M22 KNIG9404 ISM C22 - media.physicsisbeautiful.com

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22-1

Conceptual Questions

22.1.

The initial light pattern is a double-slit interference pattern. It is centered behind the midpoint of the slits. The slight decrease in intensity going outward from the middle indicates that the light from each of the individual slits is not uniform but slowly decreases toward the edges of the screen. If the right slit is covered, light comes through only the left slit. Without a second slit, there is no interference. Instead, we get simply the spread-out pattern of light diffracting through a single slit, such as in the center of the photograph of Figure 22.2. The intensity is a maximum directly behind the left slit, and—as we discerned from the intensities in the double-slit pattern—the single-slit intensity fades gradually toward the edges of the screen.

22.2. Because mm Ly

dλ= increasing λ and L increases the fringe spacing. Increasing d decreases the fringe spacing.

Submerging the experiment in water decreases λ and decreases the fringe spacing. So the answers are (a) and (c).

22.3. The fringe separation for the light intensity pattern of a double-slit is determined by Δy = λL/d. (a) If λ increases, yΔ will decrease. (b) If d decreases, yΔ will increase. (c) If L decreases, yΔ will decrease. (d) Since the dot is in the 2m = bright fringe, the path length difference from the two slits is 2 1000 nmλ = .

22.4. (a) The equation for gratings does not contain the number of slits, so increasing the number of slits can’t affect the angles at which the bright fringes appear as long as d is the same. So the number of fringes on the screen stays the same. (b) The number of slits does not appear in the equation for the fringe spacing, so the spacing stays the same. (c) Decreases; the fringes become narrower. (d) The equation for intensity does contain the number of slits, so each fringe becomes brighter: 2

max 1I N I= .

22.5. .aλ < Several secondary maxima appear. For sin ,pa pθ λ= the first minima from the central maximum

require 1sin ,a θ λ= which must be less than 1.

WAVE OPTICS

22

22-2 Chapter 22


22.6. (a) The width increases because 11 22 .

Dλθ .= (b) The width decreases because 1

1 22 .D

λθ .= (c) Almost

uniformly gray with no minima.

22.7. When the experiment is immersed in water the frequency of the light stays the same. But the speed is slower,

so the wavelength is smaller. When the wavelength is smaller the fringes get closer together because Lyd

λΔ = .

22.8. (a) Because the hole is so large we do not use the wave model of light. The ray model says we treat the light as if it travels in straight lines, so increasing the hole diameter will increase the size of the spot on the screen. (b) When the hole is smaller than 1 mm we need to consider diffraction effects; making the hole 20% larger will decrease the diameter of the spot on the screen.

22.9. Moving a mirror by 200 nm increases or decreases the path length by 400 nm. Since one mirror is moved in and the other out each increases the path length difference by 400 nm. The total is a path length difference of 800 nm, so the spot will still be bright.

22.10. If the wavelength is changed to /2λ there will still be crests everywhere there were crests before (plus new crests halfway between, where there used to be troughs). If the interferometer was set up to display constructive interference originally, then crests were arriving from each arm of the interferometer together at the detector. That will still be true when the wavelength is halved. So the central spot remains bright.

Exercises and Problems

Section 22.2 The Interference of Light

22.1. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). It is symmetrical with the 3m = fringes on both sides of and equally distant from the central maximum. Solve: The bright fringes occur at angles mθ such that

sin 0, 1, 2, 3, md m mθ λ= = … 9

3 363(600 10 m) 180sin 0 0225 0 023 rad 0 023 rad 1 3

rad(80 10 m)θ θ

π

−

−× °

⇒ = = . ⇒ = . = . × = . °×

22.2. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The bright fringes are located at positions given by Equation 22.4, sin .md mθ λ= For the m = 3 bright orange

fringe, the interference condition is 93sin 3(600 10 m).d θ −= × For the m = 4 bright fringe the condition is 4sin 4d θ λ= .

Because the position of the fringes is the same, 9 93

3 4 4sin sin 4 3(600 10 m) (600 10 m) 450 nmd dθ θ λ λ− −= = = × ⇒ = × =

22.3. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). It is symmetrical, with the 2m = fringes on both sides of and equally distant from the central maximum. Solve: The two paths from the two slits to the 2m = bright fringe differ by 2 1,r r rΔ = − where

2 2(500 nm) 1000 nmr mλ λΔ = = = = Thus, the position of the 2m = bright fringe is 1000 nm farther away from the more distant slit than from the nearer slit.

Wave Optics 22-3


22.4. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The formula for fringe spacing is

3 91 8 10 m (600 10 m) 3000L L Lyd d d

λ − −Δ = ⇒ . × = × ⇒ =

The wavelength is now changed to 400 nm, and / ,L d being a part of the experimental setup, stays the same. Applying the above equation once again,

9(400 10 m)(3000) 1 2 mmLyd

λ −Δ = = × = .

22.5. Visualize: The fringe spacing for a double slit pattern is Lyd

λΔ = . We are given L = 2.0 m and 600 nmλ = .

We also see from the figure that 13 cmyΔ = .

Solve: Solve the equation for d. 9

213

(600 10 m)(2 0 m) 0 36 mm10 m

Ldy

λ −

−× .= = = .

Δ ×

Assess: 0.36 mm is a typical slit spacing.

22.6. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The fringe spacing is

9 2

3(589 10 m)(150 10 m) 0 22 mm

4 0 10 mL Ly d

d yλ λ − −

−× ×Δ = ⇒ = = = .

Δ . ×

22.7. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference fringes are equally spaced on both sides of the central maximum. The interference pattern looks like Figure 22.3(b). Solve: In the small-angle approximation

1 ( 1)m m m md d dλ λ λθ θ θ+Δ = − = + − =

Since 200 ,d λ= we have 1 rad 0 286

200dλθΔ = = = . °

22.8. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The dark fringes are located at positions given by Equation 22.9:

( )12 0, 1, 2, 3, m

Ly m md

λ′ = + = …

( ) ( )2

31 15 1 2 2 3

4 (60 10 m)5 1 6 0 10 m 500 nm0 20 10 m

L Ly yd d

λ λ λ λ−

−−

×′ ′⇒ − = + − + ⇒ . × = ⇒ =. ×

Section 22.3 The Diffraction Grating 22.9. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8. Solve: The bright constructive-interference fringes are given by Equation 22.15:

sin 0, 1, 2, md m mθ λ= = … 9

1 12(1)(550 10 m)sin 0 0275 1 6

(4 0 10 m)/2000θ θ

−

−×

⇒ = = . ⇒ = . °. ×

Likewise, 2 2sin 0 055 and 3 2 .θ θ= . = . °

22-4 Chapter 22


22.10. Model: A diffraction grating produces a series of constructive-interference fringes at values of mθ

determined by Equation 22.15. Solve: We have

2sin 0, 1, 2, 3, sin 20 0 1 and sin 2md m m d dθ λ λ θ λ= = …⇒ . ° = =

Dividing these two equations,

2 2sin 2sin 20 0 0 6840 43 2θ θ= . ° = . ⇒ = . °

22.11. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8. Solve: The bright constructive-interference fringes are given by Equation 22.15:

