ch33

Chapter 33 Interference and Diffraction Conceptual Problems 1 • A phase difference due to path-length difference is observed for monochromatic visible light. Which phase difference requires the least (minimum) path length difference? (a) 90o

(b) 180o (c) 270o (d) the answer

depends on the wavelength of the light. Determine the Concept The phase difference δ due to a path difference Δr are related according to λπδ rΔ=2 . Therefore, the least path length difference

corresponds to the smallest phase difference. ( )a is correct.

2 • Which of the following pairs of light sources are coherent: (a) two candles, (b) one point source and its image in a plane mirror, (c) two pinholes uniformly illuminated by the same point source, (d) two headlights of a car, (e) two images of a point source due to reflection from the front and back surfaces of a soap film. Determine the Concept Coherent sources have a constant phase difference. The pairs of light sources that satisfy this criterion are (b), (c), and (e).

3 • [SSM] The spacing between Newton’s rings decreases rapidly as the diameter of the rings increases. Explain qualitatively why this occurs. Determine the Concept The thickness of the air space between the flat glass and the lens is approximately proportional to the square of d, the diameter of the ring. Consequently, the separation between adjacent rings is proportional to 1/d. 4 • If the angle of a wedge-shaped air film, such as that in Example 32-2 is too large, fringes are not observed. Why? Determine the Concept There are two possible reasons that fringes might not be observed. (1) The distance between adjacent fringes is so small that the fringes are not resolved by the eye. (2) Twice the thickness of the air space is greater than the coherence length of the light. If this is the case, fringes would be observed in the region close to the point where the thickness of the air space approaches zero. 5 • Why must a film that is used to observe interference colors be thin? Determine the Concept Colors are observed when the light reflected off the front and back surfaces of the film interfere destructively for some wavelengths and

3047

Chapter 33

3048

constructively for other wavelengths. For this interference to occur, the phase difference between the light reflected off the front and back surfaces of the film must be constant. This means that twice the thickness of the film must be less than the coherence length of the light. The film is called a thin film if twice its thickness is less than the coherence length of the light. 6 • A loop of wire is dipped in soapy water and held up so that the soap film is vertical. (a) Viewed by reflection with white light, the top of the film appears black. Explain why. (b) Below the black region are colored bands. Is the first band red or violet? (a) The phase change due to reflection from the front surface of the film is 180°; the phase change due to reflection from the back surface of the film is 0°. As the film thins toward the top, the phase change due to the path length difference between the two reflected waves (the phase difference associated with the film’s thickness) becomes negligible and the two reflected waves interfere destructively. (b) The first constructive interference will arise when twice the thickness of the film is equal to half the wavelength of the color with the shortest wavelength. Therefore, the first band will be violet (shortest visible wavelength). 7 • [SSM] A two-slit interference pattern is formed using monochromatic laser light with a wavelength of 640 nm. At the second maximum from the central maximum, what is the path-length difference between the light coming from each of the slits? (a) 640 nm (b) 320 nm (c) 960 nm (d) 1280 nm. Determine the Concept For constructive interference, the path difference is an integer multiple of λ; that is, λmr =Δ . For m = 2, ( )nm 6402=Δr . ( )d is correct.

8 • A two-slit interference pattern is formed using monochromatic laser light with a wavelength of 640 nm. At the first minimum from the central maximum, what is the path-length difference between the light coming from each of the slits? (a) 640 nm (b) 320 nm (c) 960 nm (d) 1280 nm. Determine the Concept For destructive interference, the path difference is an odd-integer multiple of λ2

1 ; that is ( ) ...,5,3,1,21 ==Δ mmr λ . For the first

minimum, m = 1 and ( ) nm 320nm 64021 ==Δr . ( )b is correct.

9 • A two-slit interference pattern is formed using monochromatic laser light with a wavelength of 450 nm. What happens to the distance between the first maximum and the central maximum as the two slits are moved closer together?

Interference and Diffraction

3049

(a) The distance increases. (b) The distance decreases. (c) The distance remains the same. Determine the Concept The relationship between the slit separation d and the angular position θm of each maximum is given by ...,2,1,0,sin == mmd m λθ (Equation 33-2). Because d and sinθm are inversely proportional for a given wavelength and interference maximum (value of m), decreasing d increases sinθm and θm. ( )a is correct.

10 • A two-slit interference pattern is formed using two different monochromatic lasers, one green and one red. Which color light has its first maximum closer to the central maximum? (a) Green, (b) red, (c) both maxima are in the same location. Determine the Concept The relationship between the slit separation d, the angular position θm of each maximum, and the wavelength of the light illuminating the slits is given by ...,2,1,0 ,sin == mmd m λθ (Equation 33-2). Because λ and sinθm are directly proportional for a given interference maximum (value of m) and the wavelength of green light is shorter than the wavelength of red light, ( )a is correct.

11 • A single slit diffraction pattern is formed using monochromatic laser light with a wavelength of 450 nm. What happens to the distance between the first maximum and the central maximum as the slit is made narrower? (a) The distance increases. (b) The distance decreases. (c) The distance remains the same. Determine the Concept The relationship between the slit width a, the angular position θm of each maximum, and the wavelength of the light illuminating the slit is given by ...,3,2,1 ,sin == mma m λθ (Equation 33-11). Because a and sinθm are inversely proportional for a given diffraction maximum (value of m), narrowing the slit increases mθsin and θm. ( )a is correct.

12 • Equation 33-2 which is d sin θm = mλ, and Equation 33-11, which is a sin θm = mλ, are sometimes confused. For each equation, define the symbols and explain the equation’s application. Determine the Concept Equation 33-2 expresses the condition for an intensity maximum in two-slit interference. Here d is the slit separation, λ the wavelength of the light, m an integer, and θm the angle at which the interference maximum appears. Equation 33-11 expresses the condition for an intensity minimum in single-slit diffraction. Here a is the width of the slit, λ the wavelength of the light, and θ m the angle at which the minimum appears, and m is

Chapter 33

3050

a nonzero integer. 13 • When a diffraction grating is illuminated by white light, the first-order maximum of green light (a) is closer to the central maximum than the first-order maximum of red light. (b) is closer to the central maximum than the first-order maximum of blue light. (c) overlaps the second-order maximum of red light. (d) overlaps the second-order maximum of blue light. Picture the Problem We can solve λθ md =sin for θ with m = 1 to express the location of the first-order maximum as a function of the wavelength of the light.

λθ md =sin where d is the separation of the slits and m = 0, 1, 2, …

The interference maxima in a diffraction pattern are at angles θ given by:

Solve for the angular location θ1 of the first-order maximum : ⎟

⎠⎞

⎜⎝⎛= −

dλθ 1

1 sin

Because λgreen light < λred light: light redlightgreen θθ < and )(a is correct.

14 • A double-slit interference experiment is set up in a chamber that can be evacuated. Using light from a helium-neon laser, an interference pattern is observed when the chamber is open to air. As the chamber is evacuated, one will note that (a) the interference fringes remain fixed. (b) the interference fringes move closer together. (c) the interference fringes move farther apart. (d) the interference fringes disappear completely. Determine the Concept The distance on the screen to mth bright fringe is given by:

dLmy n

mλ

=

where L is the distance from the slits to the screen, nλ is the wavelength of the light in a medium whose index of refraction is n, and d is the separation of the slits.

The separation of the interference fringes is given by:

( )d

Ld

Lmd

Lmyy nnnmm

λλλ=−+=−+ 11

Because the index of refraction of a vacuum is slightly less than the index of refraction of air, the removal of air increases nλ and, hence, ym − ym−1. )(c is

correct.


3051

15 • [SSM] True or false: (a) When waves interfere destructively, the energy is converted into heat

energy. (b) Interference patterns are observed only if the relative phases of the waves

that superimpose remain constant. (c) In the Fraunhofer diffraction pattern for a single slit, the narrower the slit,

the wider the central maximum of the diffraction pattern. (d) A circular aperture can produce both a Fraunhofer diffraction pattern and a

Fresnel diffraction pattern. (e) The ability to resolve two point sources depends on the wavelength of the

light. (a) False. When destructive interference of light waves occurs, the energy is no longer distributed evenly. For example, light from a two-slit device forms a pattern with very bright and very dark parts. There is practically no energy at the dark fringes and a great deal of energy at the bright fringe. The total energy over the entire pattern equals the energy from one slit plus the energy from the second slit. Interference re-distributes the energy. (b) True. (c) True. The width of the central maximum in the diffraction pattern is given by

am

mλθ 1sin −= where a is the width of the slit. Hence, the narrower the slit, the

wider the central maximum of the diffraction pattern. (d) True. (e) True. The critical angle for the resolution of two sources is directly

proportional to the wavelength of the light emitted by the sources (Dλα 22.1c = ).

16 • You observe two very closely-spaced sources of white light through a circular opening using various filters. Which color filter is most likely to prevent your resolving the images on your retinas as coming from two distinct sources? (a) red (b) yellow (c) green (d) blue (e) The filter choice is irrelevant.

Chapter 33

3052

Determine the Concept The condition for the resolution of the two sources is given by Rayleigh’s criterion: Dc λα 22.1= (Equation 33-25), where αc is the critical angular separation and D is the diameter of the aperture. The larger the critical angle required for resolution, the less likely it is that you can resolve the sources as being two distinct sources. Because αc and λ are directly proportional, the filter that passes the shorter wavelength light would be most likely to resolve the sources. ( )d is correct. 17 •• Explain why the ability to distinguish the two headlights of an oncoming car, at a given distance, is easier for the human eye at night than during daylight hours. Assume the headlights of the oncoming car are on during both daytime and nighttime hours. Determine the Concept The condition for the resolution of the two sources is given by Rayleigh’s criterion: Dc λα 22.1= (Equation 33-25), where αc is the critical angular separation, D is the diameter of the aperture, and λ is the wavelength of the light coming from the objects, in this case headlights, to be resolved. Because the diameter of the pupils of your eyes are larger at night, the critical angle is smaller at night, which means that at night you can resolve the light as coming from two distinct sources when they are at a greater distance. Estimation and Approximation 18 • It is claimed that the Great Wall of China is the only man made object that can be seen from space with the naked eye. Check to see if this claim is true, based on the resolving power of the human eye. Assume the observers are in low-Earth orbit that has an altitude of about 250 km. Picture the Problem We’ll assume that the diameter of the pupil of the eye is 5.0 mm and use the best-case scenario (the minimum resolvable width varies directly with the wavelength of the light reflecting from the object) that the wavelength of light is 400 nm (the lower limit for the human eye). Then we can use the expression for the minimum angular separation of two objects than can be resolved by the eye and the relationship between this angle and the width of an object and the distance from which it is viewed to support the claim. Relate the width w of an object that can be seen at an altitude h to the critical angular separation αc:

hw

=ctanα ⇒ ctanαhw =


3053

The minimum angular separation αc of two point objects that can just be resolved by an eye depends on the diameter D of the eye and the wavelength λ of light:

Dλα 22.1c =

Substitute for αc in the expression for w to obtain: ⎟

⎠⎞

⎜⎝⎛=

Dhw λ22.1tan

Substitute numerical values and evaluate wmin for an altitude of 250 km:

( ) m24mm0.5nm40022.1tankm250min ≈⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=w

This claim is probably false. Because the minimum width that is resolvable from low-Earth orbit (250 km) is 24 m and the width of the Great Wall is 5 to 8 m high and 5 m wide, so this claim is likely false. However, it is easily seen using binoculars, and pictures can be taken of it using a camera. This is because both binoculars and cameras have apertures that are larger than the pupil of the human eye. (The Chinese astronaut Yang Liwei reported that he was not able to see the wall with the naked eye during the first Chinese manned space flight in 2003.) 19 •• [SSM] (a) Estimate how close an approaching car at night on a flat, straight stretch of highway must be before its headlights can be distinguished from the single headlight of a motorcycle. (b) Estimate how far ahead of you a car is if its two red taillights merge to look as if they were one. Picture the Problem Assume a separation of 1.5 m between typical automobile headlights and tail lights, a nighttime pupil diameter of 5.0 mm, 550 nm for the wavelength of the light (as an average) emitted by the headlights, 640 nm for red taillights, and apply the Rayleigh criterion. (a) The Rayleigh criterion is given by Equation 33-25:

Dcλα 22.1=

where D is the separation of the headlights (or tail lights).

The critical angular separation is also given by:

Ld

=α

where d is the separation of head lights (or tail lights) and L is the distance to approaching or receding automobile.

