Wave Optics
337
2.
With phase reversals occurring in the reflections at both the air-oil boundary and the oil-water
boundary, the condition for constructive interference in the reflected light is 2noilt = mλ where
m is any integer. Thus, the minimum nonzero thickness of the oil which will strongly reflect the
530-nm light is t min = λ 2 noil = (530 nm) 2(1.25) = 212 nm, and (d) is the proper choice.
3.
In a single slit diffraction pattern, formed on a screen at distance L from the slit, dark fringes
occur where ydark L = tan θ dark ≈ sin θ dark = m(λ a ) and m is any nonzero integer. Thus, the width
of the slit, a, in the described situation must be
a=
(1) λ L
( ydark )1
=
( 5.0 × 10
−7
)
m (1.0 m )
5.0 × 10
−3
m
= 1.0 × 10 −4 m = 0.10 mm
and the correct answer is seen to be choice (b).
4.
From Malus’s law, the intensity of the light transmitted through a polarizer having its transmission axis oriented at angle q to the plane of polarization of the incident polarized
light is I = I 0 cos 2 θ . Therefore, the intensity transmitted through the first polarizer having
θ = 45° − 0 = 45° is I1 = I 0 cos 2 (45°) = 0.50 I 0, and the intensity passing through the second
polarizer having θ = 90° − 45° = 45° is I 2 = (0.50 I 0 ) cos 2 (45°) = 0.25 I 0. The fraction of the original intensity making it through both polarizers is then I 2 I 0 = 0.25, which is choice (b).
5.
The spacing between successive bright fringes in a double-slit interference pattern is given by
∆ybright = ( λ L d ) ( m + 1) − ( λ L d ) m = λ L d , where d is the slit separation. As d decreases, the
spacing between the bright fringes will increase and choice (b) is the correct answer.
6.
As discussed in question 5 above, the fringe spacing in a double slit interference pattern is
∆ybright = λ L d . Therefore, as the distance L between the screen and the plane of the slits is
increased, the spacing between the bright fringes will increase, and (a) is the correct choice.
7.
As discussed in question 5 above, the fringe spacing in a double slit interference pattern is
∆ybright = λ L d . Note that this spacing is directly proportional to the wavelength of the light
illuminating the slits. Thus, with λ2 > λ1 , the spacing is greater in the second experiment and
choice (c) gives the correct answer.
8.
The bright colored patterns are the result of interference between light reflected from the upper
surface of the oil and light reflected from the lower surface of the oil film. Thus, the best answer
is choice (e).
9.
The reflected light tends to be partially polarized, with the degree of polarization depending on
the angle of incidence on the reflecting surface. Only if the angle of incidence equals the polarizing angle (or Brewster’s angle) will the reflected light be completely polarized. The better answer
for this question is choice (d).
10.
In a single slit diffraction pattern, dark fringes occur where ydark L ≈ sin θ dark = m(λ a ) . The width
of the central maximum is the distance between the locations of the first dark fringes on either
side of the center (i.e., between the m = ±1 dark fringes), giving
width of central maximum = ( +1)
λ
λ 2λ
− ( −1) =
a
a
a
Thus, if the width of the slit, a, is cut in half, the width of the central maximum will double,
making (d) the correct choice.
56157_24_ch24_p335-360.indd 337
3/20/08 1:38:16 AM
338
Chapter 24
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
The wavelength of light is extremely small in comparison to the dimensions of your hand, so the
diffraction of light around obstacles the size of your hand is totally negligible. However, sound
waves have wavelengths that are comparable to the dimensions of the hand or even larger. Therefore, significant diffraction of sound waves occurs around hand-sized obstacles.
4.
The wavelength of light traveling in water would decrease, since the wavelength of light in a
medium is given by λn = λ n , where λ is the wavelength in vacuum and n is the index of refraction of the medium. Since the positions of the bright and dark fringes are proportional to the
wavelength, the fringe separations would decrease.
6.
Every color produces its own interference pattern, and we see them superimposed. The central
maximum is white. The first maximum is a full spectrum with violet on the inside and red on the
outside. The second maximum is also a full spectrum, with red in it overlapping with violet in the
third maximum. At larger angles, the light soon starts mixing to white again.
8.
The skin on the tip of a finger has a series of closely spaced ridges and swirls on it. When the finger touches a smooth surface, the oils from the skin will be deposited on the surface in the pattern
of the closely spaced ridges. The clear spaces between the lines of deposited oil can serve as the
slits in a crude diffraction grating and produce a colored spectrum of the light passing through or
reflecting from the glass surface.
10.
Suppose the index of refraction of the coating is intermediate between vacuum and the glass.
When the coating is very thin, light reflected from its top and bottom surfaces will interfere
constructively, so you see the surface white and brighter. Once the thickness reaches one-quarter
of the wavelength of violet light in the coating, destructive interference for violet light will make
the surface look red. Then other colors in spectral order (blue, green, yellow, orange, and red)
will interfere destructively, making the surface look red, violet, and then blue. As the coating gets
thicker, constructive interference is observed for violet light and then for other colors in spectral
order. Even thicker coatings give constructive and destructive interference for several visible
wavelengths, so the reflected light starts looking white again.
12.
The reflected light is partially polarized, with the component parallel to the reflecting surface
being the most intense. Therefore, the polarizing material should have its transmission axis
oriented in the vertical direction in order to minimize the intensity of the reflected light from
horizontal surfaces.
14.
Due to gravity, the soap film tends to sag in its holder, being quite thin at the top and becoming
thicker as one moves toward the bottom of the holding ring. Because light reflecting from the
front surface of the film experiences a phase reversal, and light reflecting from the back surface
of the film does not (see Figure 24.7 in the textbook), the film must be a minimum of a half
wavelength thick before it can produce constructive interference in the reflected light. Thus, the
light must be striking the film at some distance from the top of the ring before the thickness is
sufficient to produce constructive interference for any wavelength in the visible portion of the
spectrum.
56157_24_ch24_p335-360.indd 338
3/20/08 1:38:17 AM
Wave Optics
339
PROBLEM SOLUTIONS
24.1
The location of the bright fringe of order m (measured from the position of the central maximum)
is ( ybright )m = (λ L d )m, m = 0, ± 1, ± 2, . . .. Thus, the spacing between successive bright fringes is
(
∆ybright = ybright
)
m +1
(
− ybright
) = ( λ L d )( m + 1) − ( λ L d ) m = λ L d
m
Thus, the wavelength of the laser light must be
λ=
24.2
(a)
( ∆y ) d = (1.58 × 10
bright
−2
)
cm ( 0.200 × 10 −3 m
5.00 m
L
) = 6.32 × 10
−7
m = 632 nm
For a bright fringe of order m, the path difference is δ = mλ , where m = 0, 1, 2,K. At the
location of the third order bright fringe, m = 3 and
δ = 3λ = 3 ( 589 nm ) = 1.77 × 10 3 nm = 1.77 µ m
(b)