96(2)(600 10 m)sin 1 89 10 m

sin sin(39 5 )mm

md m d λθ λθ

−−×= ⇒ = = = . ×

. °

The number of lines in per millimeter is 3 6(1 10 m)/(1 89 10 m) 530.− −× . × =

22.12. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram of Figure 22.8. Solve: The bright interference fringes are given by

sin m 0, 1, 2, 3, md mθ λ= = …

The slit spacing is 61 mm/500 2 00 10 md −= = . × and 1.m = For the red and blue light, 9 9

1 11 red 1 blue6 6

656 10 m 486 10 msin 19 15 sin 14 062 00 10 m 2 00 10 m

θ θ− −

− −− −

⎛ ⎞ ⎛ ⎞× ×= = . = = . °⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟. × . ×⎝ ⎠ ⎝ ⎠

The distance between the fringes, then, is 1 red 1 bluey y yΔ = − where

1 red

1 blue

(1 5 m) tan19 15 0 521 m(1 5 m) tan14 06 0 376 m

yy

= . . ° = .= . . ° = .

So, 0 145 m 14 5 cmyΔ = . = . .

22.13. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8.

Solve: The bright fringes are given by Equation 22.15: 1 1sin 0, 1, 2, 3, sin (1) / sinmd m m d dθ λ θ λ λ θ= = …⇒ = ⇒ =

Wave Optics 22-5


The angle 1θ can be obtained from geometry as follows:

11 1

(0 32 m)/2tan 0 080 tan (0.080) 4.572 0 m

θ θ −.= = . ⇒ = °.

Using 1sin sin 4 57 0 07968,θ = . ° = . 9633 10 m 7 9 m

0 07968d μ

−×= = ..

22.14. Model: A diffraction grating produces a series of constructive-interference fringes at values of mθ that are

determined by Equation 22.15. Solve: For the m = 3 maximum of the red light and the m = 5 maximum of the unknown wavelength, Equation 22.15 gives

93 5 unknownsin 3(660 10 m) sin 5d dθ θ λ−= × =

The 5m = fringe and the 3m = fringe have the same angular positions. This means 5 3.θ θ= Dividing the two

equations, 9

unknown unknown5 3(660 10 m) 396 nmλ λ−= × ⇒ =

Section 22.4 Single-Slit Diffraction

22.15. Model: A narrow single slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: The minima occur at positions

pLy paλ=

94

2 12 1 (633 10 m)(1 5 m)So 2 0 10 m 0 20 mm

0 00475 mL L L Ly y y a

a a a yλ λ λ λ −

−× .Δ = − = − = ⇒ = = = . × = .Δ .

22.16. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: The width of the central maximum for a slit of width 200a λ= is

2 2 (2 0 m) 20 mm200

Lwaλ λ

λ.= = =

22.17. Visualize: We are given 1 0 mL = . and 4000 .w λ=

Solve: Solve 2 Lwaλ= for a.

2 2 (1 0 m) 1 0 m 0 50 mm4000 2000

Lawλ λ

λ. .= = = = .

Assess: This is a typical slit width.

22.18. Visualize: Use Equation 22.22: 2 Lwaλ= . We are given 600 nmλ = and L = 2.0 m. We see from the figure

that 1 0 cmw = . . Solve: Solve the equation for a.

92 2(600 10 m)(2 0 m) 0 24 mm0 010 m

Lawλ −× .= = = .

.

Assess: 0.24 mm is in the range of typical slit widths.

22-6 Chapter 22


22.19. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: The width of the central maximum for a slit of width 200a λ= is

92 2(500 10 m)(2 0 m) 0 0040 m 4 0 mm0 0005 m

Lwaλ −× .= = = . = .

.

22.20. Model: The spacing between the two buildings is like a single slit and will cause the radio waves to be diffracted. Solve: Radio waves are electromagnetic waves that travel with the speed of light. The wavelength of the 800 MHz waves is

8

63 10 m/s 0 375 m

800 10 Hzλ ×= = .

×

To investigate the diffraction of these waves through the spacing between the two buildings, we can use the general condition for complete destructive interference: sin 1, 2, 3, pa p pθ λ= ( = …) where a is the spacing between the buildings. Because the width of the central maximum is defined as the distance between the two 1p = minima on either side of the central maximum, we will use 1p = and obtain the angular width 12Δθ θ= from

1 11 1

0 375 msin sin sin 1 4315 m

aaλθ λ θ − − .⎛ ⎞ ⎛ ⎞= ⇒ = = = . °⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Thus, the angular width of the wave after it emerges from between the buildings is 2(1 43 ) 2 86 2 9θΔ = . ° = . ° ≈ . °.

22.21. Model: The crack in the cave is like a single slit that causes the ultrasonic sound beam to diffract. Visualize:

Solve: The wavelength of the ultrasound wave is 340 m/s 0 0113 m30 kHz

λ = = .

Using the condition for complete destructive interference with 1,p =

11 1

0 0113 msin sin 2 1650 30 m

a θ λ θ − .⎛ ⎞= ⇒ = = . °⎜ ⎟.⎝ ⎠

From the geometry of the diagram, the width of the sound beam is 1 12 2(100 m tan ) 200 m tan 2 165 7 6 mw y θ= = × = × . ° = .

Assess: The small-angle approximation is almost always valid for the diffraction of light, but may not be valid for the diffraction of sound waves, which have a much larger wavelength.

Section 22.5 Circular-Aperture Diffraction

22.22. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes.

Wave Optics 22-7


Solve: The width of the central maximum for a circular aperture of diameter D is 9

32 44 (2 44)(500 10 m)(2 0 m) 4.9 mm

0 50 10 mLw

Dλ −

−. . × .= = =

. ×

22.23. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Visualize: The intensity pattern will look like Figure 22.15. Solve: According to Equation 22.23, the angle that locates the first minimum in intensity is

6

1 31 22 (1 22)(2.5 10 m) 0.01525 rad 0.874

0 20 10 mDλθ

−

−. . ×= = = = °

. ×

These should be rounded to 0.015 rad 0.87 .= °

22.24. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Visualize: The intensity pattern will look like Figure 22.15. Solve: From Equation 22.24,

3 2

9(0 12 10 m)(1 0 10 m) 78 cm

2 44 2 44(633 10 m)DwL

λ

− −

−. × . ×= = =

. . ×

22.25. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Visualize: The intensity pattern will look like Figure 22.15. Solve: From Equation 22.24, the diameter of the circular aperture is

9

22 44 2 44(633 10 m)(4 0 m) 0 25 mm

2 5 10 mLD

wλ −

−. . × .= = = .