Chapter 33

3054

Equate these expressions for αc to obtain:

DLd λ22.1= ⇒

λ22.1DdL =

Substitute numerical values and evaluate L:

( )( )( ) km 11

nm 55022.1m 5.1mm 0.5

≈=L

(b) For red light: ( )( )( ) km .69

nm 64022.1m 5.1mm 0.5

≈=L

20 •• A small loudspeaker is located at a large distance to the east from you. The loudspeaker is driven by a sinusoidal current whose frequency can be varied. Estimate the lowest frequency for which your ears would receive the sound waves exactly out of phase when you are facing north. Picture the Problem If your ears receive the sound exactly out of phase, the waves arriving at your ear that is farthest from the speaker must be traveling one-half wavelength farther than the waves arriving at your ear that is nearest the speaker. The lowest frequency corresponds to the longest wavelength. Assume that the speaker and your ears are on the same line and let the distance between your ears be about 20 cm. Take the speed of sound in air to be 343 m/s. The frequency received by your ears is given by: λ

vf =

Let Δr be the distance between your ears (the path difference) to obtain:

λ21=Δr ⇒ rΔ= 2λ

Substituting for λ yields: r

vfΔ

=2

Substitute numerical values and evaluate f: ( ) kHz 86.0

cm 202m/s 343

==f

or between 0.80 and 0.90 kHz. 21 •• [SSM] Estimate the maximum distance a binary star system could be resolvable by the human eye. Assume the two stars are about fifty times further apart than the Earth and Sun are. Neglect atmospheric effects. (A test similar to this ″eye test″ was used in ancient Rome to test for eyesight acuity before entering the army. A normal eye could just barely resolve two well-known close-together stars in the sky. Anyone who could not tell there were two stars was rejected. In this case, the stars were not a binary system, but the principle is the same.)


3055

Picture the Problem Assume that the diameter of a pupil at night is 5.0 mm and that the wavelength of light is in the middle of the visible spectrum at about 550 nm. We can use the Rayleigh criterion for the separation of two sources and the geometry of the Earth-to-binary star system to derive an expression for the distance to the binary stars. If the distance between the binary stars is represented by d and the Earth-star distance by L, then their angular separation is given by:

Ld

=α

The critical angular separation of the two sources is given by the Rayleigh criterion:

Dcλα 22.1=

For α = αc: DL

d λ22.1= ⇒λ22.1

DdL =


( )( )( )( )

y 5.9m 109.461

y 1km 1059.5

nm 55022.1m 105.150mm 0.5

1513

11

⋅×⋅

××≈

×=

c

c

L

Phase Difference and Coherence 22 • Light of wavelength 500 nm is incident normally on a film of water 1.00 μm thick. (a) What is the wavelength of the light in the water? (b) How many wavelengths are contained in the distance 2t, where t is the thickness of the film? (c) What is the phase difference between the wave reflected from the top of the air–water interface and the wave reflected from the bottom of the water–air interface in the region where the two reflected waves superpose? Picture the Problem The wavelength of light in a medium whose index of refraction is n is the ratio of the wavelength of the light in air divided by n. The number of wavelengths of light contained in a given distance is the ratio of the distance to the wavelength of light in the given medium. The difference in phase between the two waves is the sum of a π phase shift in the reflected wave and a phase shift due to the additional distance traveled by the wave reflected from the bottom of the water−air interface. (a) Express the wavelength of light in water in terms of the wavelength of light in air:

nm3761.33

nm500

water

airwater ===

nλ

λ

Chapter 33

3056

(b) Relate the number of wavelengths N to the thickness t of the film and the wavelength of light in water:

( ) 32.5nm376

m 00.122

water

===μ

λtN

(c) Express the phase difference as the sum of the phase shift due to reflection and the phase shift due to the additional distance traveled by the wave reflected from the bottom of the water−air interface:

Nt πππλ

π

δδδ

222

water

traveleddistance additionalreflection

+=+=

+=

( )rad6.11

rad64.11rad32.52rad

π

πππδ

=

=+=

Substitute for N and evaluate δ:

or, subtracting 11.64π rad from 12π rad,rad 1.1rad4.0 == πδ

23 •• [SSM] Two coherent microwave sources both produce waves of wavelength 1.50 cm. The sources are located in the z = 0 plane, one at x = 0, y = 15.0 cm and the other at x = 3.00 cm, y = 14.0 cm. If the sources are in phase, find the difference in phase between these two waves for a receiver located at the origin. Picture the Problem The difference in phase depends on the path difference

according to πλ

δ 2rΔ= . The path difference is the difference in the distances of

(0, 15.0 cm) and (3.00 cm, 14.0 cm) from the origin. Relate a path difference Δr to a phase shift δ:

πλ

δ 2rΔ=

The path difference Δr is: ( ) ( )cm682.0

cm0.14cm3.00cm0.15 22

=

+−=rΔ

Substitute numerical values and evaluate δ: rad .922

cm50.1cm682.0

≈⎟⎟⎠

⎞⎜⎜⎝

⎛= πδ

Interference in Thin Films 24 • A wedge-shaped film of air is made by placing a small slip of paper between the edges of two flat plates of glass. Light of wavelength 700 nm is


3057

incident normally on the glass plates, and interference fringes are observed by reflection. (a) Is the first fringe near the point of contact of the plates dark or bright? Why? (b) If there are five dark fringes per centimeter, what is the angle of the wedge? Picture the Problem Because the mth fringe occurs when the path difference 2t equals m wavelengths, we can express the additional distance traveled by the light in air as an mλ. The thickness of the wedge, in turn, is related to the angle of the wedge and the distance from its vertex to the mth fringe. (a) The first fringe is dark because the phase difference due to reflection by the bottom surface of the top plate and the top surface of the bottom plate is 180° (b) The mth fringe occurs when the path difference 2t equals m wavelengths:

λmt =2

xt

=θ ⇒ θxt =

where we’ve used a small-angle approximation to replace an arc length by the length of a chord.

Relate the thickness of the air wedge to the angle of the wedge:

Substitute for t to obtain: λθ mx =2 ⇒ λλθxm

xm

21

2==

Substitute numerical values and evaluate θ :

( ) rad1075.1nm700cm5

21 4−×=⎟

⎠⎞

⎜⎝⎛=θ

25 •• [SSM] The diameters of fine fibers can be accurately measured using interference patterns. Two optically flat pieces of glass of length L are arranged with the wire between them, as shown in Figure 33-40. The setup is illuminated by monochromatic light, and the resulting interference fringes are observed. Suppose that L is 20.0 cm and that yellow sodium light (wavelength of 590 nm) is used for illumination. If 19 bright fringes are seen along this 20.0-cm distance, what are the limits on the diameter of the wire? Hint: The nineteenth fringe might not be right at the end, but you do not see a twentieth fringe at all. Picture the Problem The condition that one sees m fringes requires that the path difference between light reflected from the bottom surface of the top slide and the top surface of the bottom slide is an integer multiple of a wavelength of the light.

Chapter 33

3058

The mth fringe occurs when the path difference 2d equals m wavelengths:

λmd =2 ⇒ 2λmd =

Because the nineteenth (but not the twentieth) bright fringe can be seen, the limits on d must be:

( ) ( )22 2

121 λλ

+<<− mdm

where m = 19

( ) ( )2nm59019

2nm59019 2

121 +<<− d Substitute numerical values to

obtain: or

m8.5m5.5 μμ << d 26 •• Light that has a wavelength equal to 600 nm is used to illuminate two glass plates at normal incidence. The plates are 22 cm in length, touch at one end, and are separated at the other end by a wire that has a radius of 0.025 mm. How many bright fringes appear along the total length of the plates? Picture the Problem The light reflected from the top surface of the bottom plate (wave 2 in the diagram) is phase shifted relative to the light reflected from the bottom surface of the top plate (wave 1 in the diagram). This phase difference is the sum of a phase shift of π (equivalent to a λ/2 path difference) resulting from reflection plus a phase shift due to the additional distance traveled.

t

1 2

wireglass plate

glass plate

Relate the extra distance traveled by wave 2 to the distance equivalent to the phase change due to reflection and to the condition for constructive interference:

...,3,2,2 21 λλλλ =+t

or ...,,,2 2

523

21 λλλ=t

and ( )λ2

12 += mt where m = 0, 1, 2, …, 0 ≤ t ≤ 2r and λ is the wavelength of light in air.


3059

Solving for m gives:

m =2tλ−

12

where m = 0, 1, 2, …and 0 ≤ t ≤ 2r

Solve for the highest value of m:

mmax = int2 2r( )λ

−12

⎛

⎝⎜⎞

⎠⎟= int

4rλ−

12

⎛⎝⎜

⎞⎠⎟

where r is the radius of the wire.

Substitute numerical values and evaluate m: mmax = int

4 0.025mm( )600nm

−12

⎛

⎝⎜⎞

⎠⎟

= int 166.2( )= 166

Because we start counting from m = 0, the number of bright fringes is mmax + 1.

N = mmax + 1 = 167

27 •• A thin film having an index of refraction of 1.50 is surrounded by air. It is illuminated normally by white light. Analysis of the reflected light shows that the wavelengths 360, 450, and 602 nm are the only missing wavelengths in or near the visible portion of the spectrum. That is, for these wavelengths, there is destructive interference. (a) What is the thickness of the film? (b) What visible wavelengths are brightest in the reflected interference pattern? (c) If this film were resting on glass with an index of refraction of 1.60, what wavelengths in the visible spectrum would be missing from the reflected light? Picture the Problem (a) We can use the condition for destructive interference in a thin film to find the thickness of the film. (b) and (c) Once we’ve found the thickness of the film, we can use the condition for constructive interference to find the wavelengths in the visible portion of the spectrum that will be brightest in the reflected interference pattern and the condition for destructive interference to find the wavelengths of light missing from the reflected light when the film is placed on glass with an index of refraction greater than that of the film.

Chapter 33

3060

...,,,2 25

23

21

21 ''''t λλλλ =+ (a) Express the condition for

destructive interference in the thin film:

or ...,3,2,2 '''t λλλ=

or

n

m'mt λλ ==2 (1)

where m = 1, 2, 3, … and λ′ is the wavelength of the light in the film.

Solving for λ yields: mnt2

=λ

Substitute for the missing wavelengths to obtain: m

nt2nm450 = and 1

2nm360+

=m

nt

Divide the first of these equations by the second and simplify to obtain:

mm

mntmnt

1

12

2

nm360nm450 +

=

+

=

Solving for m gives:

nm450for 4 == λm

Solve equation (1) for t to obtain: n

mt2λ

=

Substitute numerical values and evaluate t:

( )( ) nm600

50.12nm4504

==t

,...3,2,2 2

1 ''''t λλλλ =+ (b) The condition for constructive interference in the thin film is: or

( ) 'm'''t λλλλ 21

25

23

21 ,...,,2 +==

where λ′ is the wavelength of light in the oil and m = 0, 1, 2, …

Substitute for λ′ to obtain: ( )n

mt λ212 += ⇒

21

2+

=m

ntλ

where n is the index of refraction of the film.

Substitute numerical values and simplify to obtain:

( )( )21

21

nm1800nm60050.12+

=+

=mm

λ


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Substitute numerical values for m and evaluate λ to obtain the following table:

m 0 1 2 3 4 5 λ (nm) 3600 1200 720 514 400 327

From the table, we see that the only wavelengths in the visible spectrum are 720 nm, 514 nm, and 400 nm.

...,,,2 25

23

21 '''t λλλ= (c) Because the index of refraction

of the glass is greater than that of the film, the light reflected from the film-glass interface will be shifted by λ2

1 (as is the wave reflected from the top surface) and the condition for destructive interference becomes:

or

( )n

mt λ212 +=

where n is the index of refraction of the film and m = 0, 1, 2, …

Solving for λ yields:

21

2+

=m

ntλ


( )( )21

21

nm1800nm6005.12+

=+

=mm

λ

Substitute numerical values for m and evaluate λ to obtain the following table:

m 0 1 2 3 4 5 λ (nm) 3600 1200 720 514 400 327

From the table we see that the missing wavelengths in the visible spectrum are 720 nm, 514 nm, and 400 nm. 28 •• A drop of oil (refractive index of 1.22) floats on water (refractive index of 1.33). When reflected light is observed from above, as shown in Figure 33-41 what is the thickness of the drop at the point where the second red fringe, counting from the edge of the drop, is observed? Assume red light has a wavelength of 650 nm. Picture the Problem Because there is a λ2

1 phase change due to reflection at both the air-oil and oil-water interfaces, the condition for constructive interference is that twice the thickness of the oil film equal an integer multiple of the wavelength of light in the film.

Chapter 33

3062

,...3,2,2 '''t λλλ= The condition for constructive interference is: or

'mt λ=2 (1) where λ′ is the wavelength of light in the oil and m = 1, 2, 3, …

Substitute for λ′ to obtain: n

mt λ=2 ⇒

nmt2λ

=


( )( )( ) nm533

22.12nm6502

==t

29 •• [SSM] A film of oil that has an index of refraction of 1.45 rests on an optically flat piece of glass with an index of refraction of 1.60. When illuminated by white light at normal incidence, light of wavelengths 690 nm and 460 nm is predominant in the reflected light. Determine the thickness of the oil film. Picture the Problem Because there is a λ2

1 phase change due to reflection at both the air-oil and oil-glass interfaces, the condition for constructive interference is that twice the thickness of the oil film equal an integer multiple of the wavelength of light in the film. Express the condition for constructive interference:

'm'''t λλλλ == ,...3,2,2 (1) where λ′ is the wavelength of light in the oil and m = 0, 1, 2, …

nmt λ

=2 ⇒mnt2

=λ

where n is the index of refraction of the oil.

