UNIT 4 REVIEW
(Pages 552–555)
Understanding Concepts
1. (a) reflection, rectilinear propagation, refraction, dispersion
(b) diffraction, partial reflection/partial refraction, interference, polarization
(c) transmission through a vacuum
2. ng = 1.60
λa = 415 nm
λg = ?
λ
n= a
λg
λa
n
415 nm
=
1.60
λg = 259 nm
The wavelength of the violet light in glass is 259 nm.
3. The wavelengths of sound are very much longer than light waves. There is little diffraction, if the wavelength is smaller
than the opening. For sound waves, the wavelengths are typically larger than opening and are easily attracted through an
opening or around a corner.
4. T = 0.10 s
distance between the first and sixth crests is 5 wavelengths.
(a) λ = ?
5λ = 12.0 cm
12.0 cm
λ=
5
λ = 2.4 cm
The wavelength is 2.4 cm.
(b) v = ?
λ
v=
T
2.4 cm
=
0.10 s
v = 24 cm/s
The speed of the wave is 24 cm/s.
5. n = 2
v = 7.5 cm/s
path difference = 29.5 cm – 25.0 cm = 4.5 cm
(a) λ = ?
1

 n −  λ = path difference
2

1

 2 −  λ = 4.5 cm
2

1.5λ = 4.5 cm
λ = 3.0 cm
The wavelength is 3.0 cm.
λg =
Copyright © 2003 Nelson
Unit 4 Review 623
(b) f = ?
v= fλ
v
λ
7.5 cm/s
=
3.0 cm
f = 2.5 Hz
The frequency is 2.5 Hz.
6. Double-slit interference requires a null result (destructive interference), which is definitive, whereas diffraction could be
open to interpretation at least before wave analysis was possible.
7. 6∆x = 6.0 cm
6.0 cm
∆x =
= 1.0 cm = 1.0 × 10–2 m
6
L = 3.00 m
d = 220 µm = 2.0 × 10–4 m
(a) ∆x = ?
λ
∆x = L  
d 
( ∆x ) d
λ=
L
(1.0 × 10−2 m)(2.0 × 10−4 m)
=
3.00 m
λ = 6.7 × 10−7 m
The wavelength is 6.7 × 10–7 m.
(b) The colour of the light is red.
8. θ = 5.2°
n=2
d
=?
λ
1λ

sin θ n =  n − 
2d

f =
1

n− 
d 
2
=
sin θ n
λ
1

2− 
2
=
sin 5.2°
9.
d
= 16.6
λ
The ratio of the slit separation d to the wavelength λ of the light is 16.6:1.
d = 0.018 mm = 1.8 × 10–5 m
θ = 8.2°
n=5
λ=?
624 Unit 4 The Wave Nature of Light
Copyright © 2003 Nelson
λ=
=
d sin θ
1
n−
2
×
(1.8 10−5 m ) (sin 8.2°)
1
2
λ = 5.70 × 10−7 m
The wavelength is 5.70 × 10–7 m.
10. λ = 638 nm = 6.38 × 10–7 m
n=3
θ = 8.0°
d=?
1

n − λ
2

d=
sin θ
1

−7
 3 −  (6.38 ×10 m )
2

=
sin 8.0°
d = 1.15 × 10−5 m
The distance between the slits is 1.15 × 10-5 m.
11. λ = 633 nm = 6.33 ¯ 10–7 m
d = 0.100 mm = 1.00 ¯ 10–4 m
L = 2.10 m
x1 = ?
1 λ

x1 =  n −  L
2 d

5−
 6.33 × 10 −7 m 
 1
=  1 −  ( 2.10 m ) 