. ×

Section 22.6 Interferometers

22.26. Model: An interferometer produces a new maximum each time 2L increases by 12 λ causing the path-length

difference rΔ to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the number of fringe shifts is

22

92 2(1 00 10 m) 30,467

656 45 10 mLm

λ

−

−Δ . ×Δ = = =

. ×


difference rΔ to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the wavelength is

6722 2(100 10 m) 4 0 10 m 400 nm

500Lm

λ−

−Δ ×= = = . × =Δ


difference rΔ to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the distance the mirror moves is

9

2(33,198)(602 446 10 m) 0 0100000 m 1 00000 cm

2 2mL λ −Δ . ×Δ = = = . = .

Assess: Because the wavelength is known to 6 significant figures and the fringes are counted exactly, we can determine LΔ to 6 significant figures.

22-8 Chapter 22



difference rΔ to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: For sodium light of the longer wavelength 1( )λ and of the shorter wavelength 2( ),λ

1 2( 1)2 2

L m L mλ λ

Δ = Δ = +

We want the same path difference 2 12( )L L− to correspond to one extra wavelength for the sodium light of shorter wavelength 2( ).λ Thus, we combine the two equations to obtain:

1 2 21 2 2

1 2

589 0 nm( 1) ( ) 981 67 9822 2 589 6 nm 589 0 nm

m m m mλ λ λ

λ λ λλ λ

.= + ⇒ − = ⇒ = = = . ≅− . − .

Thus, the distance by which 2M is to be moved is

1 589 6 nm982 0 2895 mm2 2

L mλ .⎛ ⎞Δ = = = .⎜ ⎟

⎝ ⎠

22.30. Model: Two closely spaced slits produce a double-slit interference pattern with the intensity graph looking like Figure 22.3(b). The intensity pattern due to a single slit diffraction looks like Figure 22.14. Both the spectra consist of a central maximum flanked by a series of secondary maxima and dark fringes. Solve: (a) The light intensity shown in Figure P22.30 corresponds to a double-slit aperture. This is because the fringes are equally spaced and the decrease in intensity with increasing fringe order occurs slowly. (b) From Figure P22.30, the fringe spacing is 21 0 cm 1 0 10 my −Δ = . = . × . Therefore,

9(6 00 10 m)(2 5 m) 0 15 mm0 010 m

L Ly dd y

λ λ −. × .Δ = ⇒ = = = .Δ .

22.31. Model: Two closely spaced slits produce a double-slit interference pattern with the intensity graph looking like Figure 22.3(b). The intensity pattern due to a single slit diffraction looks like Figure 22.14. Both the spectra consist of a central maximum flanked by a series of secondary maxima and dark fringes. Solve: (a) The light intensity shown in Figure P22.31 corresponds to a single slit aperture. This is because the central maximum is twice the width and much brighter than the secondary maximum. (b) From Figure P22.31, the separation between the central maximum and the first minimum is

21 1.0 cm 1.0 10 m.y −= = × Therefore, using the small-angle approximation, Equation 22.21 gives the condition for

the dark minimum: 9

21

(2 5 m)(600 10 m) 0 15 mm1 0 10 mp

pL Ly ad y

λ λ −

−. ×= ⇒ = = = .

. ×

22.32. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: In a span of 12 fringes, there are 11 gaps between them. The formula for the fringe spacing is

3 952 10 m (633 10 m)(3 0 m) d 0.40 mm11

Lyd d

λ − −⎛ ⎞× × .Δ = ⇒ = ⇒ =⎜ ⎟⎜ ⎟⎝ ⎠

Assess: This is a reasonable distance between the slits, ensuring 4/ 1 34 10 1.d L −= . ×

22.33. Solve: According to Equation 22.7, the fringe spacing between the m fringe and the m + 1 fringe is / .y L dλΔ = Δy can be obtained from Figure P22.33. The separation between the 2m = fringes is 2.0 cm, implying

that the separation between the two consecutive fringes is 14 (2.0 cm) 0.50 cm.= Thus,

3 22

9(0 20 10 m)(0 50 10 m)0 50 10 m 167 cm

600 10 mL d yy L

dλ

λ

− −−

−Δ . × . ×Δ = . × = ⇒ = = =

×

Assess: A distance of 167 cm from the slits to the screen is reasonable.

Wave Optics 22-9


22.34. Solve: According to Equation 22.7, the fringe spacing between the m fringe and the m + 1 fringe is / .y L dλΔ = Δy can be obtained from Figure P22.33. Because the separation between the 2m = fringes is 2.0 cm, two

consecutive fringes are 14 (2 0 cm) 0 50 cmyΔ = . = . apart. Thus,

3 22 (0 20 10 m)(0 50 10 m)0 50 10 m 500 nm

2 0 mL d yy

d Lλ λ

− −− Δ . × . ×Δ = . × = ⇒ = = =

.

22.35. Solve: The intensity of light of a double-slit interference pattern at a position y on the screen is 2

double 14 cos dI I yL

πλ

⎛ ⎞= ⎜ ⎟⎝ ⎠

where 1I is the intensity of the light from each slit alone. At the center of the screen, that is, at 0 m,y = 11 double4 .I I=

From Figure P22.33, doubleI at the central maximum is 212 mW/m . So, the intensity due to a single slit

is 21 3 mW/mI = .

22.36. Prepare: The path length difference is affected by this rotation. See accompanying figure.

We are given 1ϕ = °. The new part of the path length difference is 1 sinr d ϕΔ = while the old piece is still

2 sin ,r θΔ = so the total path length difference must obey (sin sin )d mϕ θ λ+ =

for constructive interference. However, looking at the central maximum, 0,m = so sin sinθ ϕ= − . This is independent of λ and d. Solve:

1 1tan tan[sin ( sin )] (1.00 m)tan[sin ( sin 1.0 )] 1 75 cmy L Lθ ϕ− −= = − = − ° = − . So the central maximum moves by 1.75 cm on the screen. Assess: This seems like a large shift for a 1° rotation, but is probably reasonable for 1 m.L =

22.37. Model: Two closely spaced slits produce a double-slit interference pattern. The interference pattern is symmetrical on both sides of the central maximum. Visualize:

22-10 Chapter 22


In figure (a), the interference of laser light from the two slits forms a constructive-interference central ( 0)m = fringe at P. The paths 1P and 2P are equal. When a glass piece of thickness t is inserted in the path 1P, the interference between the two waves yields the th10 dark fringe at P. Note that the glass piece is not present in figure (a). Solve: The number of wavelengths in the air-segment of thickness t is

1tmλ

=

The number of wavelengths in the glass piece of thickness t is

2glass /t t ntm

nλ λ λ= = =

The path length has thus increased by mΔ wavelengths, where

2 1 ( 1) tm m m nλ

Δ = − = −

From the th10 dark fringe to the st1 dark fringe is 9 fringes and from the st1 dark fringe to the th0 bright fringe is one-half of a fringe. Hence,

91 19 19 19 19 (633 10 m)9 ( 1) 12 0 m2 2 2 2 ( 1) 2 1 5 1 0

tm n tn

λ μλ

−×Δ = + = ⇒ = − ⇒ = = = . − . − .