Substitute for λ′ to obtain:

Substitute for the predominant wavelengths to obtain: m

nt2nm690 = and 1

2nm460+

=m

nt

Divide the first of these equations by the second and simplify to obtain: m

m

mntmnt

1

12

2

nm460nm690 +

=

+

= ⇒ 2=m

Solve equation (1) for t:

nmt2λ

=


3063


( )( )( ) nm476

45.12nm6902

==t

30 •• A film of oil that has an index of refraction 1.45 floats on water. When illuminated with white light at normal incidence, light of wavelengths 700 nm and 500 nm is predominant in the reflected light. Determine the thickness of the oil film. Picture the Problem Because the index of refraction of air is less than that of the oil, there is a phase shift of π rad ( λ2

1 ) in the light reflected at the air-oil interface. Because the index of refraction of the oil is greater than that of the glass, there is no phase shift in the light reflected from the oil-glass interface. We can use the condition for constructive interference to determine m for λ = 700 nm and then use this value in our equation describing constructive interference to find the thickness t of the oil film.

,...3,2,2 21 ''''t λλλλ =+ Express the condition for

constructive interference between the waves reflected from the air-oil interface and the oil-glass interface:

or ( ) 'm'''t λλλλ 2

125

23

21 ,...,,2 +== (1)


Substitute for 'λ and solve for λ to obtain: 2

1

2+

=m

ntλ

Substitute the predominant wavelengths to obtain:

21

2nm700+

=m

nt and 23

2nm500+

=m

nt

Divide the first of these equations by the second to obtain:

2123

23

21

2

2

nm500nm700

++

=

+

+=

mm

mnt

mnt

⇒ 2=m

Solve equation (1) for t: ( )

nmt

221 λ

+=


( ) ( ) nm60345.12nm7002 2

1 =+=t

Chapter 33

3064

Newton’s Rings 31 •• [SSM] A Newton’s ring apparatus consists of a plano-convex glass lens with radius of curvature R that rests on a flat glass plate, as shown in Figure 33-42. The thin film is air of variable thickness. The apparatus is illuminated from above by light from a sodium lamp that has a wavelength of 590 nm. The pattern is viewed by reflected light. (a) Show that for a thickness t the condition for a bright (constructive) interference ring is ( )λ2

12 += mt where m = 0, 1, 2, . . . (b) Show that for t << R, the radius r of a fringe is related to t by r = 2tR . (c) For a radius of curvature of 10.0 m and a lens diameter of 4.00 cm. How many bright fringes would you see in the reflected light? (d) What would be the diameter of the sixth bright fringe? (e) If the glass used in the apparatus has an index of refraction n = 1.50 and water replaces the air between the two pieces of glass, explain qualitatively the changes that will take place in the bright-fringe pattern. Picture the Problem This arrangement is essentially identical to a ″thin film″ configuration, except that the ″film″ is air. A phase change of 180° ( λ2

1 ) occurs at the top of the flat glass plate. We can use the condition for constructive interference to derive the result given in (a) and use the geometry of the lens on the plate to obtain the result given in (b). We can then use these results in the remaining parts of the problem.

,...3,2,2 21 λλλλ =+t (a) The condition for

constructive interference is: or ( )λλλλ 2

125

23

21 ,...,,2 +== mt

where λ is the wavelength of light in air and m = 0, 1, 2, …

Solving for t yields: ( ) ...,2,1,0,

221 =+= mmt λ (1)

( ) 222 RtRr =−+

or 2222 2 tRtRrR +−+=

(b) From Figure 33-42 we have:

For t << R we can neglect the last term to obtain:

RtRrR 2222 −+≈ ⇒ Rtr 2= (2)

(c) Square equation (2) and substitute for t from equation (1) to obtain:

( ) λRmr 212 += ⇒

212

−=λR

rm


3065

Substitute numerical values and evaluate m:

( )( )( )

fringes.bright 68be will thereso and

6721

nm590m0.10cm00.2 2

=−=m

(d) The diameter of the mth fringe is: ( ) λRmrD 2

122 +==

Noting that m = 5 for the sixth fringe, substitute numerical values and evaluate D:

( )( )( )cm14.1

nm590m0.1052 21

=

+=D

(e) The wavelength of the light in the film becomes λair/n = 444 nm. The separation between fringes is reduced (the fringes would become more closely spaced.) and the number of fringes that will be seen is increased by a factor of 1.33. 32 •• A plano-convex glass lens of radius of curvature 2.00 m rests on an optically flat glass plate. The arrangement is illuminated from above with monochromatic light of 520-nm wavelength. The indexes of refraction of the lens and plate are 1.60. Determine the radii of the first and second bright fringe from the center in the reflected light. Picture the Problem This arrangement is essentially identical to a ″thin film″ configuration, except that the ″film″ is air. A phase change of 180° ( λ2

1 ) occurs at the top of the flat glass plate. We can use the condition for constructive interference and the results from Problem 31(b) to determine the radii of the first and second bright fringes in the reflected light.

,...3,2,2 21 λλλλ =+t The condition for constructive

interference is: or ( )λλλλ 2

125

23

21 ,...,,2 +== mt

where λ is the wavelength of light in air and m = 0, 1, 2, …

Solving for t gives:

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

1 =+= mmt λ

In Problem 31(b) it was shown that:

tRr 2=

Substitute for t to obtain:

( ) Rmr λ21+=

Chapter 33

3066

The first fringe corresponds to m = 0:

( )( ) mm721.0m00.2nm52021 ==r

The second fringe corresponds to m = 1:

( )( ) mm25.1m00.2nm52023 ==r

33 ••• Suppose that before the lens of Problem 32 is placed on the plate, a film of oil of refractive index 1.82 is deposited on the plate. What will then be the radii of the first and second bright fringes? Picture the Problem This arrangement is essentially identical to a ″thin film″ configuration, except that the ″film″ is oil. A phase change of 180° ( λ2

1 ) occurs at lens-oil interface. We can use the condition for constructive interference and the results from Problem 31(b) to determine the radii of the first and second bright fringes in the reflected light.

,...3,2,2 21 ''''t λλλλ =+ The condition for constructive

interference is: or ( ) 'm'''t λλλλ 2

125

23

21 ,...,,2 +==


Substitute for λ′ and solve for t:

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

1 =+= mn

mt λ

where λ is the wavelength of light in air.

In Problem 31(b) it was shown that:

tRr 2=

Substitute for t to obtain: ( )

nRmr λ

21+=

The first fringe corresponds to m = 0:

( )( ) mm535.082.1

m00.2nm52021

==r

The second fringe corresponds to m = 1:

( )( ) mm926.082.1

m00.2nm52023

==r

Two-Slit Interference Patterns 34 • Two narrow slits separated by 1.00 mm are illuminated by light of wavelength 600 nm, and the interference pattern is viewed on a screen 2.00 m


3067

away. Calculate the number of bright fringes per centimeter on the screen in the region near the center fringe. Picture the Problem The number of bright fringes per unit distance is the reciprocal of the separation of the fringes. We can use the expression for the distance on the screen to the mth fringe to find the separation of the fringes. Express the number N of bright fringes per centimeter in terms of the separation of the fringes:

yN

Δ=

1 (1)

Express the distance on the screen to the mth and (m + 1)st bright fringe:

dLmym

λ= and ( )

dLmym

λ11 +=+

Subtract the first of these equations from the second to obtain: d

Ly λ=Δ

Substitute in equation (1) to obtain: L

dNλ

=

Substitute numerical values and evaluate N: ( )( )

1cm33.8m00.2nm600

mm00.1 −==N

35 • [SSM] Using a conventional two-slit apparatus with light of wavelength 589 nm, 28 bright fringes per centimeter are observed near the center of a screen 3.00 m away. What is the slit separation? Picture the Problem We can use the expression for the distance on the screen to the mth and (m + 1)st bright fringes to obtain an expression for the separation Δy of the fringes as a function of the separation of the slits d. Because the number of bright fringes per unit length N is the reciprocal of Δy, we can find d from N, λ, and L. Express the distance on the screen to the mth and (m + 1)st bright fringe:

dLmym

λ= and ( )

dLmym

λ11 +=+

Subtract the first of these equations from the second to obtain:

dLy λ

=Δ ⇒yLdΔ

=λ

Because the number of fringes per unit length N is the reciprocal of Δy:

LNd λ=

Chapter 33

3068

Substitute numerical values and evaluate d:

( )( )( )mm95.4

m00.3nm589cm28 1

=

= −d

36 • Light of wavelength 633 nm from a helium–neon laser is shone normally on a plane containing two slits. The first interference maximum is 82 cm from the central maximum on a screen 12 m away. (a) Find the separation of the slits. (b) How many interference maxima is it, in principle, possible to observe? Picture the Problem We can use the geometry of the setup, represented to the right, to find the separation of the slits. To find the number of interference maxima that, in principle, can be observed, we can apply the equation describing two-slit interference maxima and require that sinθ ≤ 1.

d

cm 821 =y

m 12=L

θ

θ1

λ

0

Because d << L, we can approximate sinθ1 as:

dλθ ≈1sin ⇒

1sinθλ

≈d (1)

From the right triangle whose sides are L and y1 we have:

( ) ( )06817.0

cm82m12

cm82sin221 =

+=θ

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

m3.9m29.906817.0

nm633 μμ ==≈d

(b) The equation describing two-slit interference maxima is:

...,2,1,0sin == m,md λθ

Because sinθ ≤ 1 determines the maximum number of interference fringes that can be seen:

λmaxmd = ⇒λdm =max

Substitute numerical values and evaluate mmax:

14nm633

m29.9max ==

μm because m must

be an integer.

Because there are 14 fringes on either side of the central maximum:

( ) 29114212 max =+=+= mN


3069

37 •• Two narrow slits are separated by a distance d. Their interference pattern is to be observed on a screen a large distance L away. (a) Calculate the spacing between successive maxima near the center fringe for light of wavelength 500 nm, when L is 1.00 m and d is 1.00 cm. (b) Would you expect to be able to observe the interference of light on the screen for this situation? (c) How close together should the slits be placed for the maxima to be separated by 1.00 mm for this wavelength and screen distance? Picture the Problem We can use the equation for the distance on a screen to the mth bright fringe to derive an expression for the spacing of the maxima on the screen. In (c) we can use this same relationship to express the slit separation d. (a) Express the distance on the screen to the mth and (m + 1)st bright fringe:

dLmym

λ= and ( )

dLmym

λ11 +=+

Subtract the first of these equations from the second to obtain:

dLy λ

=Δ (1)

Substitute numerical values and evaluate Δy:

( )( ) m0.50cm00.1

m00.1nm500 μ==Δy

(b) According to the Raleigh criterion you could resolve them, but not by much. (c) Solve equation (1) for d to obtain:

yLdΔ

=λ

Substitute numerical values and evaluate d:

( )( ) mm500.0mm00.1

m00.1nm500==d

38 •• Light is incident at an angle φ with the normal to a vertical plane containing two slits of separation d (Figure 33-43). Show that the interference maxima are located at angles θm given by sin θm + sin φ = mλ/d. Picture the Problem Let the separation of the slits be d. We can find the total path difference when the light is incident at an angle φ and set this result equal to an integer multiple of the wavelength of the light to obtain the given equation. Express the total path difference:

msinsin θφ dd +=Δl

The condition for constructive interference is:

λm=Δl where m is an integer.

Chapter 33

3070

Substitute to obtain: λθφ mdd =+ msinsin

Divide both sides of the equation by d to obtain: d

mλθφ =+ msinsin

39 •• [SSM] White light falls at an angle of 30º to the normal of a plane containing a pair of slits separated by 2.50 μm. What visible wavelengths give a bright interference maximum in the transmitted light in the direction normal to the plane? (See Problem 38.) Picture the Problem Let the separation of the slits be d. We can find the total path difference when the light is incident at an angle φ and set this result equal to an integer multiple of the wavelength of the light to relate the angle of incidence on the slits to the direction of the transmitted light and its wavelength. Express the total path difference:

θφ sinsin dd +=Δl

The condition for constructive interference is:

λm=Δl where m is an integer.

Substitute to obtain: λθφ mdd =+ sinsin

Divide both sides of the equation by d to obtain: d

mλθφ =+ sinsin

Set θ = 0 and solve for λ:

md φλ sin

=


( )mm

m25.130sinm50.2 μμλ =°

=

Evaluate λ for positive integral values of m:

m λ (nm) 1 1250 2 625 3 417 4 313

From the table we can see that 625 nm and 417 nm are in the visible portion of the electromagnetic spectrum.