For a dark fringe, the path difference is δ =  m +

1
 λ , where m = 0, 1, 2,K.
2
At the third dark fringe, m = 2 and
1
5

δ =  2 +  λ = ( 589 nm ) = 1.47 × 10 3 nm = 1.47 µ m

2
2
24.3
(a)
The distance between the central maximum and the first order bright fringe is
∆y = ybright
m =1
− ybright
m=0
=
λL
, or
d
−9
λ L ( 546.1 × 10 m ) (1.20 m )
∆y =
=
= 2.62 × 10 −3 m = 2.62 mm
d
0.250 × 10 −3 m
(b)
The distance between the first and second dark bands is
∆y = ydark
24.4
m =1
− ydark
m=0
=
λL
= 2 .62 mm as in (a) above.
d
The location of the dark fringe of order m (measured from the position of the central maximum)
is given by ( ydark )m = (λ L d )( m + 12 ) , where m = 0, ± 1, ± 2, . . . . Thus, the spacing between the
first and second dark fringes will be
∆y = ( ydark )m =1 − ( ydark )m = 0 = ( λ L d ) (1 + 12 ) − ( λ L d ) ( 0 + 12 ) = λ L d
24.5
)
or
( 5.30 × 10
∆y =
(a)
From d sin θ = mλ, the angle for the m = 1 maximum for the sound waves is
−7
m ( 2.00 m )
0.300 × 10 −3 m
= 3.53 × 10 −3 m = 3.53 mm
m  v

 354 m s  
m
1

θ = sin −1  λ  = sin −1   sound   = sin −1 
 2 000 Hz   = 36.2°
d 


.
d
f
0
300
m




continued on next page
56157_24_ch24_p335-360.indd 339
3/20/08 4:55:44 AM
340
Chapter 24
(b)
For 3.00-cm microwaves, the required slit spacing is
d=
(c)
The wavelength is λ = d sin θ m ; and if this is light, the frequency is
f =
24.6
(a)
mλ (1) ( 3.00 m )
=
= 5.08 cm
sin θ
sin ( 36.2°)
)
(1) ( 3.00 × 10 8 m s
c
mc
= 5.08 × 1014 Hz
=
=
λ d sin θ (1.00 × 10 −6 m sin 36.2°
)
The location of the bright fringe of order m (measured from the position of the central
maximum) is ( ybright )m = (λL d )m, m = 0, ± 1, ± 2, K.
If ( ybright )m = 3 − ( ybright )m = 0 = 5.30 cm , then 3(λ L d ) − 0 = 5.30 cm , and
λ=
(b)
( 5.30 cm ) d
3L
−2
)
m ( 0.050 0 × 10 −3 m
3(1.50 m )
) = 5.89 × 10
−7
m = 589 nm
The separation between adjacent bright fringes is ∆ybright = λ L d , or
∆ybright =
24.7
( 5.30 × 10
=
( 589 × 10
−9
)
m (1.50 m )
0.050 0 × 10 −3 m
= 1.77 × 10 −2 m = 1.77 cm
Note that, with the conditions given, the small
angle approximation does not work well.
That is, sinθ , tanθ , and θ are significantly
different. The approach to be used is outlined
below.
(a)
At the m = 2 maximum, δ = d sin θ = 2 λ ,
λ=
or
λ=
(b)

d
d
y
sin θ = 

2
2  L2 + y 2 
( 300 m ) 
2


400 m
(1 000 m )2 + ( 400 m )2

 = 55.7 m


The next minimum encountered is the m = 2 minimum; and at that point,
1
5

δ = d sin θ =  m +  λ = λ


2
2
 5 ( 55.7 m ) 
 5λ 
= 27.7°
θ = sin −1   = sin −1 
 2d 
 2 ( 300 m ) 
or
Then,
y = (1 000 m ) tan 27.7° = 524 m
so the car must travel an additional 124 m .
24.8
The angular position of the bright fringe of order m is given by d sin θ = mλ . Thus, if the m = 1
bright fringe is located at θ = 12° when λ = 6.0 × 10 2 nm = 6.0 × 10 −7 m, the slit spacing is
d=
56157_24_ch24_p335-360.indd 340
mλ (1) ( 6.0 × 10
=
sin θ
sin 12°
−7
m
) = 2.9 × 10
−6
m = 2.9 µ m
3/20/08 1:38:18 AM
Wave Optics
24.9
The location of the bright fringe of order m (measured from the position of the central maximum)
is ( ybright )m = (λL d )m, m = 0, ± 1, ± 2, K. If the m = 1 bright fringe is located at y = 3.40 mm
when d = 0.500 mm and L = 3.30 m, the wavelength of the light is
λ=
24.10
341
( ybright )m d
mL
=
( 3.40 × 10
−3
)
m ( 0.500 × 10 −3 m
(1) ( 3.30 m )
) = 5.15 × 10
−7
m = 515 nm
The angular deviation from the line of the central maximum is given by
 y
 1.80 cm 
θ = tan −1   = tan −1 
= 0.737°
 L
 140 cm 
(a)
The path difference is then
δ = d sin θ = ( 0.150 mm ) sin ( 0.737°) = 1.93 × 10 −3 mm = 1.93 µ m
(b)
(c)
24.11
λ


