

81.52 nm 82 nm .(1 38)

t = = ≈.

(b) For the red light, 1min 4(720 nm)

130.4 nm 130 nm .(1 38)

t = = ≈.

82. The phase difference caused by the path difference back and forth through the coating must correspond

to half a wavelength in order to produce destructive interference.

1 1 12 4 42 (2 cm) 0.5 cmt t λ λ = → = = =

83. We first find the angular half-width for the first order, using Eq. 24–3a, sin . D

λ θ = Since this angle is

small, we may use the approximation that sin tanθ θ ≈ . The width from the central maximum to thefirst minimum is given by tan . x θ = That width is then doubled to find the width of the beam, from

the first diffraction minimum on one side to the first diffraction minimum on the other side.

8 94

tan sin

2(3 8 10 m)(633 10 m)2 2 sin 2 4.8 10 m 48 km

0.010 m

x x

x x D

θ θ

λ θ

−

= =

. × ×∆ = = = = = × =


from the top surface of the film has a phase change of

1 .φ π = With respect to the incident wave, the wave that

reflects from the glass ( 1.52)n = at the bottom surface of

the film has a phase change due to both the additional path

length and reflection, so 2film

22 .

t φ π π

λ

⎛ ⎞= +⎜ ⎟

⎝ ⎠ For

constructive interference, the net phase change must be an even nonzero integer multiple of π .

1 1net 2 1 film2 2

film film

22 2 , 1, 2, 3,

t m t m m m

n


λ

⎡ ⎤⎛ ⎞= − = + − = → = = = …⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

The minimum nonzero thickness occurs for 1m = .

minfilm

643 nm240 nm

2 2(1.34)t

n

λ = = =

The answer has 3 significant figures.

1φ π =

( )2 film2 2t φ λ π π = +

1.34n =

1.52n =


27/33




85. Because the width of the pattern is much smaller than the distance to the screen, the angles from the

diffraction pattern for this first order will be small. Thus we may make the approximation that

sin tan .θ θ = We find the angle to the first minimum from the distances, using half the width of the

full first-order pattern. Then we use Eq. 24–3b to find the slit width.

11min 1min2

4 5

(8.20 cm)tan 0.01302 sin(315 cm)

(1)(415 nm)sin 3.188 10 nm 3.19 10 m

sin 0.01302

m D m D

θ θ

λ θ λ

θ −

= = =

= → = = = × = ×

86. If the original intensity is 0 , I then the first polarizers will reduce the intensity to1

1 02. I I = Each

subsequent polarizer oriented at an angle θ to the preceding one will reduce the intensity as given by

the equation 2transmitted incident cos . I I θ = For n polarizers (including the first one),

2 1102(cos ) .nn I I θ

−= Solve for n such that the intensity is 1 05 I .

2 1 2 1 21 1

0 05 2

2

(cos 10 ) 0.4 (cos 10 ) ln(0 4) ( 1) ln(cos 10 )

ln(0 4)1 30.93

ln(cos 10 )

n n

n I I I n

n

− −= = ° → = ° → . = − ° →

.= + =

°

Thus 31 polarizers are needed for the intensity to drop below 15

of its original value.

87. The lines act like a reflection grating, similar to a CD. We assume that we see the first diffractive

order, so 1.m = Use Eq. 24–4.

(1)(480 nm)sin 580 nm

sin sin 56

md m d

λ θ λ

θ = → = = =

°

88. We assume that the sound is diffracted when it passes through the doorway and find the angles of theminima from Eq. 24–3b.

1

1 1

1 1

1 1 1

; sin sin , 1, 2, 3, ...

(1)(340 m/s)1: sin = sin 24

(0.88 m)(950 Hz)

(2)(340 m/s)2 : sin sin 54

(0.88 m)(950 Hz)

(3)(340 m/s)3 : sin sin sin 1.22 im

(0.88 m)(950 Hz)

m m D m m

f f Df

mm

Df

mm

Df

mm

Df

υ υ υ λ θ λ θ

υ θ

υ θ

υ θ

−

− −

− −

− − −

= = = → = =

= = = °

= = = = °

= = = = = possible

Thus the whistle would not be heard clearly at angles of 24° and 54° on either side of the normal.

89. We find the angles for the first order from Eq. 24�

ch24 giancoli7e manual

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