3071

40 •• Two small loudspeakers are separated by 5.0 cm, as shown in Figure 33-44. The speakers are driven in phase with a sine wave signal of frequency 10 kHz. A small microphone is placed a distance 1.00 m away from the speakers on the axis running through the middle of the two speakers, and the microphone is then moved perpendicular to the axis. Where does the microphone record the first minimum and the first maximum of the interference pattern from the speakers? The speed of sound in air is 343 m/s. Picture the Problem The diagram shows the two speakers, S1 and S2, the central-bright image and the first-order image to the left of the central-bright image. The distance y is measured from the center of the central-bright image. We can apply the conditions for constructive and destructive interference from two sources and use the geometry of the speakers and microphone to find the distance to the first interference minimum and the distance to the first interference maximum. S S2

1

θ

d

L

y

Relate the distance Δy to the first minimum from the center of the central maximum to θ and the distance L from the speakers to the plane of the microphone:

Ly

=θtan ⇒ θtanLy = (1)

Interference minima occur where:

( )λθ 21sin += md

where m = 0, 1, 2, 3, …

Solve for θ to obtain:

( )⎥⎦⎤

⎢⎣⎡ +

= −

dm λθ 2

11sin

Relate the wavelength λ of the sound waves to the speed of sound v and the frequency f of the sound:

fv

=λ

Substitute for λ in the expression for θ to obtain:

( )⎥⎦

⎤⎢⎣

⎡ += −

dfvm 2

11sinθ

Substituting for θ in equation (1) yields:

( )⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ += −

dfvmLy 2

11sintan (2)

Chapter 33

3072

Noting that the first minimum corresponds to m = 0, substitute numerical values and evaluate Δy:

( ) ( )( )( )( ) cm37

kHz10cm0.5m/s343

sintanm00.1 21

1min1st =

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡= −y

The maxima occur where:

λθ md =sin where m = 1, 2, 3, …

For diffraction maxima, equation (2) becomes: ⎭

⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡= −

afmvLy 1sintan

Noting that the first maximum corresponds to m = 1, substitute numerical values and evaluate y:

( ) ( )( )( )( ) cm94

kHz10cm0.5m/s3431sintanm00.1 1

max1st =⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡= −y

Diffraction Pattern of a Single Slit 41 • Light that has a 600-nm wavelength is incident on a long narrow slit. Find the angle of the first diffraction minimum if the width of the slit is (a) 1.0 mm, (b) 0.10 mm, and (c) 0.010 mm. Picture the Problem We can use the expression locating the first zeroes in the intensity to find the angles at which these zeroes occur as a function of the slit width a. The first zeroes in the intensity occur at angles given by:

aλθ =sin ⇒ ⎟

⎠⎞

⎜⎝⎛= −

aλθ 1sin

(a) For a = 1.0 mm: mrad60.0

mm0.1nm600sin 1 =⎟⎟

⎠

⎞⎜⎜⎝

⎛= −θ

(b) For a = 0.10 mm: mrad0.6

mm10.0nm600sin 1 =⎟⎟

⎠

⎞⎜⎜⎝

⎛= −θ

(c) For a = 0.010 mm: mrad60

mm010.0nm600sin 1 =⎟⎟

⎠

⎞⎜⎜⎝

⎛= −θ


3073

42 • Plane microwaves are incident on the thin metal sheet that has a long, narrow slit of width 5.0 cm in it. The microwave radiation strikes the sheet at normal incidence. The first diffraction minimum is observed at θ = 37º. What is the wavelength of the microwaves? Picture the Problem We can use the expression locating the first zeroes in the intensity to find the wavelength of the radiation as a function of the angle at which the first diffraction minimum is observed and the width of the plate. The first zeroes in the intensity occur at angles given by:

aλθ =sin ⇒ θλ sina=

Substitute numerical values and evaluate λ:

( ) cm0.337sincm0.5 =°=λ

43 ••• [SSM] Measuring the distance to the moon (lunar ranging) is routinely done by firing short-pulse lasers and measuring the time it takes for the pulses to reflect back from the moon. A pulse is fired from Earth. To send the pulse out, the pulse is expanded so that it fills the aperture of a 6.00-in-diameter telescope. Assuming the only thing spreading the beam out is diffraction and that the light wavelength is 500 nm, how large will the beam be when it reaches the Moon, 3.82 × 105 km away? Picture the Problem The diagram shows the beam expanding as it travels to the moon and that portion of it that is reflected from the mirror on the moon expanding as it returns to Earth. We can express the diameter of the beam at the moon as the product of the beam divergence angle and the distance to the moon and use the equation describing diffraction at a circular aperture to find the beam divergence angle.

L Ddtelescope dmirror

Relate the diameter D of the beam when it reaches the moon to the distance to the moon L and the beam divergence angle θ :

LD θ≈ (1)

Chapter 33

3074

The angle θ subtended by the first diffraction minimum is related to the wavelength λ of the light and the diameter of the telescope opening dtelescope by:

telescope

22.1sind

λθ =

Because θ << 1, sinθ ≈ θ and:

telescope

22.1d

λθ ≈

Substitute for θ in equation (1) to obtain:

telescope

22.1d

LD λ=

Substitute numerical values and evaluate D:

( ) ( ) km53.1

cm10m1

incm2.54in00.6

nm50022.1m1082.32

8 =

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

×××=D

Interference-Diffraction Pattern of Two Slits 44 • How many interference maxima will be contained in the central diffraction maximum in the interference–diffraction pattern of two slits if the separation of the slits is exactly 5 times their width? How many will there be if the slit separation is an integral multiple of the slit width (that is d ) for any value of n?

= na

Picture the Problem We need to find the value of m for which the mth interference maximum coincides with the first diffraction minimum. Then there will be fringes in the central maximum. 12 −= mN The number of fringes N in the central maximum is:

12 −= mN (1)

Relate the angle θ1 of the first diffraction minimum to the width a of the slits of the diffraction grating:

aλθ =1sin

Express the angle θm corresponding to the mth interference maxima in terms of the separation d of the slits:

dm

mλθ =sin


3075

Because we require that θ1 = θm, we can equate these expressions to obtain:

adm λλ

= ⇒ adm =

Substituting for d and simplifying yields:

55==

aam

Substitute for m in equation (1) to obtain:

( ) 9152 =−=N

If d = na: na

naadm ===

and 12 −= nN

45 •• [SSM] A two-slit Fraunhofer interference–diffraction pattern is observed using light that has a wavelength equal to 500 nm. The slits have a separation of 0.100 mm and an unknown width. (a) Find the width if the fifth interference maximum is at the same angle as the first diffraction minimum. (b) For that case, how many bright interference fringes will be seen in the central diffraction maximum? Picture the Problem We can equate the sine of the angle at which the first diffraction minimum occurs to the sine of the angle at which the fifth interference maximum occurs to find a. We can then find the number of bright interference fringes seen in the central diffraction maximum using .12 −= mN (a) Relate the angle θ1 of the first diffraction minimum to the width a of the slits of the diffraction grating:

aλθ =1sin

Express the angle θ5 corresponding to the mth fifth interference maxima maximum in terms of the separation d of the slits:

dλθ 5sin 5 =

Because we require that θ1 = θm5, we can equate these expressions to obtain:

adλλ

=5 ⇒

5da =

Substituting the numerical value of d yields:

m0.205

mm100.0 μ==a

(b) Because m = 5: ( ) 915212 =−=−= mN

Chapter 33

3076

46 •• A two-slit Fraunhofer interference–diffraction pattern is observed using light that has a wavelength equal to 700 nm. The slits have widths of 0.010 mm and are separated by 0.20 mm. How many bright fringes will be seen in the central diffraction maximum? Picture the Problem We can equate the sine of the angle at which the first diffraction minimum occurs to the sine of the angle at which the mth interference maximum occurs to find m. We can then find the number of bright interference fringes seen in the central diffraction maximum using .12 −= mN The number of fringes N in the central maximum is:

12 −= mN (1)


aλθ =1sin


dm

mλθ =sin


adm λλ

= ⇒adm =

Substitute for m in equation (1) to obtain:

12−=

adN

Substitute numerical values and evaluate N:

( ) 391mm010.0mm20.02

=−=N

47 •• Suppose that the central diffraction maximum for two slits has 17 interference fringes for some wavelength of light. How many interference fringes would you expect in the diffraction maximum adjacent to one side of the central diffraction maximum? Picture the Problem There are 8 interference fringes on each side of the central maximum. The secondary diffraction maximum is half as wide as the central one. It follows that it will contain 8 interference maxima.


3077

48 •• Light that has a wavelength equal to 550 nm illuminates two slits that both have widths equal to 0.030 mm and separations equal to 0.15 mm. (a) How many interference maxima fall within the full width of the central diffraction maximum? (b) What is the ratio of the intensity of the third interference maximum to one side of the center interference maximum to the intensity of the center interference maximum? Picture the Problem We can equate the sine of the angle at which the first diffraction minimum occurs to the sine of the angle at which the mth interference maximum occurs to find m. We can then find the number of bright interference fringes seen in the central diffraction maximum using .12 −= mN In (b) we can use the expression relating the intensity in a single-slit diffraction pattern to phase

constant θλπφ sin2 a= to find the ratio of the intensity of the third interference

maximum to one side of the center interference maximum. (a) The number of fringes N in the central maximum is:

12 −= mN (1)


aλθ =1sin


dm

mλθ =sin


adm λλ

= ⇒adm =

Substitute in equation (1) to obtain: 12−=

adN

Substitute numerical values and evaluate N:

( ) 91mm030.0mm15.02

=−=N

(b) Express the intensity for a single-slit diffraction pattern as a function of the phase difference φ:

2

21

21

0sin

⎟⎟⎠

⎞⎜⎜⎝

⎛=

φφII (2)

where θλπφ sin2 a=

Chapter 33

3078

dλθ 3sin 3 = For m = 3:

and

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛==

da

daa πλ

λπθ

λπφ 632sin2

3

Substitute numerical values and evaluate φ: 5

6mm15.0mm030.06 ππφ =⎟⎟

⎠

⎞⎜⎜⎝

⎛=

Solve equation (2) for the ratio of I3 to I0:

2

21

21

0

3 sin⎟⎟⎠

⎞⎜⎜⎝

⎛=

φφ

II

Substitute numerical values and evaluate I3/I0:

25.0

56

21

56

21sin

2

0

3 =

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=π

π

II

Using Phasors to Add Harmonic Waves 49 • [SSM] Find the resultant of the two waves whose electric fields at a given location vary with time as follows: iE ˆsin2 01 tA ω=

rand

( )iE ˆsin3 23

02 πω += tAr

. Picture the Problem Chose the coordinate system shown in the phasor diagram. We can use the standard methods of vector addition to find the resultant of the two waves.

y

xδ

1E

r

2E

r

Er

The resultant of the two waves is of the form:

( )iEE δω += tsinrr

(1)

The magnitude of Er

is: ( ) ( ) 02

02

0 6.332 AAA =+=Er

The phase angle δ is:

rad 98.023tan

0

01 −=⎟⎟⎠

⎞⎜⎜⎝

⎛ −= −

AA

δ


3079

Substitute for Er

and δ in equation

(1) to obtain:

( )iE rad 98.0sin6.3 0 −= tA ωr

50 • Find the resultant of the two waves whose electric fields at a given location vary with time as follows: iE ˆsin4 01 tA ω=

rand ( )iE ˆsin3 6

102 πω += tA

r.

Picture the Problem Chose the coordinate system shown in the phasor diagram. We can use the standard methods of vector addition to find the resultant of the two waves.

y

xδ °60

1E

r

2E

r

Er

The resultant of the two waves is of the form:

( )iEE δω += tsinrr

(1)

°−+= 120cos2 1

22

1

2

22 EEEEErrrrr

or

°−+= 120cos2 1

22

1 22 EEEEErrrrr

Apply the law of cosines to the magnitudes of the scalars to obtain:

Substitute for 1Er

and 2Er

and evaluate Er

to obtain:

( ) ( ) ( )( ) 000

20

20 08.6120cos34234 AAAAA =°−+=E

r

Applying the law of sines yields:

EErr

°=

120sinsin

2

δ

Solve for δ to obtain:

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡ °= −

E

Er

r120sin

sin21δ

Substitute numerical values and evaluate δ: rad 43.0

08.6120sin3sin

0

01 =⎥⎦

⎤⎢⎣

⎡ °= −

AA

δ

Chapter 33

3080

Substitute for Er

and δ in equation

(1) to obtain:

( )iE rad 43.0sin1.6 0 += tA ωr

Remarks: We could have used the law of cosines to find R and the law of sines to find δ. 51 •• Monochromatic light is incident on a sheet with a long narrow slit (Figure 33-45). Let I0 be the intensity at the central maximum of the diffraction pattern on a distant screen, and let I be the intensity at the second intensity maximum from the central intensity maximum. The distance from this second intensity maximum to the far edge of the slit is longer than the distance from the second intensity maximum to the near edge of the slit by approximately 2.5 wavelengths. What is the ratio if I to I0? Picture the Problem We can evaluate the expression for the intensity for a single-slit diffraction pattern at the second secondary maximum to express I2 in terms of I0. The intensity at the second secondary maximum is given by:

⇒⎥⎦

⎤⎢⎣

⎡=

2

21

21

0sin

φφ

II2

21

21

0

sin⎥⎦

⎤⎢⎣

⎡=

φφ

II

where

θλπφ sin2 a=

At this second secondary maximum: λθ25sin =a and πλ

λπφ 5

252

=⎟⎠⎞

⎜⎝⎛=

Substitute for φ and evaluate 0II :

0162.0

25

25sin

2

0

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡⎟⎠⎞

⎜⎝⎛

=π

π

II

52 •• Monochromatic light is incident on a sheet that has three long narrow parallel equally spaced slits a distance d apart. (a) Show that the positions of the interference minima on a screen a large distance L away from the sheet that has the three equally spaced slits (spacing d, with d >> λ) are given approximately by

dLmym 3λ= where m = 1, 2, 4, 5, 7, 8, 10, . . . that is, m is not a multiple of 3. (b) For a screen distance of 1.00 m, a light wavelength of 500 nm, and a source spacing of 0.100 mm, calculate the width of the principal interference maxima (the distance between successive minima) for three sources.