−4
 2
 1.00 × 10 m 
x1 = 6.65 × 10 −3 m
The distance of the first dark fringe is 6.65 ¯ 10–3 m, or 6.65 mm.
12. d = 0.42 mm = 4.2 ¯ 10–5 m
L = 4.00 m
∆x = 5.5 cm = 5.5 ¯ 10–2 m
λ=?
f=?
To calculate wavelength:
d 
λ = ∆x  
L
=
(5.5 ×10
(minima)
−2
m )( 4.2 × 10−5 m )
4.00 m
λ = 5.78 × 10 −7 m, or 5.8 ×10−7 m
To calculate frequency:
c
λ
3.00 × 108 m/s
=
5.78 ×10 −7 m
f = 5.2 × 1014 Hz
The wavelength is 5.8 ¯ 10–7 m. The frequency is 5.2 ¯ 1014 Hz.
f =
Copyright © 2003 Nelson
Unit 4 Review 625
13. λ = 639 nm = 6.39 ¯ 10–7 m
d = 0.048 mm = 4.8 ¯ 10–5 m
L = 2.80 m
x1 = ?
xn 
1λ
= n− 
2d
L 
(minima)
−7
 1  (6.39 × 10 m ) ( 2.80 m )
x1 =  1 − 
4.8 × 10−5 m
 2
x1 = 1.86 × 10 −2 m, or 1.9 × 10−2 m
The first dark fringe is 1.9 ¯ 10–2 m, or 1.9 cm away from the centre of the pattern.
14. λ = 656 nm = 6.56 ¯ 10–7 m
L = 1.50 m
n=4
x = 48.0 mm = 4.80 ¯ 10–3 m
d=?
nλ
sin θ =
(maxima)
d
x nλ
=
L d
4 ( 6.56 × 10−7 m ) (1.50 m )
d=
4.80 × 10−3 m
d = 8.20 × 10−4 m
The separation of the two slits is 8.20 ¯ 10–4 m.
15. λ1 = 4.80 ¯ 102 nm = 4.80 ¯ 10–7 m
λ2 = 6.20 ¯ 102 nm = 6.20 ¯ 10–7 m
d = 0.68 mm = 6.8 ¯ 10–4 m
L = 1.6 m
n=2
∆x = ?
nL
nL
λ1
λ2
x1 =
and
x2 =
d
d
( x2 − x1 ) =
=
nL
(λ2 − λ1 )
d
2 (1.60 m )
6.8 × 10 −4 m
( x2 − x1 ) = 6.6 ×10−4 m
(6.20 ×10
−7
m − 4.80 × 10−7 m )
The second order maxima are 6.6 ¯ 10–4 m, or 0.66 mm apart.
16. Let the subscript a represent air, and w represent water.
λa = 4.00 ¯ 102 nm = 4.00 ¯ 10–7 m
d = 5.00 ¯ 10–4 m
nw = 1.33
L = 50.0 cm = 5.00 ¯ 10–1 m
∆x = ?
First we must find the wavelength of the light in water:
λ
λw = a
nw
4.00 × 10−7 m
1.33
λw = 3.00 × 10−7 m
=
626 Unit 4 The Wave Nature of Light
Copyright © 2003 Nelson
To calculate the distance between the fringes:
Lλ
∆x =
d
5.00 × 10 −1 m )(3.00 × 10−7 m )
(
=
5.00 × 10 −4 m
∆x = 3.00 × 10 −4 m
The fringes are 3.00 ¯ 10–4 m apart on the screen.
17. Dispersion requires double refraction to be visible. Since the sides of window glass are parallel, little refraction occurs.
18. Polarization was important because it showed the waves of light were transverse.
19. Light travels in all directions until it is reflected off a horizontal surface (such as the hood of a car, or a puddle on the
road). This reflected light become horizontally polarized—the light is only vibrating in one plane. Polaroid sunglasses
have polarizing filters that are arranged in the vertical plane to absorb these horizontally polarized light waves. As a
result, Polaroid sunglasses reduce the effect of glare.
20. w = 5.60 ¯ 10–4 m
L = 3.00 m
y1 = 3.5 mm = 3.5 ¯ 10–3 m
λ=?
yn 
1λ
= n− 
2 d
L 
y1d
λ1 =
1