22.38. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8. Solve: 500 lines per mm on the diffraction grating gives a spacing between the two lines of 1 mm/500d = =

3 6(1 10 m)/500 2 0 10 m.− −× = . × The wavelength diffracted at angle 30mθ = ° in order m is 6sin (2 0 10 m)sin30 1000 nmmd

m m mθ

λ−. × °= = =

We’re told it is visible light that is diffracted at 30 ,° and the wavelength range for visible light is 400–700 nm. Only 2m = gives a visible light wavelength, so 500 nm.λ =

22.39. Model: Each wavelength of light is diffracted at a different angle by a diffraction grating. Solve: Light with a wavelength of 501.5 nm creates a first-order fringe at y = 21.90 cm. This light is diffracted at angle

11

21 90 cmtan 23 6550 00 cm

θ − .⎛ ⎞= = . °⎜ ⎟.⎝ ⎠

We can then use the diffraction equation sin ,md mθ λ= with 1,m = to find the slit spacing:

1

501 5 nm 1250 nmsin sin(23 65 )

d λθ

.= = =. °

The unknown wavelength creates a first order fringe at 31.60 cm,y = or at angle

11

31 60 cmtan 32 2950 00 cm

θ − .⎛ ⎞= = . °⎜ ⎟.⎝ ⎠

With the split spacing now known, we find that the wavelength is 1sin (1250 nm)sin(32 29 ) 667 8 nmdλ θ= = . ° = .

Assess: The distances to the fringes and the first wavelength were given to 4 significant figures. Consequently, we can determine the unknown wavelength to 4 significant figures.

22.40. Model: Assume the incident light is coherent and monochromatic. Solve: (a) Where the path length difference is λ between adjacent slits there will be constructive interference from all three slits, and 2

max 1I N I= applies. 2

max 1 13 9I I I= =

Wave Optics 22-11


(b) For the case where the path length difference is /2λ adjacent slits will produce destructive interference, so the third slit will just give a total intensity of 1.I

22.41. Model: We will assume that the listeners did not hear any other loud spots between the center and 1.4 m on each side, 1.m = We’ll use the diffraction grating equations. First solve Equation 22.16 for 1θ and insert it into

Equation 22.15. We need the wavelength of the sound waves, and we’ll use the fundamental relationship for periodic waves to get it.

340 m/s 3 4 cm 0 034 m10000 Hz

vf

λ = = = . = .

We are also given 10 mL = and 1 1 4 my = . . Solve:

1 111

1 4 mtan tan 0 14 rad10 m

yL

θ − − ⎛ ⎞.⎛ ⎞= = = .⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

(This may be small enough to use the small-angle approximation, but we are almost finished with the problem without it, so maybe we can save the approximation and try it in the assess step.)

1

(1) 0 034 m 0 25 m 25 cmsin sin(0 14 rad)

d λθ

.= = = . =.

Assess: These numbers seem reasonable given the size (wavelength) of sound waves. With the small-angle approximation, sin tan ,θ θ θ≈ ≈

1(0 034 m) 24 cm/ 1 4 m/10 mm

mdy L

λ .= = =.

This is almost the same, but rounded down to two significant figures rather than up. The 1θ we computed seemed

almost small enough to use the approximation, and it would probably be OK for many applications when the angle is this small.

22.42. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram of Figure 22.8. Solve: (a) A grating diffracts light at angles sin / .m m dθ λ= The distance between adjacent slits is 1 mm/600d = =

61 667 10 m 1667 nm−. × = . The angle of the 1m = fringe is

11

500 nmsin 17 461667 nm

θ − ⎛ ⎞= = . °⎜ ⎟⎝ ⎠

The distance from the central maximum to the 1m = bright fringe on a screen at distance L is o

1 1tan (2 m) tan17 46 0 629 my L θ= = . = .

(Note that the small angle approximation is not valid for the maxima of diffraction gratings, which almost always have angles 10 .> ° ) There are two 1m = bright fringes, one on either side of the central maximum. The distance between them is 12 1 258 m 1 3 my yΔ = = . ≈ . . (b) The maximum number of fringes is determined by the maximum value of m for which sin mθ does not exceed 1 because there are no physical angles for which sin 1.θ > In this case,

(500 nm)sin1667 nmm

m mdλθ = =

We can see by inspection that m = 1, m = 2, and m = 3 are acceptable, but m = 4 would require a physically impossible 4sin 1θ > . Thus, there are three bright fringes on either side of the central maximum plus the central

maximum itself for a total of seven bright fringes.

22.43. Model: Assume the screen is centered behind the slit. We actually want to solve for m, but given the other data, it is unlikely that we will get an integer from the equations for the edge of the screen, so we will have to truncate our answer to get the largest order fringe on the screen.

22-12 Chapter 22


Visualize: Refer to Figure 22.7. Use Equation 22.15: sin ,md mθ λ= and Equation 22.16: tanm my L θ= . We are

given 510 nm,λ = 2 0 m,L = . and 1500 mmd = . As mentioned above, we are not guaranteed that a bright fringe will

occur exactly at the edge of the screen, but we will kind of assume that one does and set 1 0 m;my = . if we do not get an integer for m then the fringe was not quite at the edge of the screen and we will truncate our answer to get an integer m. Solve: Solve Equation 22.16 for mθ and insert it in Equation 22.15.

1tan mm

yL

θ −=

Solve Equation 22.15 for m. 1

1 1500 mm 1 0 msin sin tan sin tan 1 8510 nm 2 0 m

mm

d d ymL

θλ λ

− − .⎛ ⎞ ⎛ ⎞= = = = .⎜ ⎟ ⎜ ⎟.⎝ ⎠ ⎝ ⎠

Indeed, we did not get an integer, so truncate 1.8 to get 1m = . This means we will see three fringes, one for 0,m = and one on each side for 1m = ± . Assess: Our answer fits with the statement in the text: “Practical gratings, with very small values for ,d display only a few orders.”

22.44. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. The visible spectrum spans the wavelengths 400 nm to 700 nm. Solve: According to Equation 22.16, the distance my from the center to the mth maximum is tan .m my L θ= The

angle of diffraction is determined by the constructive-interference condition sin ,md mθ λ= where 0, 1, 2, 3, m = …

The width of the rainbow for a given fringe order is thus red violet .w y y= − The slit spacing is 3

61 mm 1 0 10 m 1 6667 10 m600 600

d−

−. ×= = = . ×

For the red wavelength and for the 1m = order, 9

1 11 1 6

700 10 msin (1) sin sin 24 831 6667 10 m

ddλθ λ θ

−− −

−×= ⇒ = = = . °

. ×

From the equation for the distance of the fringe, red 1tan (2 0 m) tan(24 83 ) 92 56 cmy L θ= = . . ° = .