3081

Picture the Problem We can use phasor concepts to find the phase angle δ in terms of the number of phasors N (three in this problem) forming a closed polygon of N sides at the minima and then use this information to express the path difference Δr for each of these locations. Applying a small angle approximation, we can obtain an expression for y that we can evaluate for enough of the path differences to establish the pattern given in the problem statement. (a) Express the phase angle δ in terms of the number of phasors N forming a closed polygon of N sides:

⎟⎠⎞

⎜⎝⎛=

Nm πδ 2

where m = 1, 2, 3, 4, 5, 6, ,7, …

For three equally spaced sources, the phase angle is:

⎟⎠⎞

⎜⎝⎛=

32πδ m

Express the path difference corresponding to this phase angle to obtain:

32λδ

πλ mr =⎟⎠⎞

⎜⎝⎛=Δ (1)

Interference maxima occur for:

m = 3, 6, 9, 12, …

Interference minima occur for:

m = 1, 2, 4, 5, 7, 8, … (Note that m is not a multiple of 3.)

θsindr =Δ or, provided the small angle approximation is valid,

Lydr =Δ ⇒ r

dLy Δ=

Express the path difference Δr in terms of sinθ and the separation d of the slits:

Substituting for Δr from equation (1) yields:

...,8,7,5,4,2,1 ,3min == mdLmy λ

(b) For L = 1.00 m, λ = 500 nm, and d = 0.100 mm:

( )( )( ) mm33.3

mm0.1003m00.1nm50022 min ==y

53 •• [SSM] Monochromatic light is incident on a sheet that has four long narrow parallel equally spaced slits a distance d apart. (a) Show that the positions of the interference minima on a screen a large distance L away from four equally spaced sources (spacing d, with d >> λ) are given approximately by

dLmym 4λ= where m = 1, 2, 3, 5, 6, 7, 9, 10, . . . that is, m is not a multiple of 4. (b) For a screen distance of 2.00 m, light wavelength of 600 nm, and a source

Chapter 33

3082

spacing of 0.100 mm, calculate the width of the principal interference maxima (the distance between successive minima) for four sources. Compare this width with that for two sources with the same spacing. Picture the Problem We can use phasor concepts to find the phase angle δ in terms of the number of phasors N (four in this problem) forming a closed polygon of N sides at the minima and then use this information to express the path difference Δr for each of these locations. Applying a small angle approximation, we can obtain an expression for y that we can evaluate for enough of the path differences to establish the pattern given in the problem statement. (a) Express the phase angle δ in terms of the number of phasors N forming a closed polygon of N sides:

⎟⎠⎞

⎜⎝⎛=

Nm πδ 2

where m = 1, 2, 3, 4, 5, 6, ,7, …

For four equally spaced sources, the phase angle is:

⎟⎠⎞

⎜⎝⎛=

2πδ m

Express the path difference corresponding to this phase angle to obtain:

42λδ

πλ mr =⎟⎠⎞

⎜⎝⎛=Δ (1)

Interference maximum occur for: m = 3, 6, 9, 12, …

Interference minima occur for:

m = 1, 2, 4, 5, 7, 8, … (Note that m is not a multiple of 3.)

θsindr =Δ or, provided the small angle approximation is valid,

Lydr =Δ ⇒ r

dLy Δ=

Express the path difference Δr in terms of sinθ and the separation d of the slits:

Substituting for Δr from equation (1) yields:

,...9,7,6,5,3,2,1 ,4min == m

dLmy λ

(b) For L = 2.00 m, λ = 600 nm, d = 0.100 mm, and n = 1:

( )( )( ) mm00.6

mm0.1004m00.2nm60022 min ==y

For two slits: ( )d

Lmy λ21

min22 +

=


3083

For L = 2.00 m, λ = 600 nm, d = 0.100 mm, and m = 0:

( )( ) mm0.12mm0.100

m00.2nm6002 min ==y

The width for four sources is half the width for two sources. 54 •• Light of wavelength 480 nm falls normally on four slits. Each slit is 2.00 μm wide and the center-to-center separation between it and the next slit is 6.00 μm. (a) Find the angular width of the of the central intensity maximum of the single-slit diffraction pattern on a distant screen. (b) Find the angular position of all interference intensity maxima that lie inside the central diffraction maximum. (c) Find the angular width of the central interference. That is, find the angle between the first intensity minima on either side of the central intensity maximum. (d) Sketch the relative intensity as a function of the sine of the angle. Picture the Problem We can use aλθ =sin to find the first zeros in the intensity pattern. The four-slit interference maxima occur at angles given by

,md λθ =sin where m = 0,1,2, … . In (c) we can use the result of Problem 53 to find the angular spread between the central interference maximum and the first interference minimum on either side of it. In (d) we’ll use a phasor diagram for a four-slit grating to find the resultant amplitude at a given point in the intensity pattern as a function of the phase constant δ, that, in turn, is a function of the angle θ that determines the location of a point in the interference pattern. (a) The first zeros in the intensity occur at angles given by: a

λθ =sin ⇒ ⎟⎠⎞

⎜⎝⎛= −

aλθ 1sin

Substitute numerical values and evaluate θ : mrad242

m00.2nm480sin 1 =⎟⎟

⎠

⎞⎜⎜⎝

⎛= −

μθ

(b) The four-slit interference maxima occur at angles given by:

λθ md =sin ⇒ ⎥⎦⎤

⎢⎣⎡= −

dm

mλθ 1sin

where m = 0, 1, 2, 3, …

Substitute numerical values to obtain:

( )

( )m

mm

0800.0sin

m00.6nm480sin

1

1

−

−

=

⎥⎦

⎤⎢⎣

⎡=

μθ

Chapter 33

3084

( )[ ] 00800.00sin 10 == −θ Evaluate θm for m = 0, 1, 2, and 3:

( )[ ] mrad1.800800.01sin 11 ±== −θ

( )[ ] mrad1610800.02sin 12 ±== −θ

( )[ ] rad.24200800.03sin 13 ±== −θ

where θ3 will not be seen as it coincides with the first minimum in the diffraction pattern.

(c) From Problem 53: d

n4minλθ =

For n = 1:

( ) mrad20m00.64

nm480min ==

μθ

(d) Use the phasor method to show the superposition of four waves of the same

amplitude A0 and constant phase difference .sin2 θλπδ d=

A0

A0

A

A0

A0

f

α

δ δ

δ

δ

δ

"

"

δ '

φ

φ

α

Express A in terms of δ ′ and δ ′′:

( )'A''AA δδ coscos2 00 += (1)

Because the sum of the external angles of a polygon equals 2π:

πδα 232 =+


3085

Examining the phasor diagram we see that:

πδα =+ ''

Eliminate α and solve for ''δ to obtain:

δδ 23=''

Because the sum of the internal angles of a polygon of n sides is (n − 2)π :

πδφ 323 =+ ''

From the definition of a straight angle we have:

πδδφ =+− '

Eliminate φ between these equations to obtain:

δδ 21='

Substitute for ''δ and 'δ in equation (1) to obtain:

( )δδ 21

23

0 coscos2 += AA

Because the intensity is proportional to the square of the amplitude of the resultant wave:

( )221

23

0 coscos4 δδ += II

The following graph of I/I0 as a function of sinθ was plotted using a spreadsheet

program. The diffraction envelope was plotted using ,sin42

21

21

2

0⎟⎟⎠

⎞⎜⎜⎝

⎛=

φφ

II where

.sin2 θλπφ a= Note the excellent agreement with the results calculated in (a), (b)

and (c).

Chapter 33

3086

-2

0

2

4

6

8

10

12

14

16

18

-0.3 -0.2 -0.1 0 0.1 0.2 0.3

sin(theta)

I /I 0

intensitydiffraction envelope

55 ••• [SSM] Three slits, each separated from its neighbor by 60.0 μm, are illuminated at the central intensity maximum by a coherent light source of wavelength 550 nm. The slits are extremely narrow. A screen is located 2.50 m from the slits. The intensity is 50.0 mW/m2. Consider a location 1.72 cm from the central maximum. (a) Draw a phasor diagram suitable for the addition of the three harmonic waves at that location. (b) From the phasor diagram, calculate the intensity of light at that location. Picture the Problem We can find the phase constant δ from the geometry of the diagram to the right. Using the value of δ found in this fashion we can express the intensity at the point 1.72 cm from the centerline in terms of the intensity on the centerline. On the centerline, the amplitude of the resultant wave is 3 times that of each individual wave and the intensity is 9 times that of each source acting separately.

m 50.2=L

cm .721y

θ

(a) Express δ for the adjacent slits:

θλπδ sin2 d=

For θ << 1, θθθ ≈≈ tansin :

Ly

=≈ θθ tansin


3087

Substitute to obtain: L

dyλπ

δ2

=

Substitute numerical values and evaluate δ :

( )( )( )( )

°==

=

270rad2

3m50.2nm550

cm72.1m0.602

π

μπδ

The three phasors, 270° apart, are shown in the diagram to the right. Note that they form three sides of a square. Consequently, their sum, shown as the resultant R, equals the magnitude of one of the phasors.

δ

δ

A

A

A

0

0

0

0R = A

(b) Express the intensity at the point 1.72 cm from the centerline:

2RI ∝

Because I0 ∝ 9R2: 2

2

0 9RR

II= ⇒

90I

I =

Substitute for I0 and evaluate I: 2

2

mW/m56.59

mW/m0.50==I

56 ••• In single-slit Fraunhofer diffraction, the intensity pattern (Figure 33-11) consists of a broad central maximum with a sequence of secondary maxima to

either side of the central maximum. This intensity is given by 2

21

21

0

sin⎟⎟⎠

⎞⎜⎜⎝

⎛=

φφ

II ,

where φ is the phase difference between the wavelets arriving from the opposite edges of the slits. Calculate the values of φ for the first three secondary maxima to one side of the central maximum by finding the values of φ for which dI/dφ is equal to zero. Check your results by comparing your answers with approximate values for φ of 3π, 5π and 7π. (That these values for φ are approximately correct at the secondary intensity maxima is discussed in the discussion surrounding Figure 33-27.)

Chapter 33

3088

Picture the Problem We can use the phasor diagram shown in Figure 33-26 to determine the first three values of φ that produce secondary intensity maxima. Setting the derivative of Equation 33-19 equal to zero will yield a transcendental equation whose roots are the values of φ corresponding to the intensity maxima in the diffraction pattern. Referring to Figure 33-26 we see that the first subsidiary maximum occurs where:

πφ 3=

An intensity minimum occurs where:

πφ 4=

Another intensity maximum occurs where:

πφ 5=

Thus, secondary intensity maxima occur where:

( ) ...,3,2,112 =+= n,n πφ and the first three secondary intensity maxima are at πππφ 7 and ,5,3=

The intensity in the single-slit diffraction pattern is given by:

2

21

21

0

sin⎟⎟⎠

⎞⎜⎜⎝

⎛=

φφ

II

Set the derivative of this expression equal to zero for extreme values (relative minima and maxima):

( )extremafor 0

sincossin2 2

21

21

21

21

41

21

21

0 =⎥⎥⎦

⎤

⎢⎢⎣

⎡ −⎟⎟⎠

⎞⎜⎜⎝

⎛=

φφφφ

φφ

φI

ddI

Simplify to obtain the transcendental equation:

φφ 21

21tan =

Solve this equation numerically (use the ″Solver″ function of your calculator or trial-and-error methods) to obtain:

πππφ 6.94 and 92.4 86.2 ,,=

At the three intensity minima φ = 2π, 4π, and 6π , and at the three intensity maxima φ = 2.86π, 4.92π, and 6.94π. At the intensity maxima φ ≈ 3π, 5π, and 7π.