n − L
2

(3.5 ×10
=
−3
(minima)
m )(5.60 × 10−4 m )
(1)(3.00 m )
λ1 = 6.53 × 10−7 m, or 6.5 × 10−7 m
The wavelength of the light is 6.5 ¯ 10–7 m.
21. λ = 675 nm = 6.75 ¯ 10–7 m
θ=?
(a) w = 1.80 ¯ 10–4 m
nλ
sin θ n =
w
(1) (6.75 ×10−7 m )
sin θ1 =
1.80 × 10−4 m
θ1 = 0.21°
The angle that locates the first dark fringe is 0.21°.
(b) w = 1.80 ¯ 10–6 m
nλ
sin θ n =
w
(1) (6.75 ×10−7 m )
sin θ1 =
1.80 × 10−6 m
θ1 = 22°
The angle that locates the first dark fringe is 22°.
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Unit 4 Review 627
22. λ = 638 nm = 6.38 ¯ 10–7 m
w = 4.40 ¯ 10–4 m
L = 1.45 m
y1 = ?
y1 λ
=
(minima)
L w
Lλ
y1 =
w
(1.45 m ) (6.38 × 10 −7 m )
=
4.40 × 10−4 m
y1 = 2.10 × 10−3 m
central maxima = 2 y1
= 2 ( 2.10 × 10 −3 m )
central maxima = 4.20 × 10−3 m
The width of the central fringe is 4.20 ¯ 10–3 m.
23. (a) λ = 589 nm = 5.89 ¯ 10–7 m
w = 1.08 ¯ 10–6 m
n=1
θ=?
nλ
sin θ n =
(minima)
w
(1) (5.89 ×10−7 m )
=
1.08 × 10 −6 m
θ = 33°
The first minimum is at an angle of 33°.
2λ
(b) Since sin θ 2 =
= 1.09 , sin θ > 1, and θ is undefined. Therefore, there is no second minimum.
w
24. λ = 1.15 ¯ 10–7 m
θ = 8.4°
w=?
nλ
sin θ n =
(minima)
w
nλ
w=
sin θ n
=
(8 ) (1.15 × 10 −7 m )
sin 8.4°
w = 6.29 × 10−6 m, or 6.3 × 10 −6 m
The width of the slit is 6.3 ¯ 10–6 m.
25. λ = 451 nm = 4.51 ¯ 10–7 m
w = 0.10 mm = 1.0 ¯ 10–4 m
L = 3.50 m
y1 = ?
Lλ
y1 =
w
4.5 ×10−7 m ) (3.50 m )
(
=
1.0 ×10 −4 m
y1 = 1.575 ×10 −2 m
628 Unit 4 The Wave Nature of Light
Copyright © 2003 Nelson
central maximum = 2 y1
= 2 (1.575 × 10−2 m )
central maximum = 3.15 × 10−2 m, or 3.2 × 10−2 m
The width of the central maximum is 3.2 ¯ 10–7 m, or 32 mm.
26. λ = 639 nm = 6.39 ¯ 10–7 m
w = 4.2 ¯ 10–4 m
L = 3.50 m
∆y = ?
Lλ
∆y =
w
(3.50 m ) (6.39 × 10−7 m )
=
4.2 ×10 −4 m
∆y = 5.32 × 10 −3 m
The maxima are 5.32 ¯ 10–3 m, or 5.3 mm far apart.
λ
27.
=?
w
Since 2 y1 = L , therefore:
y1 λ
=
L w
y1
λ
=
2 y1 w
λ 1
=
w 2
λ
1
is , or 0.5.
2
w
28. Spacing between fringes is much greater. Bright fringes are much brighter and sharper.
29. Immersed in water, the wavelength of the light will decrease.
mλ
sin θ m =
d
sin θ m ∝ λ
Since θ will decrease, the smaller the wavelength, the smaller the distances between the bright lines.
30. d = 2.2 ¯ 10–6 m
λ1 = 412 nm = 4.12 ¯ 10–7 m
λ2 = 661 nm = 6.61 ¯ 10–7 m
L = 3.10 m
∆y = ?
mλ
y
Since sin θ m =
and sin θ m = :
L
d
mLλ1
mLλ2
y1 =
and y2 =
d
d
The ratio
y2 − y1 =
=
mL
(λ2 − λ1 )
d
(1)(3.10 m ) ( 6.61× 10−7 m − 4.12 × 10−7 m )
2.2 ×10 −6 m
y2 − y1 = 3.5 × 10−1 m
The first order spectrum is 3.5 ¯ 10–1 m, or 35 cm wide.
Copyright © 2003 Nelson
Unit 4 Review 629
31. diffraction grating =
1 cm
= 2.00 × 10−4 cm = 2.00 × 10−6 m
5000 line
m=2
θ = 35.0°
λ=?
mλ
d
d sin θ m
λ=
m
( 2.00 ×10−6 m ) (sin 35.0°)
=
2
−7
λ = 5.74 × 10 m
The wavelength of the light is 5.74 ¯ 10–7 m, or 574 nm.
1 cm
32. (a) d =
= 3.33 ×10−5 cm = 3.33 × 10−7 m
30 000 line
The largest wavelength of visible light would be red at approximately 710 nm, and the shortest would be violet at
approximately 410 nm.
mλ
sin θ m =
d
However, the largest value would be for n = 1, therefore:
(1) λ
sin θ m =
3.33 × 10−7 m
Whether you substitute for red or violet light, the value for sin θm > 1, therefore no wavelength of visible light can be
diffracted.
mλ
(b) Since sin θ ≤ 1,
≤ 1 . For the first nodal to show:
d
mλ
=1
d
λ = d = 3.33 × 10−7 m
Therefore, the longest wavelength that will show is 3.33 ¯ 10–7 m.
1 cm
= 1.18 × 10−4 cm = 1.18 × 10−6 m
33. d =
8500 line
θ1 = 26.6°, 26.8°
