1. Comparing the light speeds in sapphire and diamond, we obtain

FG 1 − 1 IJ

Hn n K

F 1 − 1 IJ = 4.55 × 10 m s.

= c2.998 × 10 m sh G

H 177

.

2.42 K

∆v = vs − vd = c

s

8

d

7

2. (a) The frequency of yellow sodium light is

c 2.998 × 108 m s

f = =

= 5.09 × 1014 Hz.

−9

λ

589 × 10 m

(b) When traveling through the glass, its wavelength is

λn =

λ 589 nm

=

= 388 nm.

n

152

.

(c) The light speed when traveling through the glass is

v = f λ n = ( 5.09 ×1014 Hz )( 388 ×10−9 m ) = 1.97 ×108 m s.

3. The index of refraction is found from Eq. 35-3:

c 2.998 × 108 m s

. .

n= =

= 156

. × 108 m s

v 192

4. Note that Snell’s Law (the law of refraction) leads to θ1 = θ2 when n1 = n2. The graph

indicates that θ2 = 30° (which is what the problem gives as the value of θ1) occurs at n2 =

1.5. Thus, n1 = 1.5, and the speed with which light propagates in that medium is

c

v = 1.5 = 2.0 × 108 m/s.

5. The fact that wave W2 reflects two additional times has no substantive effect on the

calculations, since two reflections amount to a 2(λ/2) = λ phase difference, which is

effectively not a phase difference at all. The substantive difference between W2 and W1 is

the extra distance 2L traveled by W2.

(a) For wave W2 to be a half-wavelength “behind” wave W1, we require 2L = λ/2, or L =

λ/4 = 155 nm using the wavelength value given in the problem.

(b) Destructive interference will again appear if W2 is

this case, 2 L ′ = 3λ 2 , and the difference is

L′ − L =

3

2

λ “behind” the other wave. In

3λ λ λ

− = = 310 nm .

4 4 2

6. In contrast to the initial conditions of problem 30, we now consider waves W2 and W1

with an initial effective phase difference (in wavelengths) equal to 21 , and seek positions

of the sliver which cause the wave to constructively interfere (which corresponds to an

integer-valued phase difference in wavelengths). Thus, the extra distance 2L traveled by

W2 must amount to 21 λ , 23 λ , and so on. We may write this requirement succinctly as

L=

2m + 1

λ

4

where m = 0, 1, 2,! .

(a) Thus, the smallest value of L / λ that results in the final waves being exactly in phase

is when m =0, which gives L / λ = 1/ 4 = 0.25 .

(b) The second smallest value of L / λ that results in the final waves being exactly in

phase is when m =1, which gives L / λ = 3 / 4 = 0.75 .

(c) The third smallest value of L / λ that results in the final waves being exactly in phase

is when m =2, which gives L / λ = 5 / 4 = 1.25 .

7. (a) We take the phases of both waves to be zero at the front surfaces of the layers. The

phase of the first wave at the back surface of the glass is given by φ1 = k1L – ωt, where k1

(= 2π/λ1) is the angular wave number and λ1 is the wavelength in glass. Similarly, the

phase of the second wave at the back surface of the plastic is given by φ2 = k2L – ωt,

where k2 (= 2π/λ2) is the angular wave number and λ2 is the wavelength in plastic. The

angular frequencies are the same since the waves have the same wavelength in air and the

frequency of a wave does not change when the wave enters another medium. The phase

difference is

§ 1

1 ·

− ¸ L.

© λ1 λ 2 ¹

φ1 − φ2 = ( k1 − k2 ) L = 2π ¨

Now, λ1 = λair/n1, where λair is the wavelength in air and n1 is the index of refraction of

the glass. Similarly, λ2 = λair/n2, where n2 is the index of refraction of the plastic. This

means that the phase difference is

φ1 – φ2 = (2π/λair) (n1 – n2)L.

The value of L that makes this 5.65 rad is

bφ − φ gλ = 5.65c400 × 10 mh = 3.60 × 10

L=

2 πb n − n g

2 πb1.60 −1.50g

−9

1

2

1

air

−6

m.

2

(b) 5.65 rad is less than 2π rad = 6.28 rad, the phase difference for completely

constructive interference, and greater than π rad (= 3.14 rad), the phase difference for

completely destructive interference. The interference is, therefore, intermediate, neither

completely constructive nor completely destructive. It is, however, closer to completely

constructive than to completely destructive.