3089

Remarks: Note that our results in (b) are smaller than the approximate values found in (a) by 4.9%, 1.6%, and 0.86% and that the agreement improves as n increases. Diffraction and Resolution 57 • [SSM] Light that has a wavelength equal to 700 nm is incident on a pinhole of diameter 0.100 mm. (a) What is the angle between the central maximum and the first diffraction minimum for a Fraunhofer diffraction pattern? (b) What is the distance between the central maximum and the first diffraction minimum on a screen 8.00 m away? Picture the Problem We can use

Dλθ 22.1= to find the angle between

the central maximum and the first diffraction minimum for a Fraunhofer diffraction pattern and the diagram to the right to find the distance between the central maximum and the first diffraction minimum on a screen 8 m away from the pinhole.

L

minyθ

D

(a) The angle between the central maximum and the first diffraction minimum for a Fraunhofer diffraction pattern is given by:

Dλθ 22.1=

Substitute numerical values and evaluate θ : mrad54.8

mm100.0nm70022.1 =⎟⎟

⎠

⎞⎜⎜⎝

⎛=θ

(b) Referring to the diagram, we see that:

θtanmin Ly =

Substitute numerical values and evaluate ymin:

( ) ( )cm83.6

mrad54.8tanm00.8min

=

=y

58 • Two sources of light that both have wavelengths equal to 700 nm are 10.0 m away from the pinhole of Problem 57. How far apart must the sources be for their diffraction patterns to be resolved by Rayleigh’s criterion?

Chapter 33

3090

Picture the Problem We can apply Rayleigh’s criterion to the overlapping diffraction patterns and to the diameter D of the pinhole to obtain an expression that we can solve for Δy.

Pinhole

α

αΔy

L

c

c

Rayleigh’s criterion is satisfied provided:

Dλα 22.1c =

Relate αc to the separation Δy of the light sources:

LyΔ

≈cα provided αc << 1.

Equate these expressions to obtain: DL

y λ22.1=Δ⇒

DLy λ22.1=Δ


( )( ) cm54.8mm100.0

m0.10nm70022.1 ==Δy

59 • Two sources of light that both have wavelengths of 700 nm are separated by a horizontal distance x. They are 5.00 m from a vertical slit of width 0.500 mm. What is the smallest value of x for which the diffraction pattern of the sources can be resolved by Rayleigh’s criterion? Picture the Problem We can use Rayleigh’s criterion for slits and the geometry of the diagram to the right showing the overlapping diffraction patterns to express x in terms of λ, L, and the width a of the slit.

c

cLα

α

x


3091

Referring to the diagram, relate αc, L, and x: L

x≈cα

For slits, Rayleigh’s criterion is: a

λα =c

Equate these two expressions to obtain:

aLx λ= ⇒

aLx λ

=

Substitute numerical values and evaluate x:

( )( ) mm00.7mm500.0

m00.5nm700==x

60 •• The ceiling of your lecture hall is probably covered with acoustic tile, which has small holes separated by about 6.0 mm. (a) Using light that has a wavelength of 500 nm, how far could you be from this tile and still resolve these holes? Assume the diameter of the pupil of your eye is about 5.0 mm. (b) Could you resolve these holes better using red light or using violet light? Explain your answer. Picture the Problem We can use Rayleigh’s criterion for circular apertures and the geometry of the diagram to the right showing the overlapping diffraction patterns to express L in terms of λ, x, and the diameter D of your pupil.

Your pupil

α

α

x

L

c

c

(a) Referring to the diagram, relate αc, L, and x: L

x≈cα provided α << 1

For circular apertures, Rayleigh’s criterion is:

Dλα 22.1c =


DLx λ22.1= ⇒

λ22.1xDL =

Chapter 33

3092


( )( )( ) m49

nm50022.1mm0.5mm0.6

==L

(b) Because L is inversely proportional to λ, the holes can be resolved better with violet light which has a shorter wavelength. The critical angle for resolution is proportional to the wavelength. Thus, the shorter the wavelength the farther away you can be and still resolve the two images. 61 •• [SSM] The telescope on Mount Palomar has a diameter of 200 in. Suppose a double star were 4.00 light-years away. Under ideal conditions, what must be the minimum separation of the two stars for their images to be resolved using light that has a wavelength equal to 550 nm? Picture the Problem We can use Rayleigh’s criterion for circular apertures and the geometry of the diagram to obtain an expression we can solve for the minimum separation Δx of the stars.

Your pupil

Δα

α

x

L

c

c


Dλα 22.1c =

Relate αc to the separation Δx of the light sources:

LxΔ

≈cα because αc << 1

Equate these expressions to obtain: DL

x λ22.1=Δ ⇒DLx λ22.1=Δ

Substitute numerical values and evaluate Δx:

( )m1000.5

in1cm2.54in200

y1m10461.9y4nm550

22.1 9

15

×=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

×

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅×

×⋅=Δ

cc

x


3093

62 •• The star Mizar in Ursa Major is a binary system of stars of nearly equal magnitudes. The angular separation between the two stars is 14 seconds of arc. What is the minimum diameter of the pupil that allows resolution of the two stars using light that has a wavelength equal to 550 nm? Picture the Problem We can use Rayleigh’s criterion for circular apertures and the geometry of the diagram to obtain an expression we can solve for the minimum diameter D of the pupil that allows resolution of the binary stars.

Your pupil

Δα

α

x

L

c

cD


Dλα 22.1c = ⇒

c

22.1αλ

=D


cm1mm9.9180

rad3600

114

nm55022.1

≈=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

°×

°×

=π

''''

D

Diffraction Gratings 63 • [SSM] A diffraction grating that has 2000 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. (a) At what angles in the first-order spectrum would you expect to find the two violet lines that have wavelengths of 434 nm and 410 nm? (b) What are the angles if the grating has 15 000 slits per centimeter? Picture the Problem We can solve λθ md =sin for θ with m = 1 to express the location of the first-order maximum as a function of the wavelength of the light.

λθ md =sin where d is the separation of the slits and m = 0, 1, 2, …

(a) The interference maxima in a diffraction pattern are at angles θ given by:

Chapter 33

3094

Solve for the angular location θm of the maxima : ⎟

⎠⎞

⎜⎝⎛= −

dm

mλθ 1sin

Relate the number of slits N per centimeter to the separation d of the slits:

dN 1=

Substitute for d to obtain:

( )λθ mNm1sin−= (1)

Evaluateθ for λ = 434 nm and m =1: ( )( )[ ]mrad9.86

nm434cm2000sin 111

=

= −−θ

Evaluateθ for λ = 410 nm and m = 1: ( )( )[ ]

mrad1.82

nm410cm2000sin 111

=

= −−θ

(b) Use equation (1) to evaluateθ for λ = 434 nm and m = 1:

( )( )[ ]mrad 709

nm434cm15000sin 111

=

= −−θ

Evaluate θ for λ = 410 nm and m = 1: ( )( )[ ]

mrad 662

nm410cm15000sin 111

=

= −−θ

64 • Using a diffraction grating that has 2000 lines per centimeter, two other lines in the first-order hydrogen spectrum are found at angles of 9.72 × 10–2 rad and 1.32 × 10–1 rad. What are the wavelengths of these lines? Picture the Problem We can solve λθ md =sin for λ with m = 1 to express the location of the first-order maximum as a function of the angles at which the first-order images are found. The interference maxima in a diffraction pattern are at angles θ given by:

λθ md =sin ⇒m

d θλ sin=

where d is the separation of the slits and m = 0, 1, 2, …

Relate the number of slits N per centimeter to the separation d of the slits:

dN 1=


3095

Let m = 1 and substitute for d to obtain: N

θλ sin=

Substitute numerical values and evaluate λ1 for θ1 = 9.72 ×10–2 rad:

( ) nm485cm2000

rad1072.9sin1

2

1 =×

= −

−

λ

Substitute numerical values and evaluate λ1 for θ 2 = 1.32 ×10–1 rad:

( ) nm658cm2000

rad1032.1sin1

1

1 =×

= −

−

λ

65 • The colors of many butterfly wings and beetle carapaces are due to effects of diffraction. The Morpho butterfly has structural elements on its wings that effectively act as a diffraction grating with spacing 880 nm. At what angle will the first diffraction maximum occur for normally incident light diffracted by the butterfly’s wings? Assume the light is blue with a wavelength of 440 nm. Picture the Problem We can use the grating equation to find the angle at which normally incident blue light will be diffracted by the butterfly’s wings. The grating equation is:

λθ md =sin , where m = 1, 2, 3, …

Solve for θ to obtain: ⎥⎦

⎤⎢⎣⎡= −

dmλθ 1sin

Substitute numerical values and evaluate θ1:

( )( )°=⎥

⎦

⎤⎢⎣

⎡= − 0.30

nm880nm4401sin 1θ

66 •• A diffraction grating that has 2000 slits per centimeter is used to analyze the spectrum of mercury. (a) Find the angular separation in the first-order spectrum of the two lines of wavelength 579 nm and 577 nm. (b) How wide must the beam on the grating be for these lines to be resolved? Picture the Problem We can use the grating equation to find the angular separation of the first-order spectrum of the two lines. In Part (b) we can apply the definition of the resolving power of the grating to find the width of the grating that must be illuminated for the lines to be resolved. (a) Express the angular separation in the first-order spectrum of the two lines:

577579 θθθ −=Δ

Solve the grating equation for θ : ⎟⎠⎞

⎜⎝⎛= −

dmλθ 1sin

Chapter 33

3096

Substitute for θ579 and θ577 to obtain:

( ) ( ) [ ] [ mmmm 1154.0sin1158.0sin

cm20001

nm577sin

cm20001

nm579sin 11

1

1

1

1 −−

−

−

−

− −=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=Δθ ]

For m = 1:

[ ] [ ]°=°−°=

−=Δ −−

02.063.665.6

1154.0sin1158.0sin 11 mmθ

(b) The width of the beam necessary for these lines to be resolved is given by:

Ndw = (1)

Relate the resolving power of the diffraction grating to the number of slits N that must be illuminated in order to resolve these wavelengths in the mth order:

mN=Δλλ

For m = 1: λλΔ

=N

Substitute for N in equation (1) to obtain: λ

λΔ

=dw

Letting λ be the average of the two wavelengths, substitute numerical values and evaluate w:

( )mm1

nm2cm2000

1nm578 1

=⎟⎟⎠

⎞⎜⎜⎝

⎛

=−

w

67 •• [SSM] A diffraction grating that has 4800 lines per centimeter is illuminated at normal incidence with white light (wavelength range of 400 nm to 700 nm). How many orders of spectra can one observe in the transmitted light? Do any of these orders overlap? If so, describe the overlapping regions. Picture the Problem We can use the grating equation ... 3, 2, 1, sin == m,md λθ to express the order number in terms of the slit separation d, the wavelength of the light λ, and the angle θ.


3097

The interference maxima in the diffraction pattern are at angles θ given by:

λθ md =sin ⇒λθsindm =

where m = 1, 2, 3, …

If one is to see the complete spectrum, it must be true that:

1sin ≤θ ⇒λdm ≤

Evaluate mmax:

98.2nm700cm4800

1cm4800

11

max

1

max ===−−

λm

Because mmax = 2.98, one can see the complete spectrum only for m = 1 and 2.

Express the condition for overlap:

2211 λλ mm ≥

One can see the complete spectrum for only the first and second order spectra. That is, only for m = 1 and 2. Because 700 nm < 2 × 400 nm, there is no overlap of the second-order spectrum into the first-order spectrum; however, there is overlap of long wavelengths in the second order with short wavelengths in the third-order spectrum. 68 •• A square diffraction grating that has an area of 25.0 cm2 has a resolution of 22 000 in the fourth order. At what angle should you look to see a wavelength of 510 nm in the fourth order? Picture the Problem We can use the grating equation and the resolving power of the grating to derive an expression for the angle at which you should look to see a wavelength of 510 nm in the fourth order. The interference maxima in the diffraction pattern are at angles θ given by:

... 3, 2, 1, where sin

==m

,md λθ (1)

The resolving power R is given by:

mNR = where N is the number of slits and m is the order number.