θ2 = 41.1°, 41.3°
λ1 = ?
λ2 = ?
26.6° + 26.8°
average θ1 =
= 26.7°
2
41.1° + 41.3°
= 41.2°
average θ 2 =
2
sin θ m =
To calculate wavelength:
λ1 =
=
d sin θ1
m
(1.18 ×10−6 m ) sin 26.7°
1
−7
λ1 = 5.30 × 10 m
λ2 =
=
d sin θ 2
m
(1.18 ×10−6 m ) sin 41.2°
1
λ2 = 7.77 × 10 m
−7
The wavelengths are 5.30 ¯ 10–7 m, or 530 nm, and 7.77 ¯ 10–7 m, or 777 nm.
630 Unit 4 The Wave Nature of Light
Copyright © 2003 Nelson
34. λblue = 482 nm
noil = 1.40
t=?
First we must calculate the wavelength of the light in oil:
λ
λoil = air
n
482 nm
=
1.40
λoil = 362 nm
For bright reflection, the path difference must be
λ
.
2
λ
4
362 nm
=
4
t = 90.6 nm
The minimum thickness of the oil slick is 90.6 nm.
35. Let the subscript c represent the coating, g represent the glass, and a represent the air.
nc = 1.38
ng = 1.52
λa = 565 nm
t=?
t=
First we must determine the wavelength of the light in the coating:
λ
λc = a
n
565 nm
=
1.38
λc = 409 nm
The minimum path difference for cancellation is
λ
.
2
λ
4
409 nm
=
4
t = 102 nm
The minimum thickness of the coating is 102 nm.
36. Let the subscript c represent the coating, g represent the glass, and a represent the air.
nc = 1.61
ng = 1.52
λa = 589 nm
t=?
t=
Copyright © 2003 Nelson
Unit 4 Review 631
Since the coating appears black there is no reflection (i.e., there is cancellation). Therefore, the thickness is 0,
λ
, and λ.
2
First we must determine the wavelength of the light in the coating:
λ
λc = a
n
589 nm
=
1.61
λc = 366 nm
Therefore, the two smallest non-zero thicknesses are 183 nm, and 366 nm.
37. Let the subscript f represent the film, g represent the glass, and a represent air.
nf = 1.36
ng = 1.52
λa = 525 nm
t=?
First we must calculate the wavelength of the light in film:
λ
λf = a
n
5.25 × 10 −7 m
=
1.36
λf = 3.86 × 10−7 m
To calculate the thickness:
λ
t=
2
3.86 × 10−7 m
=
2
t = 1.93 × 10−7 m
The thickness is 1.93 ¯ 10–7 m.
38. Let the subscript s represent soap, and a represent air.
t = 112 nm
θ = 90°
λ=?
λ
The reflected bright path difference is , therefore:
2
λs
t=
4
λ = 4t
= 4 (112 nm )
λs = 448 nm
This is the wavelength in the soap. To calculate the wavelength in air:
λ
n= a
λs
λa = nλs
= (1.33)( 448 nm )
λa = 596 nm
The wavelength of the reflected light is 596 nm, and the colour is yellowish-orange.
632 Unit 4 The Wave Nature of Light
Copyright © 2003 Nelson
39. (a) The film is thin enough that there is destructive interference. The two rays are out-of-phase, as illustrated.
(b) The pattern will change as the soap moves under gravity creating a larger wedge of soap. The top will be a larger,
darker area and the distance between the areas of dark (or bright) will come closer together as the bottom thickens.
Eventually, the bubble breaks since the surface tension is not sufficient to hold it in place.
λ 3 5
λ 3
(c) The path differences for bright reflection are , λ , λ , etc. The thickness will be , λ , etc. Therefore, the
2 2 2
4 4
λ
difference in thickness between adjacent bands would be .
2
(d) λa = 588 nm
ts= ?
First we must calculate the wavelength of the light in the soap film:
λ
n= a
λs
λa
n
588 nm
=
1.33
λs = 442 nm
The lowest dark band is the second fringe, so the path difference would be 2λ, and the thickness would be λ. Therefore,
the thickness is 442 nm.
40. λ = 639 nm = 6.39 ¯ 10–7 m
m = 38
t=?
λs =
Thirty-eight dark fringes with one at each end is equal to 37 spaces. Each dark fringe represents a change in thickness of
λ
, therefore:
2
λ
t = m 
2
 6.39 × 10−7 m 
= 37 