δ = (1.93 × 10 −6 m ) 
= 3.00 λ
 643 × 10 −9 m 
Since the path difference for this position is a whole number of wavelengths, the waves
interfere constructively and produce a maximum at this spot.
The distance between the central maximum (position of A) and the first minimum is
y=
λL 
1
λL
=
 m + 
d
2 m=0 2 d
Thus, d =
24.12
(a)
λ L ( 3.00 m ) (150 m )
=
= 11.3 m
2y
2 ( 20.0 m )
The angular position of the bright fringe of order m is given by d sin θ = mλ , or
θ m = sin −1 ( mλ d ) . If d = 25 λ , the first three bright fringes are found at
 1
θ1 = sin −1   = 2.3° ,
 25 
(b)
 3
θ 3 = sin −1   = 6.9°
 25 
The angular position of the dark fringe of order m is given by d sin θ = ( m + 12 )λ , or
θ m = sin −1 [ (m + 12 )λ d ] , m = 0, ± 1, ± 2,K. If d = 25 λ , the first three dark fringes are
found at
 1 
θ 0 = sin −1  2  = 1.1° ,
 25 
(c)
 2
θ 2 = sin −1   = 4.6° , and
 25 
 3 
θ1 = sin −1  2  = 3.4° , and
 25 
5 
θ 2 = sin −1  2  = 5.7°
 25 
The answers are evenly spaced because the angles are small and θ ≈ sin θ . At larger
angles, the approximation breaks down and the spacing isn’t so regular.
24.13
As shown in the figure at the right, the path
difference in the waves reaching the telescope is
δ = d2 − d1 = d2 (1 − sin α ) . If the first minimum
(δ = λ 2 ) occurs when θ = 25.0°, then
α = 180° − (θ + 90.0° + θ ) = 40.0°, and
d2 =
( 250 m 2 ) = 350 m
δ
=
1 − sin α 1 − sin 40.0°
Thus, h = d2 sin 25.0° = 148 m .
56157_24_ch24_p335-360.indd 341
3/20/08 4:44:43 AM
342
24.14
Chapter 24
(a)
y1 4.52 × 10 −3 m
=
= 2.51 × 10 −3 = 0.002 51
L
1.80 m
(b)
tan θ1 =
(c)
θ1 = tan −1 ( 2.51 × 10 −3 ) = 0.144° and sin θ1 = sin ( 0.144°) = 2.51 × 10 −3
The sine and the tangent are very nearly the same, but only because the angle is small.
(d)
From δ = d sin θ = mλ for the order m bright fringe,
λ=
24.15
)
−4
d sin θ1 ( 2.40 × 10 m sin ( 0.144°)
=
= 6.03 × 10 −7 m = 603 nm
1
1
(e)
 5 ( 6.03 × 10 −7 m ) 
 5λ 
θ 5 = sin −1   = sin −1 
 = 0.720°
−4
 d 
 2.40 × 10 m 
(f)
y5 = L tan θ 5 = (1.80 m ) tan ( 0.720°) = 2.26 × 10 −2 m = 2.26 cm
The path difference in the two waves received at the home is δ = 2 d , where d is the distance
from the home to the mountain. Neglecting any phase change upon reflection, the condition for
destructive interference is
1

δ =  m +  λ with m = 0, 1, 2,…

2
24.16
δ min 
1  λ λ 300 m
= 0 +  = =
= 75.0 m

2
4
2 2 4
so
dmin =
(a)
With phase reversal in the reflection at the outer surface of the soap film and no reversal
on reflection from the inner surface, the condition for constructive interference in the light
reflected from the soap bubble is 2 nfilm t = ( m + 12 )λ where m = 0, 1, 2,K. For the lowest
order reflection (m = 0), and the wavelength is
λ=
2 nwater t
= 4 (1.333) (120 nm ) = 6.4 × 10 2 nm = 640 nm
(0 + 1 2)
(b)
To strongly reflect the same wavelength light, a thicker film will need to make use of a
higher order reflection, i.e., use a larger value of m .
(c)
The next greater thickness of soap film that can strongly reflect 640 nm light corresponds to
m = 1, giving
t1 =
(1 + 1 2 ) λ = 3  640 nm  = 3.6 × 10 2


2 nfilm
2  2 (1.333) 
nm = 360 nm
and the third such thickness (corresponding to m = 2) is
t2 =
56157_24_ch24_p335-360.indd 342
( 2 + 1 2 ) λ = 5  640 nm  = 6.0 × 10 2


2 nfilm
2  2 (1.333) 
nm = 600 nm
3/20/08 1:38:20 AM
Wave Optics
24.17
Light reflecting from the first (glass-iodine) interface suffers a phase reversal, but light reflecting
at the second (iodine-glass) interface does not have a phase reversal. Thus, the condition for constructive interference in the reflected light is 2 nfilm t = ( m + 12 )λ with m = 0, 1, 2, K. The smallest
film thickness capable of strongly reflecting the incident light is
t min =
24.18
343
(a)
( 0 + 1 2 ) λ = 6.00 × 10 2
nm
4 (1.756 )
2 nfilm
= 85.4 nm
With nair < noil < nwater, phase reversals are experienced by light reflecting at both surfaces of
the oil film, an upper air-oil interface and a lower oil-water interface. Under these conditions, the requirement for constructive interference is 2 nfilm t = m1λ with m1 = 0, 1, 2,K,
and the requirement for destructive interference is 2 nfilm t = ( m2 + 12 )λ with m2 = 0, 1, 2,K.
To have the thinnest film that produces simultaneous constructive interference of
λ1 = 640 nm and destructive interference of λ2 = 512 nm, it is necessary that
1

2 nfilm t = m1 ( 640 nm ) =  m2 +  ( 512 nm )

2
where both m1 and m2 are the smallest integers for which this is true. It is found that
m1 = m2 = 2 are the smallest integer values that will satisfy this condition, giving the
minimum acceptable film thickness as
t min =
(b)
m1λ1 ( m2 + 12 ) λ2 2 ( 640 nm ) ( 2.5 ) ( 512 nm )
=
=
=
= 512 nm
2 nfilm
2 nfilm
2 (1.25 )
2 (1.25 )
From the discussion above, it is seen that in order to have a film thickness that produces
simultaneous constructive interference of 640-nm light and destructive interference of
512-nm light, it is necessary that
1

m1 ( 640 nm ) =  m2 +  ( 512 nm )

2
(1.25 ) m1 = m2 +
This gives
24.19
1
2
or
or
1
 640 nm 
m1 
= m2 +
 512 nm 
2
2.5 m1 = 2 m2 + 1 .
With ncoating > nair and ncoating > nlens , light reflecting at the air-coating boundary experiences a
phase reversal, but light reflecting from the coating-lens boundary does not. Therefore, the
condition for destructive interference in the two reflected waves is
2ncoating t = mλ
where
m = 0, 1, 2,…
For finite wavelengths, the lowest allowed value of m is m = 1 . Then, if
t = 177.4 nm and ncoating = 1.55 , the wavelength associated with this lowest order destructive
interference is
λ1 =
56157_24_ch24_p335-360.indd 343
2 ncoating t
1
= 2 (1.55 ) (177.4 nm ) = 550 nm
3/20/08 4:50:57 AM
344
24.20
Chapter 24
Since nair < noil < nwater , light reflected from both top and bottom surfaces of the oil film experiences phase reversal, resulting in zero net phase difference due to reflections. Therefore, the
condition for constructive interference in reflected light is
2 t = mλ n = m
 λ 
λ
, or t = m 
where m = 0, 1, 2,…
nfilm
 2 nfilm 
Assuming that m = 1, the thickness of the oil slick is
t = (1)
24.21
600 nm
λ
=
= 2 33 nm
2 nfilm 2 (1.29 )
There will be a phase reversal of the radar waves reflecting from both surfaces of the polymer,
giving zero net phase change due to reflections. The requirement for destructive interference in
the reflected waves is then
λ
1