Likewise for the violet wavelength, 9

11 violet6

400 10 msin 13 88 (2 0 m) tan(13 88 ) 49 42 cm1 6667 10 m

yθ−

−−

⎛ ⎞×= = . °⇒ = . . ° = .⎜ ⎟⎜ ⎟. ×⎝ ⎠

The width of the rainbow is thus 92.56 cm 49.42 cm 43.14 cm 43 cm.− = ≈

22.45. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: (a) If blue light (the shortest wavelengths) is diffracted at angle ,θ then red light (the longest wavelengths) is diffracted at angle 30°.θ + In the first order, the equations for the blue and red wavelengths are

brsin sin( 30 )d

dλ

θ θ λ= + ° =

Combining the two equations we get for the red wavelength, 2

b br 2(sin cos30 cos sin30 ) (0 8660sin 0 50cos ) 0 8660 (0 50) 1d d d d

d d

λ λλ θ θ θ θ

⎛ ⎞⎛ ⎞⎜ ⎟= ° + ° = . + . = . + . −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2b 2 2 2 2

r b b r b2(0 50) 1 0 8660 (0 50) ( ) ( 0 8660 )d dd

λλ λ λ λ λ⇒ . − = − . ⇒ . − = − .

Wave Optics 22-13


2r b 2

b0 86600 50

dλ λ

λ− .⎛ ⎞

⇒ = +⎜ ⎟.⎝ ⎠

Using 9b 400 10 mλ −= × and 9

r 700 10 m,λ −= × we get 78 125 10 md −= . × . This value of d corresponds to

3

71 mm 1 0 10 m 1230 lines/mm

8 125 10 md

−

−. ×= =

. ×

(b) Using the value of d from part (a) and 9589 10 m,λ −= × we can calculate the angle of diffraction as follows: 7 9

1 1 1sin (1) (8 125 10 m)sin 589 10 m 46 5d θ λ θ θ− −= ⇒ . × = × ⇒ = . °

22.46. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: An 800 line/mm diffraction grating has a slit spacing 3 6(1 0 10 m)/800 1 25 10 md − −= . × = . × . Referring to Figure P22.46, the angle of diffraction is given by

11 1 1

0 436 mtan 0 436 23 557 sin 0 4001 0 m

yL

θ θ θ.= = = . ⇒ = . °⇒ = ..

Using the constructive-interference condition sin ,md mθ λ=

1 6sin(1 25 10 m)(0 400) 500 nm

1d θ

λ −= = . × . =

We can obtain the same value of λ by using the second-order interference fringe. We first obtain 2:θ

22 2 2

0 436 m 0 897 mtan 1 333 53 12 sin 0 8001 0 m

yL

θ θ θ. + .= = = . ⇒ = . °⇒ = ..

Using the constructive-interference condition, 6

2sin (1 25 10 m)(0 800) 500 nm2 2

d θλ

−. × .= = =

Assess: Calculations with the first-order and second-order fringes of the interference pattern give the same value for the wavelength.

22.47. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: From Figure P22.46,

1 1 10 436 mtan 0 436 23 557 sin 0 4001 0 m

θ θ θ.= = . ⇒ = . °⇒ = ..

Using the constructive-interference condition sin ,md mθ λ=

99 6600 10 msin 23 557 (1)(600 10 m) 1 50 10 m

sin(23 557 )d d

−− −×. ° = × ⇒ = = . ×

. °

Thus, the number of lines per millimeter is 3

61 0 10 m 670 lines/mm

1 50 10 m

−

−. × =

. ×

Assess: The same answer is obtained if we perform calculations using information about the second-order bright constructive-interference fringe.

22.48. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: The dark fringes are located at

0, 1, 2, 3,pp Ly p =

aλ= …

22-14 Chapter 22


The locations of the first and third dark fringes are

1 33L Ly y

a aλ λ= =

Subtracting the two equations, 9

3 1 33 1

2 2 2(589 10 m)(0 75 m)( ) 0.12 mm7 5 10 m

L Ly y aa y yλ λ −

−× .− = ⇒ = = =

− . ×

22.49. Visualize: The relationship between the diffraction grating spacing d, the angle at which a particular order of constructive interference occurs ,mθ the wavelength of the light, and the order of the constructive interference m

is sin .md mθ λ= Also note 1/N d= .

Solve: The first order diffraction angle for green light is 1 1 7 6 1

1 sin ( / ) sin (5 5 10 m/2 0 10 m) sin (0 275) 0 278 rad 16dθ λ− − − − −= = . × . × = . = . = °

Assess: This is a reasonable angle for a first order maximum.

22.50. Model: The peacock feather is acting as a reflection grating, so we may use Equation 22.15: sin ,md mθ λ=

with 1m = (we are told it is first-order diffraction), 470 nm,λ = and 1 15 0 262 radθ = ° = . .

Solve: Solve Equation 22.15 for d. 9(1)(470 10 m) 1 8 m

sin sin(0.262 rad)m

md λ μθ

−×= = = .

Assess: The answer is small, but plausible for the bands on the barbules.

22.51. Model: Assume the incident light is coherent and monochromatic. Visualize: The distances given in the table are 12y since the measurement is between the two first order fringes. Also note that 1m = in this problem.

Solve: The equation for diffraction gratings is sin md mθ λ= where d is the distance (in mm) between slits; we seek 1 ,d the number of lines per mm.

11sind

θ λ=

From tanm my L θ= and distance1 2y = , where “distance” is the distance between first order fringes as given in the

table, we arrive at

1 distance 1sin tan2L d

λ−⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

This leads us to believe that a graph of 1 distancesin tan2L

−⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠vs. λ will produce a straight line whose slope is 1

d and

whose intercept is zero.

Wave Optics 22-15


We see from the spreadsheet that the linear fit is excellent and that the slope is 1799.84 mm− and the intercept is very small. Because the number of lines per mm is always reported as an integer we round this answer to 800 lines/mm. Assess: This number of lines per mm is a typical number for a decent grating.

22.52. Model: Small angle approximations would not apply here. Visualize: We see from the figure that if 1y L= then 1 45 .θ = °

Solve: From sin md mθ λ= we get

1sindλθ =

but 1 45θ = ° , and 2sin45 .2

° =

22

dλ =

Assess: This large diffraction angle corresponds to a wavelength that is about the same size as the slit spacing.

22-16 Chapter 22


22.53. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: These are not small angles, so we can’t use equations based on the small-angle approximation. As given by Equation 22.19, the dark fringes in the pattern are located at sin ,pa pθ λ= where 1, 2, 3, p = … For the first minimum of the pattern, 1.p = Thus,

1

1sin sinp

a pλ θ θ

= =

For the three given angles the slit width to wavelength ratios are

(a): 30

1 2,sin30

aλ °

⎛ ⎞ = =⎜ ⎟ °⎝ ⎠ (b):

60

1 1 15,sin 60

aλ °

⎛ ⎞ = = .⎜ ⎟ °⎝ ⎠ (c):

90

1 1sin90

aλ °

⎛ ⎞ = =⎜ ⎟ °⎝ ⎠

Assess: It is clear that the smaller the a/λ ratio, the wider the diffraction pattern. This is a conclusion that is contrary to what one might expect.