2


t = 1.18 × 10 −5 m
The thickness of the foil is 1.18 ¯ 10–5 m.
Copyright © 2003 Nelson
Unit 4 Review 633
41. t = 7.62 ¯ 10–5 m
λ = 539 nm = 5.39 ¯ 10–7 m
m=?
λ
t = m 
2
2t
m=
λ
2 ( 7.62 × 10−7 m )
=
5.39 × 10−7 m
m = 283
There are 283 bright fringes across the wedge.
42. L = 15.8 cm = 1.58 ¯ 10–1 m
λ = 548 nm = 5.48 ¯ 10–7 m
∆x = 1.3 mm = 1.3 ¯ 10–3 m
t=?
λ
∆x = L  
 2t 
Lλ
t=
2 ∆x
(1.58 ×10−1 )(5.48 ×10−7 m )
=
2 (1.3 ×10 −3 m )
t = 3.3 × 10 −5 m
The thickness of the paper strip is 3.3 ¯ 10–5 m.
43. λ = 589 nm = 5.89 ¯ 10–7 m
number of fringes = 2000
For each fringe the path difference is λ and the distance the mirror moves is
λ
.
2
The distance moved is:
λ
2000   = 1000 (589 nm )
2
= 5.89 × 10−4 m
The distance the mirror must be moved is 5.89 ¯ 10–4 m
44. (a) “Coherent” means same wavelength and same phase. The distance between each slit and the screen is nearly equal, but
the mirror changes the phase 180°. Therefore, the image on the screen will not be coherent with the source.
(b) It is a combination of two single slit diffractions overlapping. All we see is a double slit interference, which is the
central maximum of the single slit diffraction patterns.
(c) It will be dark, since the sources are essentially 180° out-of-phase.
45.
Type
Nature of Source
radio
infrared
oscillating charge m
electron transitions in
atoms and molecules
higher energy electron
transitions in atoms
rapid deceleration of
charges
ultraviolet
X ray
Typical Means of
Detection
radio receiver
temperature
measurement
fluorescence
Nonionizing or
Ionizing
no
no
no
photo sensitive device
yes
46. w = 20.0 cm = 2.00 ¯ 10–1 m
θ1 = ?
634 Unit 4 The Wave Nature of Light
Copyright © 2003 Nelson
nλ
w
w sin θ n
λ=
n
2.0
( ×10−1 m ) (sin 36°)
=
1
−1
λ = 1.17 × 10 m
The wavelength of the microwaves is 1.17 ¯ 10–1 m, or 11.7 cm.
sin θ n =
Applying Inquiry Skills
47. In both cases, Figures 5(a) and (b) represent a single slit interference pattern. If the apparatus is the same, the wavelength
nλ
1
or, if λ is constant sin θ ∝ . Since θ is
of the light will be constant. From the single slit relationship, sin θ n =
w
w
smaller in the 5 (b) that in 5(a), then it follows that the slit width must have increased. Also, note that intensity of the
central maximum is higher in (b) than in (a), additional evidence that the width has increased. Since more light passes
through the slit, the intensity is higher.
48. f = 10.5 GHz = 10.5 ¯ 109 Hz
A = L = 50 cm = 0.5 m
B = y1 = ?
C=θ=?
First we must calculate the wavelength:
c
λ=
f
3.00 × 108 m/s
10.5 × 109 Hz
λ = 2.8 × 10−2 m
=
To calculate the value of C:
y1
L
1λ