8. (a) The time t2 it takes for pulse 2 to travel through the plastic is

t2 =

L

L

L

L

6.30 L

+

+

+

=

.

c 155

c 170

c 160

c 145

c

.

.

.

.

Similarly for pulse 1:

t1 =

L

L

2L

6.33 L

+

+

=

.

c 159

c 165

c 150

c

.

.

.

Thus, pulse 2 travels through the plastic in less time.

(b) The time difference (as a multiple of L/c) is

∆t = t 2 − t1 =

Thus, the multiple is 0.03.

6.33 L 6.30 L 0.03 L

−

=

.

c

c

c

9. (a) We wish to set Eq. 35-11 equal to 1/ 2, since a half-wavelength phase difference is

equivalent to a π radians difference. Thus,

Lmin =

λ

620 nm

=

= 1550 nm = 155

. µm.

2 n2 − n1

.

2 1.65 − 145

b

g b

(b) Since a phase difference of

g

3

(wavelengths) is effectively the same as what we

2

required in part (a), then

L=

3λ

= 3 Lmin = 3 155

. µm = 4.65 µm.

2 n2 − n1

b

g

b

g

10. (a) The exiting angle is 50º, the same as the incident angle, due to what one might call

the “transitive” nature of Snell’s law: n1 sinθ 1 = n2 sinθ 2 = n3 sinθ 3 = …

(b) Due to the fact that the speed (in a certain medium) is c/n (where n is that medium’s

index of refraction) and that speed is distance divided by time (while it’s constant), we

find

t = nL/c = (1.45)(25 × 10−19 m)/(3.0 × 108 m/s) = 1.4 × 10−13 s = 0.14 ps.

11. (a) Eq. 35-11 (in absolute value) yields

c

hb

g

c

hb

g

c

hb

g

8.50 × 10−6 m

L

160

n2 − n1 =

. − 150

. = 170

. .

500 × 10−9 m

λ

(b) Similarly,

8.50 × 10−6 m

L

= 170

172

n2 − n1 =

. − 162

.

. .

500 × 10−9 m

λ

(c) In this case, we obtain

3.25 × 10−6 m

L

179

n2 − n1 =

. − 159

. = 130

. .

500 × 10−9 m

λ

(d) Since their phase differences were identical, the brightness should be the same for (a)

and (b). Now, the phase difference in (c) differs from an integer by 0.30, which is also

true for (a) and (b). Thus, their effective phase differences are equal, and the brightness in

case (c) should be the same as that in (a) and (b).

12. (a) We note that ray 1 travels an extra distance 4L more than ray 2. To get the least

possible L which will result in destructive interference, we set this extra distance equal to

half of a wavelength:

4L = 12 λ

λ

L = 8 = 52.50 nm .

(b) The next case occurs when that extra distance is set equal to 32 λ. The result is

3λ

L = 8 = 157.5 nm .

13. (a) We choose a horizontal x axis with its origin at the left edge of the plastic.

Between x = 0 and x = L2 the phase difference is that given by Eq. 35-11 (with L in that

equation replaced with L2). Between x = L2 and x = L1 the phase difference is given by an

expression similar to Eq. 35-11 but with L replaced with L1 – L2 and n2 replaced with 1

(since the top ray in Fig. 35-36 is now traveling through air, which has index of refraction

approximately equal to 1). Thus, combining these phase differences and letting all lengths

be in µm (so λ = 0.600), we have

350

4.00 − 350

L2

L − L2

.

.

1 − n1 =

160

1 − 140

n2 − n1 + 1

. − 140

. +

. = 0.833.

λ

λ

0.600

0.600

b

g

b

g

b

g

b

g

(b) Since the answer in part (a) is closer to an integer than to a half-integer, the

interference is more nearly constructive than destructive.

14. (a) For the maximum adjacent to the central one, we set m = 1 in Eq. 35-14 and obtain

§ mλ ·

¸

© d ¹

θ1 = sin −1 ¨

ª (1)( λ ) º

= sin −1 «

» = 0.010 rad.

m =1

¬ 100λ ¼

(b) Since y1 = D tan θ1 (see Fig. 35-10(a)), we obtain

y1 = (500 mm) tan (0.010 rad) = 5.0 mm.