2 t =  m +  λn , or t = ( 2 m + 1)
where m = 0, 1, 2,…

4 nfilm
2
If the film is as thin as possible, then m = 0 and the needed thickness is
t=
λ
3.00 cm
=
= 0.500 cm
4 nfilm 4 (1.50 )
This anti-reflectance coating could be easily countered by changing the wavelength of the
radar — to 1.50 cm — now creating maximum reflection!
24.22
(a)
With nair < nwater < noil , reflections at the air-oil interface experience a phase reversal, while
reflections at the oil-water interface experience no phase reversal. Thus, with one phase
reversal at the surfaces, the condition for constructive interference in the light reflected by
the film is 2 nfilm t = ( m + 12 )λ , where m is any positive integer. Thus,
λm =
2nfilm t 2 (1.45) ( 280 nm ) 812 nm
=
=
m + 12
m + 12
m + 12
m = 0, 1, 2, 3, K
The possible wavelengths are: λ0 = 1.62 × 10 3 nm, λ1 = 541 nm, λ2 = 325nm,K
Of these, only λ1 = 541 nm (green) is in the visible portion of the spectrum.
(b)
The wavelengths that will be most strongly transmitted are those that suffer destructive
interference in the reflected light. With one phase reversal at the surfaces, the condition
for destructive interference in the light reflected by the film is 2nfilm t = mλ , where m is any
positive, nonzero, integer. The possible wavelengths are
λm =
or
2nfilm t 2 (1.45) ( 280 nm ) 812 nm
=
=
m
m
m
m = 1, 2, 3, K
λ1 = 812 nm, λ2 = 406 nm, λ3 = 271 nm,K
of which only λ2 = 406 nm (violet) is in the visible spectrum.
56157_24_ch24_p335-360.indd 344
3/20/08 1:38:22 AM
Wave Optics
24.23
(a)
345
For maximum transmission, we want destructive interference in the light reflected from the
front and back surfaces of the film.
If the surrounding glass has refractive index greater than 1.378, light reflected from the
front surface suffers no phase reversal, and light reflected from the back does undergo
phase reversal. This effect by itself would produce destructive interference, so we want the
distance down and back to be one whole wavelength in the film. Thus, we require that
2t = λn = λ nfilm
or
t=
λ
656.3 nm
=
= 238 nm
2 nfilm 2 (1.378 )
(b)
The filter will expand. As t increases in 2nfilm t = λ, so does λ increase .
(c)
Destructive interference for reflected light happens also for λ in 2 t = 2 λ nfilm , or
λ = nfilm t = (1.378 ) ( 238 nm ) = 328 nm (near ultraviolet)
24.24
Light reflecting from the lower surface of the air layer experiences phase reversal, but light
reflecting from the upper surface of the layer does not. The requirement for a dark fringe (destructive interference) is then
 λ 
= m λ , where m = 0, 1, 2,…
2 t = mλ n = m 
 nair 
At the thickest part of the film ( t = 2 .00 µ m ) , the order number is
m=
)
−6
2 t 2 ( 2 .00 × 10 m
= 7.32
=
λ
546.1 × 10 −9 m
Since m must be an integer, m = 7 is the order of the last dark fringe seen. Counting the m = 0
order along the edge of contact, a total of 8 dark fringes will be seen.
24.25
With a phase reversal upon reflection from the
lower surface of the air layer and no phase change
for reflection at the upper surface of the layer, the
condition for destructive interference is
 λ 
= m λ , where m = 0, 1, 2,…
2 t = mλ n = m 
 nair 
Counting the zeroth order along the edge of contact, the
order number of the thirtieth dark fringe observed is
m = 29. The thickness of the air layer at this point is
t = 2 r , where r is the radius of the wire. Thus,
r=
56157_24_ch24_p335-360.indd 345
)
−9
t 29 λ 29 ( 600 × 10 m
=
=
= 4.35 µ m
2
4
4
3/20/08 1:38:22 AM
346
Chapter 24
24.26
From the geometry shown in the figure, R 2 = ( R − t ) + r 2, or
2
t = R − R2 − r 2
= 3.0 m −
( 3.0 m )2 − ( 9.8 × 10 −3 m )
2
= 1.6 × 10 −5 m
With a phase reversal upon reflection at the lower surface of the air layer, but no reversal with
reflection from the upper surface, the condition for a bright fringe is
1
1 λ 
1


=  m +  λ , where m = 0, 1, 2,…
2 t =  m +  λn =  m + 


2
2  nair 
2
At the 50 th bright fringe, m = 49, and the wavelength is found to be
λ=
24.27
)
2 (1.6 × 10 −5 m
2t
=
= 6.5 × 10 −7 m = 6.5 × 10 2 nm
m +1 2
49.5
There is a phase reversal due to reflection at the bottom of the air film but not at the top of the
film. The requirement for a dark fringe is then
2 t = mλ n = m
λ
= mλ , where m = 0, 1, 2,…
nair
At the 19 th dark ring (in addition to the dark center spot), the order number is m = 19, and the
thickness of the film is
t=
24.28
mλ 19 ( 500 × 10
=
2
2
−9
m
) = 4.75 × 10
−6
m = 4.75 µ m
With a phase reversal due to reflection at each surface of the magnesium fluoride layer, there is
zero net phase difference caused by reflections. The condition for destructive interference is then
1
1 λ