Relate d to the width w of the grating:

Nwd =

Substitute for N and simplify to obtain: R

mwd =

Chapter 33

3098

Substitute for d in equation (1) to obtain:

λθ mR

mw=sin ⇒ ⎟

⎠⎞

⎜⎝⎛= −

wRλθ 1sin


( )( )°=⎥

⎦

⎤⎢⎣

⎡= − 0.13

cm00.5nm510000,22sin 1θ

69 •• Sodium light that has a wavelength equal to 589 nm falls normally on a 2.00-cm-square diffraction grating ruled with 4000 lines per centimeter. The Fraunhofer diffraction pattern is projected onto a screen a distance of 1.50 m from the grating by a 1.50-m-focal-length lens that is placed immediately in front of the grating. Find (a) the distance of the first and second order intensity maxima from the central intensity maximum, (b) the width of the central maximum, and (c) the resolution in the first order. (Assume the entire grating is illuminated.) Picture the Problem The distance on the screen to the mth bright fringe can be found using λθ md =sin (where d is the slit separation and m = 0, 1, 2, …) and the geometry of the grating and projection screen. We can use

LyNd 2min Δ== λθ to find the width of the central maximum and R = mN, where N is the number of slits in the grating, to find the resolving power in the first order. (a) The angle θ at which maxima occur is related to the slit separation d, the wavelength of the incident light λ, and the order number m according to:

λθ md =sin (1) where m = 0, 1, 2, … .

θ is also related to the distance to the screen L and the positions of the intensity maxima ym:

22sin

m

m

yLy+

=θ

Substituting for sin θ in equation (1) yields:

λmyL

dy

m

m =+ 22

⇒222 λ

λmdLmym

−=

Substituting numerical values yields:

( )( )

( )222

nm5894000

cm00.1

m50.1nm589

m

mym

−⎟⎠⎞

⎜⎝⎛

=


3099

Evaluate this expression for m = 1 and m = 2 to obtain:

cm4.361 =y and cm1.802 =y

(b) The angle θmin that locates the first minima in the diffraction pattern is given by:

Ly

Nd 2minΔ

==λθ ⇒

NdLy λ2

=Δ

where Δy is the width of the central maximum.


( )( )

( )

m4.88

cm40001lines8000

nm589m50.12

1

μ=

⎟⎟⎠

⎞⎜⎜⎝

⎛=Δ

−

y

(c) The resolving power R in the mth order is given by:

mNR =

Substitute numerical values and evaluate R:

( )( ) 800080001 ==R

70 •• The spectrum of neon is exceptionally rich in the visible region. Among the many lines are two lines at wavelengths of 519.313 nm and 519.322 nm. If light from a neon discharge tube is normally incident on a transmission grating with 8400 lines per centimeter and the spectrum is observed in second order, what must be the width of the grating that is illuminated, so that these two lines can be resolved? Picture the Problem The width of the grating w is the product of its number of lines N and the separation of its slits d. Because the resolution of the grating is a function of the average wavelength, the difference in the wavelengths, and the order number, we can express w in terms of these quantities. Express the width w of the grating as a function of the number of lines N and the slit separation d:

Ndw =

The resolving power R of the grating is given by:

mNR =Δ

=λλ ⇒

λλΔ

=m

N

Substitute for N in the expression for w to obtain:

λλΔ

=m

dw

Chapter 33

3100

Letting λ be the average of the given wavelengths, substitute numerical values and evaluate w:

( )

( ) cm3nm519.313nm322.5192

cm84001nm519.322nm313.519 12

1

=−

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=−

w

71 •• [SSM] Mercury has several stable isotopes, among them 198Hg and 202Hg. The strong spectral line of mercury, at about 546.07 nm, is a composite of spectral lines from the various mercury isotopes. The wavelengths of this line for 198Hg and 202Hg are 546.07532 nm and 546.07355 nm, respectively. What must be the resolving power of a grating capable of resolving these two isotopic lines in the third-order spectrum? If the grating is illuminated over a 2.00-cm-wide region, what must be the number of lines per centimeter of the grating? Picture the Problem We can use the expression for the resolving power of a grating to find the resolving power of the grating capable of resolving these two isotopic lines in the third-order spectrum. Because the total number of the slits of the grating N is related to width w of the illuminated region and the number of lines per centimeter of the grating and the resolving power R of the grating, we can use this relationship to find the number of lines per centimeter of the grating. The resolving power of a diffraction grating is given by:

mNR =Δ

=λλ (1)

Substitute numerical values and evaluate R: 55 1009.3100852.3

07355.54607532.54607532.546

×=×=

−=R

Express n, be the number of lines per centimeter of the grating, in terms of the total number of slits N of the grating and the width w of the grating:

wNn =

From equation (1) we have: m

RN =

Substitute for N to obtain: mw

Rn =


3101

Substitute numerical values and evaluate n: ( )( )

145

cm1014.5cm00.23100852.3 −×=

×=n

72 ••• A diffraction grating has n lines per unit length. Show that the angular separation (Δθ ) of two lines of wavelengths λ and λ + Δλ is approximately

222

1 λλθ −=mn

ΔΔ where m is the order number.

Picture the Problem We can differentiate the grating equation implicitly and approximate dθ /dλ by Δθ /Δλ to obtain an expression Δθ as a function of m, n, Δλ, and cosθ. We can use the Pythagorean identity sin2θ + cos2θ = 1 and the grating equation to write cosθ in terms of n, m, and λ. Making these substitutions will yield the given equation. The grating equation is:

... 2, 1, 0, sin == , mmd λθ (1)

( ) ( )λλ

θλ

mddd

dd

=sin Differentiate both sides of this equation with respect to λ:

or

mddd =λθθcos

Because n = 1/d:

nmdd

=λθθcos ⇒

λθθ

dd

mn cos1=

Approximate dθ /dλ by Δθ /Δλ: θ

λθ cos1

ΔΔ

=m

n

Solving for Δθ yields: θ

λθcosΔ

=Δnm

Substitute for cosθ to obtain:

θ

λθ2sin1−

Δ=Δ

nm

From equation (1): λλθ nm

dm

==sin

Substituting for sinθ yields:

2221 λ

λθmn

nm−

Δ=Δ

Chapter 33

3102

Simplify by dividing the numerator and denominator by nm to obtain:

22222

222222 1111λ

λλ

λ

λ

λθ−

Δ=

−

Δ=

−

Δ=Δ

mnmnmnmn

nm

73 ••• [SSM] For a diffraction grating in which all the surfaces are normal to the incident radiation, most of the energy goes into the zeroth order, which is useless from a spectroscopic point of view, since in zeroth order all the wavelengths are at 0º. Therefore, modern reflection gratings have shaped, or blazed, grooves, as shown in Figure 33-45. This shifts the specular reflection, which contains most of the energy, from the zeroth order to some higher order. (a) Calculate the blaze angle φm in terms of the groove separation d, the wavelength λ, and the order number m in which specular reflection is to occur for m = 1, 2, . . . . (b) Calculate the proper blaze angle for the specular reflection to occur in the second order for light of wavelength 450 nm incident on a grating with 10 000 lines per centimeter. Picture the Problem We can use the grating equation and the geometry of the grating to derive an expression for φm in terms of the order number m, the wavelength of the light λ, and the groove separation d. (a) Because θi = θr, application of the grating equation yields:

( )... 2, 1, 0, where

2sin i

==

m, md λθ

(1)

Because φ and θi have their left and right sides mutually perpendicular:

mi φθ =

Substitute for θi to obtain:

( ) λφ md =m2sin

Solving for φm yields: ⎟⎠⎞

⎜⎝⎛= −

dm λφ 1

21

m sin

(b) For m = 2:

°=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=

−

− 1.32

cm000,101

nm4502sin1

121

2φ

74 ••• In this problem, you will derive the relation mNR == λλ Δ (Equation 33-27) for the resolving power of a diffraction grating containing N slits separated by a distance d. To do this, you will calculate the angular


3103

separation between the intensity maximum and intensity minimum for some wavelength λ and set it equal to the angular separation of the mth-order maximum for two nearby wavelengths. (a) First show that the phase difference φ between

the waves from two adjacent slits is given by φ =

2πdλ

sin θ . (b) Next differentiate

this expression to show that a small change in angle dθ results in a change in

phase of dφ given by dφ =

2πdλ

cosθ dθ . (c) Then for N slits, the angular

separation between an interference maximum and an interference minimum corresponds to a phase change of dφ = 2 π/N. Use this to show that the angular separation dθ between the intensity maximum and intensity minimum for some

wavelength λ is given by dθ =

λNd cosθ

. (d) Next use the fact that the angle of

the mth-order interference maximum for wavelength λ is specified by λθ md =sin (Equation 33-26). Compute the differential of each side of this

equation to show that angular separation of the mth-order maximum for two

nearly equal wavelengths differing by dλ is given by dθ =

mdλd cosθ

. (e) According

to Rayleigh’s criterion, two wavelengths will be resolved in the mth order if the angular separation of the wavelengths, given by the Part (d) result, equals the angular separation of the interference maximum and the interference minimum given by the Part (c) result. Use this to arrive at mNR == λλ Δ (Equation 33-27) for the resolving power of a grating. Picture the Problem We can follow the procedure outlined in the problem statement to obtain mNR == λλ Δ .

(a) Express the relationship between the phase difference φ and the path difference Δr:

λπφ rΔ

=2

⇒ λπφ rΔ

=2

Because Δr = dsinθ : θ

λπφ sin2 d

=

(b) Differentiate this expression with respect to θ to obtain:

θλπθ

λπ

θθφ cos2sin2 dd

dd

dd

=⎥⎦⎤

⎢⎣⎡=

Solve for dφ: θθ

λπφ ddd cos2

=

(c) From Part (b):

θπφλθ

cos2 ddd =

Chapter 33

3104

Substitute 2π/N for dφ to obtain: θ

λθcosNd

d =

(d) Equation 33-26 is: ... 2, 1, 0, sin == m,md λθ

[ ] [ ]λλ

θλ

mddd

dd

=sin

or

mddd =λθθcos

Differentiate this expression implicitly with respect to λ to obtain:

Solve for dθ to obtain: θ

λθcosd

mdd =

(e) Equate the two expressions for dθ obtained in (c) and (d): θ

λθ

λcoscos d

mdNd

=

Approximating dλ by Δλ and allowing for the possibility that Δλ < 0 yields:

θλ

θλ

coscos dm

NdΔ

=

Solving for R = λ/Δλ yields: mNR ==

λλΔ

General Problems 75 • [SSM] Naturally occurring coronas (brightly colored rings) are sometimes seen around the Moon or the Sun when viewed through a thin cloud. (Warning: When viewing a sun corona, be sure that the entire sun is blocked by the edge of a building, a tree, or a traffic pole to safeguard your eyes.) These coronas are due to diffraction of light by small water droplets in the cloud. A typical angular diameter for a coronal ring is about 10º. From this, estimate the size of the water droplets in the cloud. Assume that the water droplets can be modeled as opaque disks with the same radius as the droplet, and that the Fraunhofer diffraction pattern from an opaque disk is the same as the pattern from an aperture of the same diameter. (This last statement is known as Babinet’s principle.) Picture the Problem We can use Dλθ 22.1sin = to relate the diameter D of the opaque-disk water droplets to the angular diameter θ of a coronal ring and to the wavelength of light. We’ll assume a wavelength of 500 nm.


3105

The angle θ subtended by the first diffraction minimum is related to the wavelength λ of light and the diameter D of the opaque-disk water droplet:

Dλθ 22.1sin =

Because of the great distance to the cloud of water droplets, θ << 1 and:

Dλθ 22.1≈ ⇒

θλ22.1

=D


( ) m5.3

180rad10

nm50022.1 μπ

=

°×°

=D

76 • An artificial corona (see Problem 75) can be made by placing a suspension of polystyrene microspheres in water. Polystyrene microspheres are small, uniform spheres made of plastic with an index of refraction equal to 1.59. Assuming that the water has an index of refractive equal to 1.33, what is the angular diameter of such an artificial corona if 5.00-μm-diameter particles are illuminated by light from a helium–neon laser with wavelength in air of 632.8 nm? Picture the Problem We can use Dnλθ 22.1sin = to relate the diameter D of a microsphere to the angular diameter θ of a coronal ring and to the wavelength of light in water. The angle θ subtended by the first diffraction minimum is related to the wavelength λn of light in water and the diameter D of the microspheres:

nDDn λλθ 22.122.1sin ==

Because θ << 1: nD

λθ 22.1≈


( )( )( )( )

°=

=≈

65.6

rad116.0m00.533.1

nm8.63222.1μ

θ

77 • Coronas (see Problem 74) can be caused by pollen grains, typically of birch or pine. Such grains are irregular in shape, but they can be treated as if they had an average diameter of about 25 μm. What is the angular diameter (in degrees) of the corona for blue light? What is the diameter (in degrees) of the corona for red light?

Chapter 33

3106

Picture the Problem We can use Dλθ 22.1sin = to relate the diameter D of a pollen grain to the angular diameter θ of a coronal ring and to the wavelength of light. We’ll assume a wavelength of 450 nm for blue light and 650 nm for red light.