The separation is ∆y = y1 – y0 = y1 – 0 = 5.0 mm.

15. The angular positions of the maxima of a two-slit interference pattern are given by

d sin θ = mλ , where d is the slit separation, λ is the wavelength, and m is an integer. If θ

is small, sin θ may be approximated by θ in radians. Then, θ = mλ/d to good

approximation. The angular separation of two adjacent maxima is ∆θ = λ/d. Let λ' be the

wavelength for which the angular separation is greater by10.0%. Then, 1.10λ/d = λ'/d. or

λ' = 1.10λ = 1.10(589 nm) = 648 nm.

16. (a) We use Eq. 35-14 with m = 3:

θ = sin

−1

FG mλ IJ = sin

HdK

(b) θ = (0.216) (180°/π) = 12.4°.

−1

LM 2c550 × 10

MN 7.70 × 10

h OP = 0.216 rad.

m P

Q

−9

−6

m

17. Interference maxima occur at angles θ such that d sin θ = mλ, where m is an integer.

Since d = 2.0 m and λ = 0.50 m, this means that sin θ = 0.25m. We want all values of m

(positive and negative) for which |0.25m| ≤ 1. These are –4, –3, –2, –1, 0, +1, +2, +3, and

+4. For each of these except –4 and +4, there are two different values for θ. A single

value of θ (–90°) is associated with m = –4 and a single value (+90°) is associated with m

= +4. There are sixteen different angles in all and, therefore, sixteen maxima.

18. (a) The phase difference (in wavelengths) is

φ = d sinθ/λ = (4.24 µm)sin(20°)/(0.500 µm) = 2.90 .

(b) Multiplying this by 2π gives φ = 18.2 rad.

(c) The result from part (a) is greater than 52 (which would indicate the third minimum)

and is less than 3 (which would correspond to the third side maximum).

19. The condition for a maximum in the two-slit interference pattern is d sin θ = mλ,

where d is the slit separation, λ is the wavelength, m is an integer, and θ is the angle made

by the interfering rays with the forward direction. If θ is small, sin θ may be

approximated by θ in radians. Then, θ = mλ/d, and the angular separation of adjacent

maxima, one associated with the integer m and the other associated with the integer m + 1,

is given by ∆θ = λ/d. The separation on a screen a distance D away is given by ∆y =

D ∆θ = λD/d. Thus,

c500 × 10 mhb5.40 mg = 2.25 × 10

∆y =

−9

−3

120

. × 10 m

−3

m = 2.25 mm.

20. In Sample Problem 35-2, an experimentally useful relation is derived: ∆y = λD/d.

Dividing both sides by D, this becomes ∆θ = λ/d with θ in radians. In the steps that

follow, however, we will end up with an expression where degrees may be directly used.

Thus, in the present case,

∆θ n =

∆θ 0.20°

λn

λ

=

=

=

= 015

. °.

d

nd

n

133

.

21. The maxima of a two-slit interference pattern are at angles θ given by d sin θ = mλ,

where d is the slit separation, λ is the wavelength, and m is an integer. If θ is small, sin θ

may be replaced by θ in radians. Then, dθ = mλ. The angular separation of two maxima

associated with different wavelengths but the same value of m is ∆θ = (m/d)(λ2 – λ1), and

their separation on a screen a distance D away is

∆y = D tan ∆θ ≈ D ∆θ =

=

LM mD OP bλ

NdQ

LM 3b10. mg OP c600 × 10

N5.0 × 10 m Q

−3

−9

2

− λ1

g

h

m − 480 × 10−9 m = 7.2 × 10−5 m.

The small angle approximation tan ∆θ ≈ ∆θ (in radians) is made.

22. (a) We use Eq. 35-14 to find d:

d sinθ = mλ

d = (4)(450 nm)/sin(90°) = 1800 nm .

For the third order spectrum, the wavelength that corresponds to θ = 90° is

λ = d sin(90°)/3 = 600 nm .

Any wavelength greater than this will not be seen. Thus, 600 nm < θ ≤ 700 nm are

absent.

(b) The slit separation d needs to be decreased.

(c) In this case, the 400 nm wavelength in the m = 4 diffraction is to occur at 90°. Thus

dnew sinθ = mλ

dnew = (4)(400 nm)/sin(90°) = 1600 nm .