2 t =  m +  λn =  m + 
, where m = 0, 1, 2,…


2
2  nfilm
For minimum thickness, m = 0 , and the thickness is
t = ( 2 m + 1)
56157_24_ch24_p335-360.indd 346
( 550 × 10 m ) = 9.96 × 10−8 m = 99.6 nm
λ
= (1)
4 nfilm
4 (1.38 )
−9
3/20/08 1:38:23 AM
Wave Optics
24.29
347
There is a phase reversal upon reflection at each surface of the film and hence zero net phase
difference due to reflections. The requirement for constructive interference in the reflected light
is then
2 t = mλ n = m
λ
, where m = 1, 2, 3, …
nfilm
With t = 1.00 × 10 −5 cm = 100 nm, and nfilm = 1.38 , the wavelengths intensified in the reflected
light are
λ=
2 nfilm t 2 (1.38 ) (100 nm )
=
, with m = 1, 2, 3, …
m
m
Thus, λ = 276 nm, 138 nm, 92.0 nm …
and none of these wavelengths are in the visiblee spectrum .
24.30
The transmitted light is brightest when the reflected light is a minimum (that is, the same conditions that produce destructive interference in the reflected light will produce constructive interference in the transmitted light). As light enters the air layer from glass, any light reflected at this
surface has zero phase change. Light reflected from the other surface of the air layer (where light
is going from air into glass) does have a phase reversal. Thus, the condition for destructive interference in the light reflected from the air film is 2 t = m λn , m = 0, 1, 2,… .
Since λn =
λ
λ
=
= λ , the minimum nonzero plate separation satisfying this condition is
nfilm 1.00
d = t = (1)
24.31
λ 580 nm
=
= 290 nm
2
2
In a single slit diffraction pattern, with the slit having width a, the dark fringe of order m occurs
at angle θ m , where sin θ m = m(λ a ) and m = ±1, ± 2, ± 3,K. The location, on a screen located
distance L from the slit, of the dark fringe of order m (measured from y = 0 at the center of the
central maximum) is ( ydark )m = L tan θ m ≈ L sin θ m = mλ ( L a ).
(a)
The central maximum extends from the m = −1 dark fringe to the m = +1 dark fringe, so
the width of this central maximum is
 λ L  2λ L
 λL
− ( −1) 
=
Central max. width = ( ydark )1 − ( ydark )−1 = 1 
 a 
 a 
a
=
(b)
)
2 ( 5.40 × 10 −7 m (1.50 m )
0.200 × 10
−3
m
= 8.10 × 10 −3 m = 8.10 mm
The first order bright fringe extends from the m = 1 dark fringe to the m = 2 dark fringe, or
( ∆y ) = ( y ) − ( y ) = 2  λaL  − 1  λaL  = λaL
( 5.40 × 10 m ) (1.50 m ) = 4.05 × 10 m =
=
bright 1
dark 2
dark 1
−7
−3
0.200 × 10 −3 m
4.05 mm
Note that the width of the first order bright fringe is exactly one-half the width of the
central maximum.
56157_24_ch24_p335-360.indd 347
3/20/08 5:03:10 AM
348
24.32
Chapter 24
(a)
Dark bands occur where sin θ = m ( λ a ) . At the first dark band, m = 1, and the distance
from the center of the central maximum is
 λ
y1 = L tan θ ≈ L sin θ = L  
 a
 600 × 10 −9 m 
= 2 .25 × 10 −3 m = 2 .3 mm
= (1.5 m ) 
 0.40 × 10 −3 m 
24.33
(b)
The width of the central maximum is 2 y1 = 2 ( 2 .25 mm ) = 4.5 mm
(a)
Dark bands (minima) occur where sin θ = m(λ a ). For the first minimum, m = 1 and the
distance from the center of the central maximum is y1 = L tan θ ≈ L sin θ = L ( λ a ) . Thus,
the needed distance to the screen is
 0.75 × 10 −3 m 
 a
= 1.1 m
L = y1   = ( 0.85 × 10 −3 m 
 λ
 587.5 × 10 −9 m 
)
(b)
24.34
The width of the central maximum is 2 y1 = 2 ( 0.85 mm ) = 1.7 mm .
Note: The small angle approximation does not
work well in this situation. Rather, you should
proceed as follows.
At the first order minimum, sin θ1 = (1) λ a or
 5.00 cm 
 λ
= 7.98°
θ1 = sin −1   = sin −1 
 36.0 cm 
 a
Then, y1 = L tan θ1 = ( 6.50 m ) tan 7.98° = 0.912 m = 91.2 cm
24.35
With the screen locations of the dark fringe of order m at
( ydark )m = L tan θ m ≈ L sin θ m = m(λ L a ) for m = ±1, ± 2, ± 3,K
the width of the central maximum is ∆ycentral
maximum
= ( ydark )m =+1 − ( ydark )m =−1 = 2(λ L a), so


a  ∆ycentral 
−3
−3
 maximum  ( 0.600 × 10 m ( 2.00 × 10 m
λ=
=
= 4.62 × 10 −7 m = 462 nm
2 (1.30 m )
2L
)
24.36
)
At the positions of the minima, sin θ m = m ( λ a ) and
ym = L tan θ m ≈ L sin θ m = m  L ( λ a ) 
Thus, tan θ p = n2 n1
and
56157_24_ch24_p335-360.indd 348
a=
)
2 ( 0.500 m ) ( 680 × 10 −9 m
2Lλ
=
= 2 .27 × 10 −4 m = 0.227 mm
y3 − y1
3.00 × 10 −3 m
3/20/08 1:38:25 AM
Wave Optics
24.37
349
The locations of the dark fringes (minima) mark the edges of the maxima, and the widths of the
maxima equals the spacing between successive minima.
At the locations of the minima, sin θ m = m(λ a ) and
ym = L tan θ m ≈ L sin θ m = m  L ( λ a ) 

 500 × 10 −9 m  
= m (1.20 m ) 
 = m (1.20 mm )
 0.500 × 10 −3 m  

Then, ∆y = ∆m(1.20 mm) and for successive minima, ∆m = 1.
Therefore, the width of each maxima, other than the central maximum, in this interference
pattern is
width = ∆y = (1) (1.20 mm ) = 1.20 mm
24.38
For diffraction by a grating, the angle at which the maximum of order m occurs is given by
d sin θ m = mλ , where d is the spacing between adjacent slits on the grating. Thus,
d=
24.39
)
(1) ( 632.8 × 10 −9 m
mλ
=
= 1.81 × 10 −6 m = 1.81 µ m
sin θ m
sin 20.5°
The grating spacing is d = (1 3 660 )cm = (1 3.66 × 10 5 ) m and d sin θ = mλ .
(a)
The wavelength observed in the first-order spectrum is λ = d sin θ , or
9
 10 4 nm 
 1 m   10 nm 
sin θ
λ=
sin θ = 
5 