= n− 
2d

sin θ n =
−2
 1   2.8 × 10 m 
sin θ1 =  1 −  

−2
 2   2.0 ×10 m 
sin θ = 7.00 × 10−1
θ = 44°
To calculate the value of B:
y1
= sin θ
L
y1 = L sin θ
= (0.50 m ) sin 44°
y1 = 0.35 m
Therefore, the value of C is 44°, and the value of B is 0.35 m, or 35 cm.
Making Connections
49. No, in fact it could be worse depending on the path difference between the two signals. Only when the path difference is
zero will you get constructive interference. At any other path difference there will be some destructive interference with a
λ
maximum with a path difference . (Assuming no reflected radiations.)
2
Copyright © 2003 Nelson
Unit 4 Review 635
50.
The path of the reflected wave is twice the hypotenuse of the triangle, therefore:
2
(2.50 ×10 m ) + (60.0 m )
2
2
2
= 514 m
Therefore, the path difference is 514 m – 500 m = 14 m. Since this is a phase inversion at Earth, a wavelength of 14 m
will interfere destructively. Therefore, the largest possible wavelength to interfere constructively will be 13 m.
51. f1 = 3.0 ¯ 104 Hz
f2 = 4.5 ¯ 104 Hz
(a) λ1 = ?
λ2 = ?
c= fλ
c
c
λ2 =
λ1 =
f2
f
1
3.00 × 108 m/s
3.00 × 108 m/s
=
4.5 × 104 Hz
3.0 × 10 4 Hz
λ2 = 6.7 × 103 m
λ1 = 1.0 × 104 m
The typical wavelengths would be 1.0 ¯ 104 m, and 6.7 ¯ 103 m.
(b) v = 3.4 ¯ 102 m/s
λ1 = ?
λ2 = ?
v= fλ
v
λ2 =
f
v
λ1 =
f
3.4 × 102 m/s
=
2
4.5 × 104 Hz
3.4 × 10 m/s
=
λ2 = 7.56 ×10−3 m, or 7.6 ×10 −3 m
3.0 × 104 Hz
−2
−2
λ1 = 1.13 × 10 m, or 1.1× 10 m
The actual wavelengths would be 1.1 ¯ 10–2 m, and 7.6 ¯ 10–3 m.
=
Extension
52. f = 14.0 kHz = 1.40 ¯ 104 Hz
v = 1.40 ¯ 102 m/s
w=?
First we must calculate the wavelength:
v
λ=
f
1.40 × 102 m
1.40 × 104 Hz
λ = 1.00 × 10−2 m
=
636 Unit 4 The Wave Nature of Light
Copyright © 2003 Nelson
For n = 1, and θ1 = 30.0°, to calculate the width of the slit for single slit maxima:
1λ

sin θ m =  m + 
2 w

 1
−2
1 +  (1.00 × 10 m )
2

sin 30.0° = 
w
w = 3.00 × 10 −2 m
The width of the slit is 3.00 ¯ 10–2 m, or 3.00 cm.
53. The white light range is 400 nm to 750 nm (red).
λR = 7.50 ¯ 10–7 m
number of lines = ?
sin θm ≤ 1
mλ
sin θ m =
d
d = mλ
= 2 ( 7.50 × 10−7 m )
d = 1.50 × 10−8 m
Since d is equal to 1.50 ¯ 10–8 m, or 1.50¯ 10–6 cm:
1
number of lines =
1.50 × 10−6 cm
number of lines = 6.67 × 10−3 lines/cm
The maximum possible number of lines per centimetre is 6.67 ¯ 103.
54. Let the subscript a represent air, and s represent soap.
n = 1.33
500 line
d=
= 2.00 × 10−6 m
mm
m=1
θ = 18°
t=?
First we must calculate the wavelength in air:
mλa
sin θ m =
d
sin θ m d
λa =
m
sin18° ( 2.00 ×10 −6 m )
=
1
λa = 6.18 ×10−7 m
Next we must calculate the wavelength in soap:
λ
λs = a
n
6.18 × 10 −7 m
=
1.33
λs = 4.65 × 10−7 m
Copyright © 2003 Nelson
Unit 4 Review 637
λ
λ
and the thickness is , therefore:
2
4
λ
t=
4
4.65 × 10 −7 m
=
4
t = 1.2 × 10−7 m
The minimum possible thickness of the soap film is 1.2 ¯ 10–7 m.
55. number of fringes = 31
λ = 589 nm = 5.89 ¯ 10–7 m
t=?
The path difference is
A dark fringe occurs at the edges where t = 0. The next dark fringe occurs when the path difference is λ, and the thickness
λ
is , therefore:
2
λ
t = number of fringes  
2
=
(31) (5.89 × 10−5 m )
2
t = 9.1× 10 m
The centre is 9.1 ¯ 10–4 m thicker than the edges.
−4
638 Unit 4 The Wave Nature of Light
Copyright © 2003 Nelson