Dλα 22.1sin =

or, because θα 21= where θ is the

angular diameter of the coronal ring,

Dλθ 22.1sin 2

1 =

The angleα subtended by the first diffraction intensity minima is related to the wavelength λ of light and to the diameter D of the microspheres:

Because θ << 1, sinθ ≈ θ and: D

λθ 22.121 ≈ ⇒

Dλθ 44.2≈

Substitute numerical values and evaluate θ for red light:

( )

°=

×=≈ −

6.3

rad10344.6m25

nm65044.2 2red μ

θ

Substitute numerical values and evaluate θ for blue light:

( )

°=

×=≈ −

5.2

rad10392.4m25

nm45044.2 2blue μ

θ

78 • Light from a He-Ne laser (632.8 nm wavelength) is directed upon a human hair, in an attempt to measure its diameter by examining the diffraction pattern. The hair is mounted in a frame 7.5 m from a wall, and the central diffraction maximum is measured to be 14.6 cm wide. What is the diameter of the hair? (The diffraction pattern of a hair with diameter d is the same as the diffraction pattern of a single slit with width a = d. See Babinet’s principle, Problem 75.) Picture the Problem The diagram shows the hair whose diameter d = a, the screen a distance L from the hair, and the separation Δy of the first diffraction peak from the center. We can use the geometry of the experiment to relate Δy to L and a and the condition for diffraction maxima to express θ1 in terms of the diameter of the hair and the wavelength of the light illuminating the hair.

yaθ Δ

L

1


3107

Relate θ to Δy: L

yΔ= 2

1

tanθ ⇒ ⎟⎠⎞

⎜⎝⎛ Δ= −

Ly

2tan 1θ

Diffraction maxima occur where: ( )λθ 2

1sin += ma ⇒( )

θλ

sin21+

=m

a

where m = 1, 2, 3, …

Substituting for θ yields:

( )

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ Δ

+=

−

Ly

ma

2tansin 1

21 λ

Substitute numerical values and evaluate a for m = 1:

( )( )

( )

m 98

m 5.72cm 6.14tansin

nm 8.6321

1

21

μ=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

+=

−

a

79 • [SSM] A long, narrow horizontal slit lies 1.00 μm above a plane mirror, which is in the horizontal plane. The interference pattern produced by the slit and its image is viewed on a screen 1.00 m from the slit. The wavelength of the light is 600 nm. (a) Find the distance from the mirror to the first maximum. (b) How many dark bands per centimeter are seen on the screen? Picture the Problem We can apply the condition for constructive interference to find the angular position of the first maximum on the screen. Note that, due to reflection, the wave from the image is 180o out of phase with that from the source. (a) Because y0 << L, the distance from the mirror to the first maximum is given by:

00 θLy = (1)

Express the condition for constructive interference:

( )λθ 21sin += md

where m = 0, 1, 2, …

Solving for θ yields:

( ) ⎥⎦⎤

⎢⎣⎡ += −

dm λθ 2

11sin

For the first maximum, m = 0 and:

⎥⎦⎤

⎢⎣⎡= −

d2sin 1

0λθ

Substitute in equation (1) to obtain: ⎥⎦

⎤⎢⎣⎡= −

dLy

2sin 1

0λ

Chapter 33

3108

Because the image of the slit is as far behind the mirror’s surface as the slit is in front of it, d = 2.00 μm. Substitute numerical values and evaluate y0:

( ) ( )cm1.15

m00.22nm600sinm00.1 1

0

=

⎥⎦

⎤⎢⎣

⎡= −

μy

(b) The separation of the fringes on the screen is given by:

dLy λ

=Δ

The number of dark bands per unit length is the reciprocal of the fringe separation:

Ld

yn

λ=

Δ=

1

Substitute numerical values and evaluate n: ( )( )

1m33.3m00.1nm600

m00.2 −==μn

80 • A radio telescope is situated at the edge of a lake. The telescope is looking at light from a radio galaxy that is just rising over the horizon. If the height of the antenna is 20 m above the surface of the lake, at what angle above the horizon will the radio galaxy be when the telescope is centered in the first intensity interference maximum of the radio waves? Assume the wavelength of the radio waves is 20 cm. Hint: The interference is caused by the light reflecting off the lake and remember that this reflection will result in a 180º phase shift. Picture the Problem The radio waves from the galaxy reach the telescope by two paths; one coming directly from the galaxy and the other reflected from the surface of the lake. The radio waves reflected from the surface of the lake are phase shifted 180°, relative to the radio waves reaching the telescope directly, by reflection from the surface of the lake. We can use the condition for constructive interference of two waves to find the angle above the horizon at which the radio waves from the galaxy will interfere constructively.

θ

d

θr

θP

Telescope

Δ

Radio waves directly from galaxy

Radio waves r

eflected fro

m the la

ke


3109

Because the reflected radio waves are phase shifted by 180°, the condition for constructive interference at point P is:

( )λ21+=Δ mr

where m = 0, 1, 2, …

Referring to the figure, note that: drΔ

≈θsin ⇒ ⎥⎦⎤

⎢⎣⎡Δ= −

dr1sinθ

Substitute for Δr to obtain: ( )

⎥⎦⎤

⎢⎣⎡ +

= −

dm λ

θ 21

1sin

Noting that m = 0 for the first interference maximum, substitute numerical values and evaluate θ0:

( )

°=

×=⎥⎦

⎤⎢⎣

⎡= −−

29.0

rad1000.5m20cm20

sin 321

10θ

81 • The diameter of the radio telescope at Arecibo, Puerto Rico, is 300 m. What is the smallest angular separation of two objects that this telescope can detect when it is tuned to detect microwaves of 3.2-cm wavelength? Picture the Problem The resolving power of a telescope is the ability of the instrument to resolve two objects that are close together. Hence we can use Rayleigh’s criterion to find the resolving power of the Arecibo telescope. Rayleigh’s criterion for resolution is: D

λα 22.1c =

Substitute numerical values and evaluate αc: mrad13.0

m300cm2.322.1c =⎟⎟

⎠

⎞⎜⎜⎝

⎛=α

82 •• A thin layer of a transparent material that has an index of refraction of 1.30 is used as a nonreflective coating on the surface of glass that has an index of refraction of 1.50. What should the minimum thickness of the material be for the material to be nonreflecting for light that has a wavelength 600 nm? Picture the Problem Note that reflection at both surfaces involves a phase shift of π rad. We can apply the condition for destructive interference to find the thickness t of the nonreflective coating.

= 600 nm

t

Air

Coating

π

π

λ

Glass

Chapter 33

3110

The condition for destructive interference is:

( ) ( )coating

air21

coating212

nmmt λλ +=+=

Solve for t to obtain: ( )coating

air21

2nmt λ+=

Evaluate t for m = 0: ( ) ( ) nm115

30.12nm600

21 ==t

83 •• [SSM] A Fabry–Perot interferometer (Figure 33-47) consists of two parallel, half-silvered mirrors that face each other and are separated by a small distance a. A half-silvered mirror is one that transmits 50% of the incident intensity and reflects 50% of the incident intensity. Show that when light is incident on the interferometer at an angle of incidence θ, the transmitted light will have maximum intensity when 2a = mλ/cos θ. Picture the Problem The Fabry-Perot interferometer is shown in the figure. For constructive interference in the transmitted light the path difference must be an integral multiple of the wavelength of the light. This path difference can be found using the geometry of the interferometer.

θ

a

Express the path difference between the two rays that emerge from the interferometer:

θcos2ar =Δ

For constructive interference we require that:

...,2,1,0==Δ m,mr λ

Equate these expressions to obtain:

θλ

cos2am =

Solve for 2a to obtain: θλ cos2 ma =

84 •• A mica sheet 1.20 μm thick is suspended in air. In reflected light, there are gaps in the visible spectrum at 421, 474, 542, and 633 nm. Find the index of refraction of the mica sheet.


3111

Picture the Problem The gaps in the spectrum of the visible light are the result of destructive interference between the incident light and the reflected light. Noting that there is a π rad phase shift at the first air-mica interface, we can use the condition for destructive interference to find the index of refraction n of the mica sheet.

Air Mica Air

n

t

π

Because there is a π rad phase shift at the first air-mica interface, the condition for destructive interference is:

...,32,1,,2 airmica === m

nmmt λλ

Solving for n yields: t

mn2

airλ= (1)

For λ = 474 nm: ( )mt nm4742 =

For λ = 421 nm: ( )( )1nm4212 += mt

Equate these two expressions and solve for m to obtain:

m = 8 for λ = 474 nm

Substitute numerical values in equation (1) and evaluate n: ( ) 58.1

m20.12nm4748 =⎟⎟

⎠

⎞⎜⎜⎝

⎛=

μn

85 •• [SSM] A camera lens is made of glass that has an index of refraction of 1.60. This lens is coated with a magnesium fluoride film (index of refraction 1.38) to enhance its light transmission. The purpose of this film is to produce zero reflection for light of wavelength 540 nm. Treat the lens surface as a flat plane and the film as a uniformly thick flat film. (a) What minimum thickness of this film will accomplish its objective? (b) Would there be destructive interference for any other visible wavelengths? (c) By what factor would the reflection for light of 400 nm wavelength be reduced by the presence of this film? Neglect the variation in the reflected light amplitudes from the two surfaces.

Chapter 33

3112

Picture the Problem Note that the light reflected at both the air-film and film-lens interfaces undergoes a π rad phase shift. We can use the condition for destructive interference between the light reflected from the air-film interface and the film-lens interface to find the thickness of the film. In (c) we can find the factor by which light of the given wavelengths is reduced by this film from .cos 2

12 δ∝I

Air

n

t

π

Film Lens

π

(a) Express the condition for destructive interference between the light reflected from the air-film interface and the film-lens interface:

( ) ( )n

mmt air21

film212 λλ +=+= (1)

where m = 0, 1, 2, …

Solving for t gives:

( )n

mt2

air21 λ

+=

Evaluate t for m = 0:

( ) nm8.97nm83.9738.12nm540

21

==⎟⎠⎞

⎜⎝⎛=t

(b) Solve equation (1) for λair to obtain: 2

1air2+

=m

tnλ

Evaluate λair for m = 1: ( )( ) nm180

138.1nm8.972

21air =

+=λ

No; because 180 nm is not in the visible portion of the spectrum.

(c) Express the reduction factor f as a function of the phase difference δ between the two reflected waves:

δ212cos=f (2)

Relate the phase difference to the path difference Δr:

film2 λπδ rΔ

= ⇒ ⎟⎟⎠

⎞⎜⎜⎝

⎛ Δ=

film

2λ

πδ r

Because Δr = 2t:

⎟⎟⎠

⎞⎜⎜⎝

⎛=

film

22λ

πδ t


3113

Substitute in equation (2) to obtain:

⎥⎦

⎤⎢⎣

⎡=

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

air

2

film

2

film212

2cos

2cos22cos

λπ

λπ

λπ

nt

ttf

Evaluate f for λ = 400 nm: ( )( )

273.0

nm400nm83.9738.12cos2

400

=

⎥⎦

⎤⎢⎣

⎡=

πf

86 •• In a pinhole camera, the image is fuzzy because of geometry (rays arrive at the film after passing through different parts of the pinhole) and because of diffraction. As the pinhole is made smaller, the fuzziness due to geometry is reduced, but the fuzziness due to diffraction is increased. The optimum size of the pinhole for the sharpest possible image occurs when the spread due to diffraction equals the spread due to the geometric effects of the pinhole. Estimate the optimum size of the pinhole if the distance from the pinhole to the film is 10.0 cm and the wavelength of the light is 550 nm. Picture the Problem As indicated in the problem statement, we can find the optimal size of the pinhole by equating the angular width of the object at the film and the angular width of the diffraction pattern.

2θD

L

Object Film

Express the angular width of the a distant object at the film in terms of the diameter D of the pinhole and the distance L from the pinhole to the object:

LD

=θ2 ⇒ L

D2

=θ

Using Rayleigh’s criterion, express the angular width of the diffraction pattern:

Dλθ 22.1ndiffractio =


DLD λ22.12

= ⇒ LD λ44.2=

Chapter 33

3114


( )( )mm366.0

cm0.10nm55044.2

=

=D

87 •• [SSM] The Impressionist painter Georges Seurat used a technique called pointillism, in which his paintings are composed of small, closely spaced dots of pure color, each about 2.0 mm in diameter. The illusion of the colors blending together smoothly is produced in the eye of the viewer by diffraction effects. Calculate the minimum viewing distance for this effect to work properly. Use the wavelength of visible light that requires the greatest distance between dots, so that you are sure the effect will work for all visible wavelengths. Assume the pupil of the eye has a diameter of 3.0 mm. Picture the Problem We can use the geometry of the dots and the pupil of the eye and Rayleigh’s criterion to find the greatest viewing distance that ensures that the effect will work for all visible wavelengths.

Dots of paint

Pupil

θ

cα

d

L

Referring to the diagram, express the angle subtended by the adjacent dots:

Ld

≈θ

Letting the diameter of the pupil of the eye be D, apply Rayleigh’s criterion to obtain:

Dλα 22.1c =

Set θ = αc to obtain: DL

d λ22.1= ⇒λ22.1

DdL =

Evaluate L for the shortest wavelength light in the visible portion of the spectrum:

( )( )( )( ) m12

nm40022.1mm0.2mm0.3

==L

ch33

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