22.54. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: As given by Equation 22.19, the dark fringes in the pattern are located at sin ,pa pθ λ= where 1, 2, 3, p = …

For the diffraction pattern to have no minima, the first minimum must be located at least at 1 90 .θ = ° From the

constructive-interference condition sin ,pa pθ λ= we have

633 nmsin sin90p

pa aλ λ λθ

= ⇒ = = =°

22.55. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The dark fringes in this diffraction pattern are given by Equation 22.21:

1, 2, 3,pp Ly p

aλ= = …

We note that the first minimum in the figure is 0.50 cm away from the central maximum. We are given a = 0.02 nm and 1 5 mL = . . Solve: Solve the above equation for λ.

2 3(0 50 10 m)(0 20 10 m) 670 nm(1)(1 5 m)

py apL

λ− −. × . ×= = =

.

Assess: 670 nm is in the visible range.

22.56. Model: A narrow slit produces a single-slit diffraction pattern. Solve: The dark fringes in this diffraction pattern are given by Equation 22.21:

1, 2, 3,pp Ly p

aλ= = …

We note from Figure P22.55 that the first minimum is 0.50 cm away from the central maximum. Thus, 3 2

9(0 15 10 m)(0 50 10 m) 1.3 nm

(1)(600 10 m)pay

Lpλ

− −

−. × . ×= = =

×

Assess: This is a typical slit to screen separation.

22.57. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: Within the small angle approximation, which is almost always valid for the diffraction of light, the width of the central maximum is

12 2 44 Lw yD

λ= = .

Wave Optics 22-17


From Figure P22.55, 1.0 cm,w = so 9

22 44 2.44(500 10 m)(1.0 m) 0.12 mm

(1.0 10 m)LD

wλ −

−. ×= = =

×

Assess: This is a typical size for an aperture to show diffraction.

22.58. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: Within the small-angle approximation, the width of the central maximum is

2 44 LwD

λ= .

Because ,w D= we have

92 44 2 44 (2 44)(633 10 m)(0 50 m) 0 88 mmLD D L DD

λ λ −= . ⇒ = . ⇒ = . × . = .

22.59. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: (a) Because the visible spectrum spans wavelengths from 400 nm to 700 nm, we take the average wavelength of sunlight to be 550 nm. (b) Within the small-angle approximation, the width of the central maximum is

92 4(550 10 m)(3 m)2 44 (1 10 m) (2 44) 4 03 10 m 0 40 mmLw D

D Dλ −

− −×= . ⇒ × = . ⇒ = . × = .

22.60. Model: The antenna is a circular aperture through which the microwaves diffract. Solve: (a) Within the small-angle approximation, the width of the central maximum of the diffraction pattern is

2.44 / .w L Dλ= The wavelength of the radiation is 8 3

93 10 m/s 2 44(0 025 m)(30 10 m)0 025 m 920 m

2 0 m12 10 Hzc wf

λ × . . ×= = = . ⇒ = =.×

That is, the diameter of the beam has increased from 2.0 m to 915 m, a factor of 458. (b) The average microwave intensity is

32

212

100 10 W 0 15 W marea (915 m)

PI /π

×= = = .⎡ ⎤⎣ ⎦

22.61. Model: The laser light is diffracted by the circular opening of the laser from which the beam emerges. Solve: The diameter of the laser beam is the width of the central maximum. We have

9 82 44 2 44 2 44(532 10 m)(3 84 10 m) 0 50 m1000 m

L Lw DD w

λ λ −. . . × . ×= ⇒ = = = .

In other words, the laser beam must emerge from a laser of diameter 50 cm.

22.62. Model: Two closely spaced slits produce a double-slit interference pattern.

22-18 Chapter 22


Visualize:

Solve: (a) The 1m = bright fringes are separated from the 0m = central maximum by 9(600 10 m)(1 0 m) 0 0030 m 3 0 mm

0 0002 mLy

dλ −× .Δ = = = . = .

.

(b) The light’s frequency is 14/ 5.00 10 Hz.f c λ= = × Thus, the period is 151/ 2.00 10 s.T f −= = × A delay of 16 15 1

45.0 10 s 0.50 10 s is .T− −× = ×

(c) The wave passing through the glass is delayed by 14 of a cycle. Consequently, the two waves are not in phase as

they emerge from the slits. The slits are the sources of the waves, so there is now a phase difference 0φΔ between

the two sources. A delay of a full cycle ( )t TΔ = would have no effect at all on the interference because it

corresponds to a phase difference 0 2 radφ πΔ = . Thus a delay of 14 of a cycle introduces a phase difference 0φΔ =

1 14 2(2 ) radπ π= .

(d) The text’s analysis of the double-slit interference experiment assumed that the waves emerging from the two slits were in phase, with 0 0φΔ = rad. Thus, there is a central maximum at a point on the screen exactly halfway between

the two slits, where 0φΔ = rad and 0 m.rΔ = Now that there is a phase difference between the sources, the central maximum—still defined as the point of constructive interference where 0φΔ = rad—will shift to one side. The wave leaving the slit with the glass was delayed by 1

4 of a period. If it travels a shorter distance to the screen, taking 14 of

a period less than the wave coming from the other slit, it will make up for the previous delay and the two waves will arrive in phase for constructive interference. Thus, the central maximum will shift toward the slit with the glass. How far? A phase difference 0 2φ πΔ = would shift the fringe pattern by 3.0 mm,yΔ = making the central maximum fall

exactly where the 1m = bright fringe had been previously. This is the point where (1) ,r λΔ = exactly compensating

for a phase shift of 2π at the slits. Thus, a phase shift of 1 10 2 4 (2 )φ π πΔ = = will shift the fringe pattern

by 14 (3 mm) 0.75 mm.= The net effect of placing the glass in the slit is that the central maximum (and the entire

fringe pattern) will shift 0.75 mm toward the slit with the glass.