 3.66 × 10   1 m 
 3.66 
(b)
This yields:
at 10.1°, λ = 479 nm ;
and
at 14.8°, λ = 698 nm
In the second order, m = 2. The second order images for the above wavelengths will be
found at angles θ 2 = sin −1 ( 2 λ d ) = sin −1 [ 2 sin θ1 ] .
This yields:
and
56157_24_ch24_p335-360.indd 349
at 13.7°, λ = 647 nm ;
for λ = 479 nm, θ 2 = 20.5° ;
for λ = 698 nm, θ 2 = 30.7°
for λ = 647 nm, θ 2 = 28.3° ;
3/20/08 4:28:07 AM
350
24.40
Chapter 24
(a)
The longest wavelength in the visible spectrum is 700 nm, and the grating spacing is
d = 1 mm 600 = 1.67 × 10 −3 mm = 1.67 × 10 −6 m.
Thus, mmax =
)
−6
d sin 90.0° (1.67 × 10 m sin 90.0°
= 2.38
=
λred
700 × 10 −9 m
so 2 complete orders will be observed.
(b)
From λ = d sin θ , the angular separation of the red and violet edges in the first order will be
−9
m
 700 × 10 −9 m 
λ 
λ

−1  400 × 10
−
sin
∆θ = sin −1  red  − sin −1  violet  = sin −1 
−6
−6



 d 
 d 
 1.67 × 10 m 
 1.67 × 10 m 
or
24.41
∆θ = 10.9°
1 cm
1m
=
. From d sin θ = mλ, the angular separation
4 500 4.50 × 10 5
between the given spectral lines will be
The grating spacing is d =
 m λred 
 m λviolet 
∆θ = sin −1 
− sin −1 

d


 d 
or
 m ( 656 × 10 −9 m ) ( 4.50 × 10 5 ) 
 m ( 434 × 10 −9 m ) ( 4.50 × 10 5 ) 
−1
∆θ = sin −1 
 − sin 

1m
1m




The results obtained are: for m = 1, ∆θ = 5.91° ; for m = 2, ∆θ = 13.2° ;
and for m = 3, ∆θ = 26.5° . Complete orders for m ≥ 4 are not visible.
24.42
(a)
1 cm
= 6.67 × 10 −4 cm = 6.67 × 10 −6 m , the highest order of λ = 500 nm that can
1 500
be observed will be
If d =
mmax =
(b)
If d =
1 cm
= 6.67 × 10 −5 cm = 6.67 × 10 −7 m, then
15 000
mmax =
56157_24_ch24_p335-360.indd 350
)
−6
d sin 90° ( 6.67 × 10 m (1)
=
= 13.3 or 13 orders
λ
500 × 10 −9 m
)
−7
d sin 90° ( 6.67 × 10 m (1)
=
= 1.33 or 1 order
λ
500 × 10 −9 m
3/20/08 1:38:26 AM
Wave Optics
24.43
351
1 cm
= 2 .00 × 10 −4 cm = 2 .00 × 10 −6 m , and d sin θ = mλ gives the
5 000
angular position of a second order spectral line as
The grating spacing is d =
sin θ =
2λ
 2λ
or θ = sin −1 
 d 
d
For the given wavelengths, the angular positions are
 2 ( 610 × 10 −9 m ) 
 2 ( 480 × 10 −9 m ) 
−1
θ
sin
=
37
and
θ1 = sin −1 
.
6°
=


 = 28.7°
2
−6
−6
 2 .00 × 10 m 
 2 .00 × 10 m 
If L is the distance from the grating to the screen, the distance on the screen from the central
maximum to a second order bright line is y = Ltanq. Therefore, for the two given wavelengths, the
screen separation is
∆y = L [ tan θ1 − tan θ 2 ]
= ( 2 .00 m )  tan ( 37.6°) − tan ( 28.7°)  = 0.445 m = 44.5 cm
24.44
With 2 000 lines per centimeter, the grating spacing is
d=
1
cm = 5.00 × 10 −4 cm = 5.00 × 10 −6 m
2 000
Then, from d sin θ = mλ , the location of the first order for the red light is
 1 ( 640 × 10 −9 m ) 
 mλ 
−1
=
θ = sin −1 
sin
 5.00 × 10 −6 m  = 7.35°
 d 


24.45
The spacing between adjacent slits on the grating is
d=
1 cm
10 −2 m
=
5 310 slits 5 310
The maximum of order m is located where d sin θ m = mλ , so
λ=
24.46

d
d
ym
10 −2 m 

sin θ m = 
=

m
m  L2 + ym2  (1) ( 5 310 ) 

 = 5.14 × 10 −7 m = 514 nm
2
+ ( 0.488 m ) 
0.488 m
(1.72 m )2
From d sin θ m = mλ , or sin θ m = mλ d , we see that
2 λviolet 800 nm
=
d
d
2λ
1 400 nm
sin θ r 2 = red =
d
d
sin θ v2 =
3λviolet 1 200 nm
=
d
d
3λ
2 100 nm
sin θ r 3 = red =
d
d
sin θ v 3 =
Since, for 0° ≤ θ ≤ 90°, θ increases as sin θ increases,
we have that
θv 2 < θv 3 < θr 2 < θr 3
so the second and third order spectra overlap in the range θ v 3 ≤ θ ≤ θ r 2 .
56157_24_ch24_p335-360.indd 351
3/20/08 1:38:27 AM
352
24.47
Chapter 24
1 cm 10 −2 m
=
= 3.636 × 10 −6 m . From d sin θ = mλ , or
2 750 2 750
θ = sin −1 ( mλ d ) , the angular positions of the red and violet edges of the second-order spectrum
are found to be
The grating spacing is d =
 2 ( 700 × 10 −9 m ) 
 2λ 
θ r = sin −1  red  = sin −1 
 = 22.65°
−6
 d 
 3.636 × 10 m 
and
 2 ( 400 × 10 −9 m ) 
 2λ

θ v = sin −1  violet  = sin −1 
 = 12.71°
−6
 d 
 3.636 × 10 m 
Screen
Note from the sketch at the right that yr = L tan θ r
and yv = L tan θ v, so the width of the spectrum on
the screen is ∆y = L ( tan θ r − tan θ v ).
Grating
Since it is given that ∆y = 1.75 cm, the distance
from the grating to the screen must be
L=
24.48
∆y
qv
∆y
1.75 cm
=
tan θ r − tan θ v tan ( 22.65° ) − tan (12.71°° )
or
L = 9.13 cm
(a)
d=
(b)
d sin θ m = mλ
(c)
y2 = L tan θ 2 = ( 2.000 m ) tan ( 29.65°) = 1.138 m
(d)
 2 ( 589.6 × 10 −9 m ) 
 2λ ′ 
= sin −1 
θ 2′ = sin −1 
 = 29.69°