This represents a change of |∆d| = d – dnew = 200 nm = 0.20 µm.

23. Initially, source A leads source B by 90°, which is equivalent to 1 4 wavelength.

However, source A also lags behind source B since rA is longer than rB by 100 m, which

is 100 m 400 m = 1 4 wavelength. So the net phase difference between A and B at the

detector is zero.

24. Imagine a y axis midway between the two sources in the figure. Thirty points of

destructive interference (to be considered in the xy plane of the figure) implies there are

7 + 1 + 7 = 15 on each side of the y axis. There is no point of destructive interference on

the y axis itself since the sources are in phase and any point on the y axis must therefore

correspond to a zero phase difference (and corresponds to θ = 0 in Eq. 35-14). In other

words, there are 7 “dark” points in the first quadrant, one along the +x axis, and 7 in the

fourth quadrant, constituting the 15 dark points on the right-hand side of the y axis. Since

the y axis corresponds to a minimum phase difference, we can count (say, in the first

quadrant) the m values for the destructive interference (in the sense of Eq. 35-16)

beginning with the one closest to the y axis and going clockwise until we reach the x axis

(at any point beyond S2). This leads us to assign m = 7 (in the sense of Eq. 35-16) to the

point on the x axis itself (where the path difference for waves coming from the sources is

simply equal to the separation of the sources, d); this would correspond to θ = 90° in Eq.

35-16. Thus,

d = ( 7 + 12 ) λ = 7.5 λ

d

λ

= 7.5 .

25. Let the distance in question be x. The path difference (between rays originating from

S1 and S2 and arriving at points on the x > 0 axis) is

FG

H

d 2 + x2 − x = m +

IJ

K

1

λ,

2

where we are requiring destructive interference (half-integer wavelength phase

differences) and m = 0, 1, 2, ". After some algebraic steps, we solve for the distance in

terms of m:

x=

b

b

g

2m + 1 λ

d2

−

.

2m + 1 λ

4

g

To obtain the largest value of x, we set m = 0:

d 2 λ ( 3.00λ ) λ

x0 = − =

− = 8.75λ = 8.75(900 nm) = 7.88 ×103 nm

λ 4

λ

4

= 7.88µ m.

2

FG 1 − 1 IJ

Hn n K

F 1 − 1 IJ = 4.55 × 10 m s.

= c2.998 × 10 m sh G

H 177

.

2.42 K

∆v = vs − vd = c

s

8

d

7

2. (a) The frequency of yellow sodium light is

c 2.998 × 108 m s

f = =

= 5.09 × 1014 Hz.

−9

λ

589 × 10 m

(b) When traveling through the glass, its wavelength is

λn =

λ 589 nm

=

= 388 nm.

n

152

.

(c) The light speed when traveling through the glass is

v = f λ n = ( 5.09 ×1014 Hz )( 388 ×10−9 m ) = 1.97 ×108 m s.

3. The index of refraction is found from Eq. 35-3:

c 2.998 × 108 m s

. .

n= =

= 156

. × 108 m s

v 192

4. Note that Snell’s Law (the law of refraction) leads to θ1 = θ2 when n1 = n2. The graph

indicates that θ2 = 30° (which is what the problem gives as the value of θ1) occurs at n2 =

1.5. Thus, n1 = 1.5, and the speed with which light propagates in that medium is

c

v = 1.5 = 2.0 × 108 m/s.

5. The fact that wave W2 reflects two additional times has no substantive effect on the

calculations, since two reflections amount to a 2(λ/2) = λ phase difference, which is

effectively not a phase difference at all. The substantive difference between W2 and W1 is

the extra distance 2L traveled by W2.

(a) For wave W2 to be a half-wavelength “behind” wave W1, we require 2L = λ/2, or L =

λ/4 = 155 nm using the wavelength value given in the problem.

(b) Destructive interference will again appear if W2 is

this case, 2 L ′ = 3λ 2 , and the difference is

L′ − L =

3

2

λ “behind” the other wave. In

3λ λ λ

− = = 310 nm .

4 4 2

6. In contrast to the initial conditions of problem 30, we now consider waves W2 and W1

with an initial effective phase difference (in wavelengths) equal to 21 , and seek positions

of the sliver which cause the wave to constructively interfere (which corresponds to an

integer-valued phase difference in wavelengths). Thus, the extra distance 2L traveled by

W2 must amount to 21 λ , 23 λ , and so on. We may write this requirement succinctly as

L=

2m + 1

λ

4

where m = 0, 1, 2,! .