22.63. Model: A diffraction grating produces an interference pattern, which looks like the diagram of Figure 22.8. Solve: (a) Nothing has changed while the aquarium is empty. The order of a bright (constructive interference) fringe is related to the diffraction angle mθ by sin ,md mθ λ= where 0, 1, 2, 3, m = … The space between the slits is

61 0 mm 1 6667 10 m600

d −.= = . ×

For 1,m = 9

11 1 6

633 10 msin sin 22 31 6667 10 md

λθ θ−

−−

⎛ ⎞×= ⇒ = = . °⎜ ⎟⎜ ⎟. ×⎝ ⎠

Wave Optics 22-19


(b) The path-difference between the waves that leads to constructive interference is an integral multiple of the wavelength in the medium in which the waves are traveling, that is, water. Thus,

77

1 16water

633 nm 633 nm 4 759 10 m4 759 10 m sin 0 2855 16 61 33 1 6667 10 mn d

λλ θ θ−

−−

. ×= = = . × ⇒ = = = . ⇒ = . °. . ×

22.64. Model: An interferometer produces a new maximum each time 2L increases by 12 λ causing the path-

length difference Δr to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: The path-length difference between the two waves is 2 12 2 .r L LΔ = − The condition for constructive interference is ,r mλΔ = hence constructive interference occurs when

( )1 12 1 2 1 22 22( ) 1200 600L L m L L mλ λ λ λ− = ⇒ − = = =

where 632.8 nmλ = is the wavelength of the helium-neon laser. When the mirror 2M is moved back and a hydrogen discharge lamp is used, 1200 fringes shift again. Thus,

( )12 1 21200 600L L λ λ′ ′ ′− = =

where 656 5 nmλ′ = . . Subtracting the two equations, 9 9

2 1 2 1( ) 600( ) 600(632 8 10 m 656 5 10 m)L L L L λ λ − −′ ′− − ( − ) = − = . × − . × 6

2 2 14 2 10 mL L −′⇒ = + . × That is, 2M is now 14.2 mμ closer to the beam splitter.

22.65. Model: The gas increases the number of wavelengths in one arm of the interferometer. Each additional wavelength causes one bright-dark-bright fringe shift. Solve: From the equation in Challenge Example 22.9, the number of fringe shifts is

2

2 1 9vac

2 (2)(2 00 10 m)( 1) (1 00028 1) 19600 10 m

dm m m nλ

−

−. ×Δ = − = − = . − =

×

22.66. Model: The piece of glass increases the number of wavelengths in one arm of the interferometer. Each additional wavelength causes one bright-dark-bright fringe shift. Solve: We can rearrange the equation in Challenge Example 22.9 to find that the index of refraction of glass is

9vac

3(500 10 nm)(200)1 1 1.50

2 2(0 10 10 m)

mn

dλ −

−

Δ ×= + = + =. ×

22.67. Model: The arms of the interferometer are of equal length, so without the crystal the output would be bright. Visualize: We need to consider how many more wavelengths fit in the electro-optic crystal than would have occupied that space (6.70 m)μ without the crystal; if it is an integer then the interferometer will produce a bright output; if it is a half-integer then the interferometer will produce a dark output. But the wavelength we need to consider is the wavelength inside the crystal, not the wavelength in air.

n nλλ =

We are told the initial n with no applied voltage is 1.522, and the wavelength in air is 1 000 mλ μ= . . Solve: The number of wavelengths that would have been in that space without the crystal is

6 70 m 6 701 000 m

μμ

. = ..

(a) With the crystal in place (and 1.522)n = the number of wavelengths in the crystal is 6.70 m 10.20

1.000 m/1.52210.20 6.70 3.50

μμ

=

− =

22-20 Chapter 22


which shows there are a half-integer number more wavelengths with the crystal in place than if it weren’t there. Consequently the output is dark with the crystal in place but no applied voltage. (b) Since the output was dark in the previous part, we want it to be bright in the new case with the voltage on. That means we want to have just one-half more extra wavelengths in the crystal (than if it weren’t there) than we did in the previous part. That is, we want 4.00 extra wavelengths in the crystal instead of 3.5, so we want 6.70 4.00 10.70+ = wavelengths in the crystal.

6 70 m 10 70(1 000 m)10 70 1 5971 000 m/ 6 70 m

nn

μ μμ μ

. . . = . ⇒ = = .. .

Assess: It seems reasonable to be able to change the index of refraction of a crystal from 1.522 to 1.597 by applying a voltage.

22.68. Model: A diffraction grating produces an interference pattern like the one shown in Figure 22.8. We also assume that the small-angle approximation is valid for this grating. Solve: (a) The general condition for constructive-interference fringes is

sin 0, 1, 2, 3, md m mθ λ= = …

When this happens, we say that the light is diffracted at an angle .mθ Since it is usually easier to measure distances

rather than angles, we will consider the distance my from the center to the mth maximum. This distance is tan .m my L θ= In the small-angle approximation, sin tan ,m mθ θ≈ so we can write the condition for constructive

interference as m

my m Ld m yL d

λλ= ⇒ =

The fringe separation is

1m mLy y y

dλ

+ − = Δ =

(b) Now the laser light falls on a film that has a series of “slits” (i.e., bright and dark stripes), with spacing Ld

dλ′ =

Applying once again the condition for constructive interference:

sin/

mm m

y m L m Ld m d m y mdL d L d

λ λθ λ λλ

′′ ′ ′= ⇒ = ⇒ = = =′

The fringe separation is 1 .m my y y d+′ ′ ′ − = Δ = That is, using the film as a diffraction grating produces a diffraction pattern whose fringe spacing is d, the spacing of the original slits.

22.69. Model: The laser beam is diffracted through a circular aperture. Visualize:

Solve: (a) No. The laser light emerges through a circular aperture at the end of the laser. This aperture causes diffraction, hence the laser beam must gradually spread out. The diffraction angle is small enough that the laser beam appears to be parallel over short distances. But if you observe the laser beam at a large distance it is easy to see that the diameter of the beam is slowly increasing.

Wave Optics 22-21


(b) The position of the first minimum in the diffraction pattern is more or less the “edge” of the laser beam. For diffraction through a circular aperture, the first minimum is at an angle

94

11 22 1 22(633 10 m) 7 72 10 rad 0 044

0 0010 mDλθ

−−. . ×= = = . × = . °

.

(c) The diameter of the laser beam is the width of the diffraction pattern: 92 44 2 44(633 10 m)(3 0 m) 4 6 mm

0 0010 mLw

Dλ −. . × .= = = .

.

(d) At 1 km 1000 m,L = = the diameter is 92 44 2 44(633 10 m)(1000 m) 1 5 m

0 0010 mLw

Dλ −. . ×= = = .

.

22.70. Visualize: To find the location y where the intensity is 1I use Equation 22.14: 2double 14 cos .dI I y

Lπλ

⎛ ⎞= ⎜ ⎟⎝ ⎠

Then divide by the distance to the first minimum 012

Lyd

λ= to get the fraction desired.

Solve: First set double 1.I I=

2double 14 cos dI I y

Lπλ

⎛ ⎞= ⎜ ⎟⎝ ⎠

21 14 cos dI I y

Lπλ

⎛ ⎞= ⎜ ⎟⎝ ⎠

21 cos4

d yL

πλ

⎛ ⎞= ⎜ ⎟⎝ ⎠

1 cos2

d yL

πλ

⎛ ⎞= ⎜ ⎟⎝ ⎠

1 1cos2

Lyd

λπ

− ⎛ ⎞= ⎜ ⎟⎝ ⎠

Now set up the ratio that will give the desired fraction. 1

1

0

1cos2 1 2 22 cos1 2 3 3

2

Ly d

Lyd

λππ

λ π π

−

−

⎛ ⎞⎜ ⎟ ⎛ ⎞⎝ ⎠= = = =⎜ ⎟′ ⎝ ⎠

Assess: The fraction must be less than 1, and 23 seems reasonable.