−6
 d 
 2.381 × 10 m 
yv
qr
yr
L
1 cm
= 2.381 × 10 −4 cm = 2.381 × 10 −6 m
4.200 × 10 3 slits
⇒
 2 ( 589..0 × 10 −9 m ) 
 2λ 
θ 2 = sin −1   = sin −1 
 = 29.65°
−6
 d 
 2.381 × 10 m 
y2′ = L tan θ 2′ = ( 2.000 m ) tan ( 29.69°) = 1.140 m
(e)
∆y = y2′ − y2 = 1.140 m − 1.138 m = 2 × 10 −3 m
(f )
 

 2λ   
 2λ ′  
− tan  sin −1    
∆y = y2′ − y2 = L  tan  sin −1 



 d  


d 
 
)
)
 

 2 ( 589.6 × 10 −9 m  
 2 ( 589.0 × 10 −9 m   
= ( 2.000 m )  tan  sin −1  −2
− tan  sin −1  −2
 
3 
10 m 4.200 × 10  
10 m 4.200 × 10 3   
 





= 1.537 × 10 −3 m
The two answers agree to only one significant figure. The calculation is sensitive to rounding at
intermediate steps.
56157_24_ch24_p335-360.indd 352
3/20/08 1:38:28 AM
Wave Optics
24.49
The grating spacing is d =
353
1.00 mm
= 2.50 × 10 −3 mm = 2.50 × 10 −6 m.
400
From d sin θ = mλ , the angle of the second-order diffracted ray is θ = sin −1 (2 λ d ) .
(a)
When the grating is surrounded by air, the wavelength is λair = λ nair ≈ λ and
 2 ( 541 × 10 −9 m ) 
 2λ 
θ a = sin −1  air  = sin −1 
 = 25.6°
−6
 d 
 2.50 × 10 m 
(b)
If the grating is immersed in water,
λn = λwater = λ nwater = λ 1.333, yielding
then
 2 ( 541 × 10 −9 m ) 
 2λ 
θb = sin −1  water  = sin −1 
 = 19.0°
−6
 d 
 ( 2.50 × 10 m ) (1.333) 
(c)
mλ n m ( λ n )
mλ
=
, we have that n sin θ =
= constant when m is kept
d
d
d
constant. Therefore nair sin θ a = nwater sin θb , or the angles of parts (a) and (b) satisfy
From sin θ =
Snell’s law.
24.50
The grating spacing is d =
1 cm
= 8.33 × 10 −4 cm = 8.33 × 10 −6 m.
1 200
mλ
and the small angle approximation, the distance from the central maximum to
d
the maximum of order m for wavelength λ is ym = L tan θ ≈ L sin θ = (λ L d )m . Therefore, the
spacing between successive maxima is ∆y = ym +1 − ym = λ L d .
Using sin θ =
The longer wavelength in the light is found to be
λlong =
( ∆y ) d
L
=
( 8.44 × 10
−3
)
m ( 8.33 × 10 −6 m
0.150 m
)=
469 nm
Since the third order maximum of the shorter wavelength falls halfway between the central
maximum and the first order maximum of the longer wavelength, we have
 1
3λshort L  0 + 1 λlong L
or λshort =   ( 469 nm ) = 78.1 nm
=

 6
 2  d
d
24.51
(a)
From Brewster’s law, the index of refraction is
n2 = tan θ p = tan ( 48.0°) = 1.11
(b)
From Snell’s law, n2 sin θ 2 = n1 sin θ1, we obtain when θ1 = θ p
 n1 sin θ p 
 (1.00 ) sin 48.0° 
θ 2 = sin −1 
= sin −1 
 = 42 .0°


1.11
 n2 
Note that when θ1 = θ p , θ 2 = 90.0° − θ p as it should.
56157_24_ch24_p335-360.indd 353
3/20/08 1:38:29 AM
354
24.52
Chapter 24
(a)
From Malus’s law, the fraction of the incident intensity of the unpolarized light that is transmitted by the polarizer is
I ′ = I 0 ( cos 2 θ
)
av
= I 0 ( 0.500 )
The fraction of this intensity incident on the analyzer that will be transmitted is
I = I ′ cos 2 ( 35.0°) = I ′ ( 0.671) = I 0 ( 0.500 ) ( 0.671) = 0.336 I 0
Thus, the fraction of the incident unpolarized light transmitted is I I 0 = 0.336 .
(b)
The fraction of the original incident light absorbed by the analyzer is
I ′ − I 0.500 I 0 − 0.336 I 0
=
= 0.164
I0
I0
24.53
The more general expression for Brewster’s angle is (see Problem 57)
tan θ p = n2 n1
24.54
(a)
n 
 1.52 
When n1 = 1.00 and n2 = 1.52 , θ p = tan −1  2  = tan −1 
= 56.7°°
 1.00 
 n1 
(b)
n 
 1.52 
When n1 = 1.333 and n2 = 1.52 , θ p = tan −1  2  = tan −1 
= 48.8°
 1.333 
 n1 
The polarizing angle for light in air striking a water surface is
n 
 1.333 
θ p = tan −1  2  = tan −1 
= 53.1°
 1.00 
 n1 
This is the angle of incidence for the incoming sunlight (that is, the angle between the incident
light and the normal to the surface). The altitude of the Sun is the angle between the incident light
and the water surface. Thus, the altitude of the Sun is
α = 90.0° − θ p = 90.0° − 53.1° = 36.9°
24.55
(a)
Brewster’s angle (or the polarizing angle) is
 nquartz 
n 
1.458 
θ p = tan −1  2  = tan −1 
= tan −1 
= 55.6°
 1.000 
 n1 
 nair 
(b)
24.56
When the angle of incidence is the polarizing angle, θ p , the angle of refraction of the transmitted light is θ 2 = 90.0° − θ p . Hence, θ 2 = 90.0° − 55.6° = 34.4° .
The critical angle for total reflection is θ c = sin −1 ( n2 n1 ) . Thus, if θ c = 34.4° as light attempts to
go from sapphire into air, the index of refraction of sapphire is
nsapphire = n1 =
n2
1.00
=
= 1.77
sin θ c sin 34.4°
Then, when light is incident on sapphire from air, the Brewster angle is
n 
 1.77 
θ p = tan −1  2  = tan −1 
= 60.5°°
 1.00 
 n1 
56157_24_ch24_p335-360.indd 354
3/20/08 5:02:45 AM
Wave Optics
24.57
355
From Snell’s law, the angles of incidence and refraction are related by n1 sin θ1 = n2 sin θ 2.
If the angle of incidence is the polarizing angle (that is, θ1 = θ p), the refracted ray is perpendicular
to the reflected ray (see Figure 24.28 in the textbook), and the angles of incidence and refraction
are also related by
θ p + θ 2 + 90° = 180°, or
θ 2 = 90° − θ p
Substitution into Snell’s law then gives
)
(
n1 sin θ p = n2 sin 90° − θ p = n2 cos θ p or sin θ p cos θ p = tan θ p = n2 n1
24.58
24.59
I = I 0 cos 2 θ
⇒
 I 
θ = cos −1 