(a) Thus, the smallest value of L / λ that results in the final waves being exactly in phase

is when m =0, which gives L / λ = 1/ 4 = 0.25 .

(b) The second smallest value of L / λ that results in the final waves being exactly in

phase is when m =1, which gives L / λ = 3 / 4 = 0.75 .

(c) The third smallest value of L / λ that results in the final waves being exactly in phase

is when m =2, which gives L / λ = 5 / 4 = 1.25 .

7. (a) We take the phases of both waves to be zero at the front surfaces of the layers. The

phase of the first wave at the back surface of the glass is given by φ1 = k1L – ωt, where k1

(= 2π/λ1) is the angular wave number and λ1 is the wavelength in glass. Similarly, the

phase of the second wave at the back surface of the plastic is given by φ2 = k2L – ωt,

where k2 (= 2π/λ2) is the angular wave number and λ2 is the wavelength in plastic. The

angular frequencies are the same since the waves have the same wavelength in air and the

frequency of a wave does not change when the wave enters another medium. The phase

difference is

§ 1

1 ·

− ¸ L.

© λ1 λ 2 ¹

φ1 − φ2 = ( k1 − k2 ) L = 2π ¨

Now, λ1 = λair/n1, where λair is the wavelength in air and n1 is the index of refraction of

the glass. Similarly, λ2 = λair/n2, where n2 is the index of refraction of the plastic. This

means that the phase difference is

φ1 – φ2 = (2π/λair) (n1 – n2)L.

The value of L that makes this 5.65 rad is

bφ − φ gλ = 5.65c400 × 10 mh = 3.60 × 10

L=

2 πb n − n g

2 πb1.60 −1.50g

−9

1

2

1

air

−6

m.

2

(b) 5.65 rad is less than 2π rad = 6.28 rad, the phase difference for completely

constructive interference, and greater than π rad (= 3.14 rad), the phase difference for

completely destructive interference. The interference is, therefore, intermediate, neither

completely constructive nor completely destructive. It is, however, closer to completely

constructive than to completely destructive.

8. (a) The time t2 it takes for pulse 2 to travel through the plastic is

t2 =

L

L

L

L

6.30 L

+

+

+

=

.

c 155

c 170

c 160

c 145

c

.

.

.

.

Similarly for pulse 1:

t1 =

L

L

2L

6.33 L

+

+

=

.

c 159

c 165

c 150

c

.

.

.

Thus, pulse 2 travels through the plastic in less time.

(b) The time difference (as a multiple of L/c) is

∆t = t 2 − t1 =

Thus, the multiple is 0.03.

6.33 L 6.30 L 0.03 L

−

=

.

c

c

c

9. (a) We wish to set Eq. 35-11 equal to 1/ 2, since a half-wavelength phase difference is

equivalent to a π radians difference. Thus,

Lmin =

λ

620 nm

=

= 1550 nm = 155

. µm.

2 n2 − n1

.

2 1.65 − 145

b

g b

(b) Since a phase difference of

g

3

(wavelengths) is effectively the same as what we

2

required in part (a), then

L=

3λ

= 3 Lmin = 3 155

. µm = 4.65 µm.

2 n2 − n1

b

g

b

g

10. (a) The exiting angle is 50º, the same as the incident angle, due to what one might call

the “transitive” nature of Snell’s law: n1 sinθ 1 = n2 sinθ 2 = n3 sinθ 3 = …

(b) Due to the fact that the speed (in a certain medium) is c/n (where n is that medium’s

index of refraction) and that speed is distance divided by time (while it’s constant), we

find

t = nL/c = (1.45)(25 × 10−19 m)/(3.0 × 108 m/s) = 1.4 × 10−13 s = 0.14 ps.

11. (a) Eq. 35-11 (in absolute value) yields

c

hb

g

c

hb

g

c

hb

g

8.50 × 10−6 m

L

160

n2 − n1 =

. − 150

. = 170

. .

500 × 10−9 m

λ

(b) Similarly,

8.50 × 10−6 m

L

= 170

172

n2 − n1 =

. − 162

.

. .