22.71. Model: A diffraction grating produces a series of constructive-interference fringes at values of mθ that are determined by Equation 22.15. Solve: (a) The condition for bright fringes is sin .d mθ λ= If λ changes by a very small amount Δλ, such that θ changes by ,θΔ then we can approximate /λ θΔ Δ as the derivative /ddλ θ.

2 22 2sin cos 1 sin 1d d d d d m d

m d m m m d mλ λ λλ θ θ θ λθ θ

Δ ⎛ ⎞ ⎛ ⎞= ⇒ ≈ = = − = − = −⎜ ⎟ ⎜ ⎟Δ ⎝ ⎠ ⎝ ⎠

22d

m

λθ

λ

Δ⇒ Δ =

⎛ ⎞ −⎜ ⎟⎝ ⎠

(b) We can now obtain the first-order and second-order angular separations for the wavelengths 589.0 nm andλ = 589.6 nm.λ λ+ Δ = The slit spacing is

361 0 10 m 1 6667 10 m

600d

−−. ×= = . ×

22-22 Chapter 22


The first-order ( 1)m = angular separation is 9 9

612 2 12 2

4

0 6 10 m 0 6 10 m1 5589 10 m2 7778 10 m 0 3476 10 m

3 85 10 rad 0 022

θ− −

−− −

−

. × . ×Δ = =. ×. × − . ×

= . × = . °

The second order ( 2)m = angular separation is 9

312 2 12 2

0 6 10 m 1 02 10 rad 0 0580 6945 10 m 0 3476 10 m

θ−

−− −

. ×Δ = = . × = . °. × − . ×

22.72. Model: The intensity in a double-slit interference pattern is determined by diffraction effects from the slits. Solve: (a) For the two wavelengths λ and λ λ+ Δ passing simultaneously through the grating, their first-order peaks are at

1 1( )L Ly y

d dλ λ λ+ Δ′= =

Subtracting the two equations gives an expression for the separation of the peaks:

1 1Ly y y

dλΔ′Δ = − =

(b) For a double-slit, the intensity pattern is

2double 14 cos dI I y

Lπλ

⎛ ⎞= ⎜ ⎟⎝ ⎠

The intensity oscillates between zero and 14 ,I so the maximum intensity is 14 .I The width is measured at the point

where the intensity is half of its maximum value. For the intensity to be 1max 12 2I I= for the 1m = peak:

2 21 1 half half half half

12 4 cos cos2 4 4

d d d LI I y y y yL L L d

π π π π λλ λ λ

⎛ ⎞ ⎛ ⎞= ⇒ = ⇒ = ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

The width of the fringe is twice half .y This means

half22

Lw yd

λ= =

But the location of the 1m = peak is 1 / ,y L dλ= so we get 112 .w y=

(c) We can extend the result obtained in (b) for two slits to 1/w y N= . The condition for barely resolving two diffraction fringes or peaks is min.w y= Δ From part (a) we have an expression for the separation of the first-order peaks and from part (b) we have an expression for the width. Thus, combining these two pieces of information,

min1 1min min

Ly y d L dyN d LN d NL N

λ λ λλΔ ⎛ ⎞= Δ = ⇒ Δ = = =⎜ ⎟

⎝ ⎠

(d) Using the result of part (c), 9

9min

656 27 10 m 3646 lines0 18 10 m

N λλ

−

−. ×= = =

Δ . ×

22.73. Model: A diffraction grating produces an interference pattern like the one shown in Figure 22.8, when the incident light is normal to the grating. The equation for constructive interference will change when the light is incident at a nonzero angle.

Wave Optics 22-23


Solve: (a) The path difference between waves 1 and 2 on the incident side is 1 sinr d φΔ = . The path difference between the waves 1 and 2 on the diffracted side, however, is 2 sinr d θΔ = . The total path difference between waves 1 and 2 is thus 1 2 (sin sin ).r r d θ φΔ + Δ = + Because the path difference for constructive interference must be equal to mλ,

(sin sin ) 0, 1, 2, d m mθ φ λ+ = = ± ± … (b) For 30φ = ° the angles of diffraction are

9

1 13(1)(500 10 m)sin sin30 0 20 11 5

(1 0 10 m)/600θ θ

−

−×= − ° = − . ⇒ = − . °

. ×

9

1 13( 1)(500 10 m)sin sin30 0 80 53 1(1 0 10 m)/600

θ θ−

− −−− ×= − ° = − . ⇒ = − . °. ×

22.74. Solve: (a)

We have two incoming and two diffracted light rays at angles φ and θ and two wave fronts perpendicular to the rays. We can see from the figure that the wave 1 travels an extra distance sinr d φΔ = to reach the reflection spot. Wave 2 travels an extra distance sinr d θΔ = from the reflection spot to the outgoing wave front. The path difference between the two waves is

1 2 (sin sin )r r r d θ φΔ = Δ − Δ = − (b) The condition for diffraction, with all the waves in phase, is still .r mλΔ = Using the results from part (a), the diffraction condition is

sin sin 2, 1, 0, 1, 2, md m d mθ λ φ= + = … − − …

Negative values of m will give a different diffraction angle than the corresponding positive values. (c) The “zeroth order” diffraction from the reflection grating is m = 0. From the diffraction condition of part (b), this implies 0sin sind dθ φ= and hence 0θ φ= . That is, the zeroth order diffraction obeys the law of reflection—the

angle of reflection equals the angle of incidence. (d) A 700 lines per millimeter grating has spacing 61

700 mm 1 429 10 m 1429 nmd −= = . × = . The diffraction angles

are given by 1 1 (500 nm)sin sin sin sin 40

1429 nmmm mdλθ φ− −⎛ ⎞ ⎛ ⎞= + = + °⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

22-24 Chapter 22


M mθ

≤ −5 not defined −4 −49.2° −3 −24.0° −2 −3.3° −1 17.0° 0 40.0° 1 83.1° 2≥ not defined

(e)

22.75. Model: Diffraction patterns from two objects can just barely be resolved if the central maximum of one image falls on the first dark fringe of the other image. Solve: (a) Using Equation 22.24 with the width equal to the aperture diameter,

72 44 2 44 (2 44)(550 10 m)(0.20 m) 0 52 mmLw D D LD

λ λ −.= = ⇒ = . = . × = .

(b) We can now use Equation 22.23 to find the angle between two distant sources that can be resolved. The angle 9

33

1 22(550 10 m)1.22 1.29 10 rad 0.0740 52 10 mD

λα−

−−

. ×= = = × = °. ×

(c) The distance that can be resolved is 3(1000 m) (1000 m)(1 29 10 rad) 1 3 mα −= . × = .

M22 KNIG9404 ISM C22 - media.physicsisbeautiful.com

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