500 × 10−9 m

λ

(c) In this case, we obtain

3.25 × 10−6 m

L

179

n2 − n1 =

. − 159

. = 130

. .

500 × 10−9 m

λ

(d) Since their phase differences were identical, the brightness should be the same for (a)

and (b). Now, the phase difference in (c) differs from an integer by 0.30, which is also

true for (a) and (b). Thus, their effective phase differences are equal, and the brightness in

case (c) should be the same as that in (a) and (b).

12. (a) We note that ray 1 travels an extra distance 4L more than ray 2. To get the least

possible L which will result in destructive interference, we set this extra distance equal to

half of a wavelength:

4L = 12 λ

λ

L = 8 = 52.50 nm .

(b) The next case occurs when that extra distance is set equal to 32 λ. The result is

3λ

L = 8 = 157.5 nm .

13. (a) We choose a horizontal x axis with its origin at the left edge of the plastic.

Between x = 0 and x = L2 the phase difference is that given by Eq. 35-11 (with L in that

equation replaced with L2). Between x = L2 and x = L1 the phase difference is given by an

expression similar to Eq. 35-11 but with L replaced with L1 – L2 and n2 replaced with 1

(since the top ray in Fig. 35-36 is now traveling through air, which has index of refraction

approximately equal to 1). Thus, combining these phase differences and letting all lengths

be in µm (so λ = 0.600), we have

350

4.00 − 350

L2

L − L2

.

.

1 − n1 =

160

1 − 140

n2 − n1 + 1

. − 140

. +

. = 0.833.

λ

λ

0.600

0.600

b

g

b

g

b

g

b

g

(b) Since the answer in part (a) is closer to an integer than to a half-integer, the

interference is more nearly constructive than destructive.

14. (a) For the maximum adjacent to the central one, we set m = 1 in Eq. 35-14 and obtain

§ mλ ·

¸

© d ¹

θ1 = sin −1 ¨

ª (1)( λ ) º

= sin −1 «

» = 0.010 rad.

m =1

¬ 100λ ¼

(b) Since y1 = D tan θ1 (see Fig. 35-10(a)), we obtain

y1 = (500 mm) tan (0.010 rad) = 5.0 mm.

The separation is ∆y = y1 – y0 = y1 – 0 = 5.0 mm.

15. The angular positions of the maxima of a two-slit interference pattern are given by

d sin θ = mλ , where d is the slit separation, λ is the wavelength, and m is an integer. If θ

is small, sin θ may be approximated by θ in radians. Then, θ = mλ/d to good

approximation. The angular separation of two adjacent maxima is ∆θ = λ/d. Let λ' be the

wavelength for which the angular separation is greater by10.0%. Then, 1.10λ/d = λ'/d. or

λ' = 1.10λ = 1.10(589 nm) = 648 nm.

16. (a) We use Eq. 35-14 with m = 3:

θ = sin

−1

FG mλ IJ = sin

HdK

(b) θ = (0.216) (180°/π) = 12.4°.

−1

LM 2c550 × 10

MN 7.70 × 10

h OP = 0.216 rad.

m P

Q

−9

−6

m

17. Interference maxima occur at angles θ such that d sin θ = mλ, where m is an integer.

Since d = 2.0 m and λ = 0.50 m, this means that sin θ = 0.25m. We want all values of m

(positive and negative) for which |0.25m| ≤ 1. These are –4, –3, –2, –1, 0, +1, +2, +3, and

+4. For each of these except –4 and +4, there are two different values for θ. A single

value of θ (–90°) is associated with m = –4 and a single value (+90°) is associated with m

= +4. There are sixteen different angles in all and, therefore, sixteen maxima.

18. (a) The phase difference (in wavelengths) is

φ = d sinθ/λ = (4.24 µm)sin(20°)/(0.500 µm) = 2.90 .

(b) Multiplying this by 2π gives φ = 18.2 rad.

(c) The result from part (a) is greater than 52 (which would indicate the third minimum)

and is less than 3 (which would correspond to the third side maximum).

19. The condition for a maximum in the two-slit interference pattern is d sin θ = mλ,

where d is the slit separation, λ is the wavelength, m is an integer, and θ is the angle made

by the interfering rays with the forward direction. If θ is small, sin θ may be

approximated by θ in radians. Then, θ = mλ/d, and the angular separation of adjacent

maxima, one associated with the integer m and the other associated with the integer m + 1,

is given by ∆θ = λ/d. The separation on a screen a distance D away is given by ∆y =

D ∆θ = λD/d. Thus,

c500 × 10 mhb5.40 mg = 2.25 × 10

∆y =

−9

−3

120

. × 10 m

−3

m = 2.25 mm.

20. In Sample Problem 35-2, an experimentally useful relation is derived: ∆y = λD/d.

Dividing both sides by D, this becomes ∆θ = λ/d with θ in radians. In the steps that

follow, however, we will end up with an expression where degrees may be directly used.

Thus, in the present case,

∆θ n =

∆θ 0.20°

λn

λ

=

=

=

= 015

. °.

d

nd

n

133

.

21. The maxima of a two-slit interference pattern are at angles θ given by d sin θ = mλ,

where d is the slit separation, λ is the wavelength, and m is an integer. If θ is small, sin θ

may be replaced by θ in radians. Then, dθ = mλ. The angular separation of two maxima

associated with different wavelengths but the same value of m is ∆θ = (m/d)(λ2 – λ1), and

their separation on a screen a distance D away is

∆y = D tan ∆θ ≈ D ∆θ =

=

LM mD OP bλ

NdQ

LM 3b10. mg OP c600 × 10

N5.0 × 10 m Q

−3

−9

2

− λ1

g

h

m − 480 × 10−9 m = 7.2 × 10−5 m.

The small angle approximation tan ∆θ ≈ ∆θ (in radians) is made.

22. (a) We use Eq. 35-14 to find d:

d sinθ = mλ

d = (4)(450 nm)/sin(90°) = 1800 nm .

For the third order spectrum, the wavelength that corresponds to θ = 90° is

λ = d sin(90°)/3 = 600 nm .

Any wavelength greater than this will not be seen. Thus, 600 nm < θ ≤ 700 nm are

absent.

(b) The slit separation d needs to be decreased.

(c) In this case, the 400 nm wavelength in the m = 4 diffraction is to occur at 90°. Thus

dnew sinθ = mλ

dnew = (4)(400 nm)/sin(90°) = 1600 nm .

This represents a change of |∆d| = d – dnew = 200 nm = 0.20 µm.

23. Initially, source A leads source B by 90°, which is equivalent to 1 4 wavelength.

However, source A also lags behind source B since rA is longer than rB by 100 m, which

is 100 m 400 m = 1 4 wavelength. So the net phase difference between A and B at the

detector is zero.

24. Imagine a y axis midway between the two sources in the figure. Thirty points of

destructive interference (to be considered in the xy plane of the figure) implies there are

7 + 1 + 7 = 15 on each side of the y axis. There is no point of destructive interference on

the y axis itself since the sources are in phase and any point on the y axis must therefore

correspond to a zero phase difference (and corresponds to θ = 0 in Eq. 35-14). In other

words, there are 7 “dark” points in the first quadrant, one along the +x axis, and 7 in the

fourth quadrant, constituting the 15 dark points on the right-hand side of the y axis. Since

the y axis corresponds to a minimum phase difference, we can count (say, in the first

quadrant) the m values for the destructive interference (in the sense of Eq. 35-16)

beginning with the one closest to the y axis and going clockwise until we reach the x axis

(at any point beyond S2). This leads us to assign m = 7 (in the sense of Eq. 35-16) to the

point on the x axis itself (where the path difference for waves coming from the sources is

simply equal to the separation of the sources, d); this would correspond to θ = 90° in Eq.

35-16. Thus,

d = ( 7 + 12 ) λ = 7.5 λ

d

λ

= 7.5 .

25. Let the distance in question be x. The path difference (between rays originating from

S1 and S2 and arriving at points on the x > 0 axis) is

FG

H

d 2 + x2 − x = m +

IJ

K

1

λ,

2

where we are requiring destructive interference (half-integer wavelength phase

differences) and m = 0, 1, 2, ". After some algebraic steps, we solve for the distance in

terms of m:

x=

b

b

g

2m + 1 λ

d2

−

.

2m + 1 λ

4

g

To obtain the largest value of x, we set m = 0:

d 2 λ ( 3.00λ ) λ

x0 = − =

− = 8.75λ = 8.75(900 nm) = 7.88 ×103 nm

λ 4

λ

4

= 7.88µ m.

2

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