 I0 
(a)
I
1
=
I 0 2.00
⇒

1 
θ = cos −1 
 = 45.0°
 2.00 
(b)
I
1
=
I 0 4.00
⇒

1 
θ = cos −1 
 = 60.0°
 4.00 
(c)
I
1
=
I 0 6.00
⇒

1 
θ = cos −1 
 = 65.9°
 6.00 
From Malus’s law, the intensity of the light transmitted by the first polarizer is I1 = I i cos 2 θ1 .
The plane of polarization of this light is parallel to the axis of the first plate and is
incident on the second plate. Malus’s law gives the intensity transmitted by the second plate
as I 2 = I1 cos 2 (θ 2 − θ1 ) = I i cos 2 θ1 cos 2 (θ 2 − θ1 ) . This light is polarized parallel to the axis of
the second plate and is incident upon the third plate. A final application of Malus’s law gives the
transmitted intensity as
I f = I 2 cos 2 (θ 3 − θ 2 ) = I i cos 2 θ1 cos 2 (θ 2 − θ1 ) cos 2 (θ 3 − θ 2 )
With θ1 = 20.0°, θ 2 = 40.0°, and θ 3 = 60.0°, this result yields
I f = (10.0 units ) cos 2 ( 20.0°) cos 2 ( 20.0°) cos 2 ( 20.0°) = 6.89 units
24.60
(a)
Using Malus’s law, the intensity of the transmitted light is found to be
(
I = I 0 cos 2 ( 45°) = I 0 1
(b)
(a)
2
From Malus’s law, I I 0 = cos 2 θ . Thus, if I I 0 = 1 3 we obtain
cos 2 θ = 1 3
24.61
)
2 , or I I 0 = 1 2
or
(
θ = cos −1 1
)
3 = 54.7°
If light has wavelength λ in vacuum, its wavelength in a medium of refractive index n is
λn = λ n . Thus, the wavelengths of the two components in the specimen are
and
λn1 =
λ 546.1 nm
=
= 413.7 nm
n1
1.320
λn2 =
λ 546.1 nm
=
= 409.7 nm
n2
1.333
continued on next page
56157_24_ch24_p335-360.indd 355
3/20/08 1:38:30 AM
356
Chapter 24
(b)
The number of cycles of vibration each component completes while passing through the
specimen are
and
N1 =
t
1.000 × 10 − 6 m
=
= 2.417
λn1 413.7 × 10 −9 m
N2 =
t
1.000 × 10 −6 m
=
= 2.441
λn2 40 9.7 × 10 −9 m
Thus, when they emerge, the two components are out of phase by N 2 − N1 = 0.024 cycles.
Since each cycle represents a phase angle of 360°, they emerge with a phase difference of
∆φ = ( 0.024 cycles ) ( 360° cycle ) = 8.6°
24.62
In a single slit diffraction pattern, the first dark fringe occurs where sin(θ dark )1 = (1) λ a. If no
diffraction minima are to be observed, it is necessary that no real solutions exist for this equation
defining the location of the first minimum. Thus, it is necessary that sin(θ dark )1 = (1) λ a > 1.00,
or λ a > 1.00 and a < λ . We then see that the maximum width the slit can have before this condition will fail and diffraction minima will start being visible is amax = λ = 632.8 nm .
24.63
The light has passed through a single slit since the central
maximum is twice the width of other maxima (the space between
the centers of successive dark fringes. In a double slit pattern,
the central maximum has the same width as all other maxima
(compare Active Figures 24.1(b) and 24.16(b) in the textbook).
Figure P24.63
In single slit diffraction the width of the central maximum on the screen is given by
∆ycentral
maximum
 (1) λ 
 λL
= 2
= 2 L tan (θ dark )m =1 ≈ 2 L sin (θ dark )m =1 = 2 L 
 a 
 a 
The width of the slit is then
a=
2λ L
∆ycentral
)
maximum
24.64
(a)
2 ( 632.8 × 10 −9 m ( 2.60 m )  10 2 cm 
−4
=
 1 m  = 1.2 × 10 m = 0.12 mm
(10.3 − 7.6 ) cm
In a double slit interference pattern, bright fringes on the screen occur where
( ybright )m = m(λ L d ) . Thus, if bright fringes of the wavelengths λ1 = 540 nm and
λ2 = 450 nm are to coincide, it is necessary that
m1
( 540 nm ) L
d
= m2
( 450 nm ) L
d
or, dividing both sides by 90 nm, 6 m1 = 5 m2 , where both m1 and m2 are integers.
(b)
The condition found above may be written as m2 = (6 5 )m1 . Trial and error reveals that the
smallest nonzero integer value of m1 that will yield an integer value for m2 is m1 = 5 ,
yielding m2 = 6 . Thus, the first overlap of bright fringes for the two given wavelengths
occurs at the screen position (measured from the central maximum)
ybright =
56157_24_ch24_p335-360.indd 356
)
5 ( 540 × 10 −9 m (1.40 m )
0.150 × 10
−3
m
=
)
6 ( 450 × 10 −9 m (1.40 m )
0.150 × 10 −3 m
= 2.52 × 10 −2 m = 2.52 cm
3/20/08 4:52:04 AM
Wave Optics
24.65
357
Dark fringes (destructive interference) occur where d sin θ = ( m + 1 2 ) λ for m = 0, 1, 2, … .
Thus, if the second dark fringe ( m = 1) occurs at
 1.00° 
= 0.300°,
θ = (18.0 min ) 
 60.0 min 
the slit spacing is
)
−9
1 λ

 3  ( 546 × 10 m
d = m+ 
= 
= 1.56 × 10 −4 m = 0.156 mm

2  sin θ  2  sin ( 0.300°)
24.66
The wavelength is λ =
vsound 340 m s
=
= 0.170 m .
f
2 000 Hz
Maxima occur where d sin θ = mλ , or θ = sin −1  m ( λ d )  for m = 0, 1, 2, … .
Since d = 0.350 m , λ d = 0.486, which gives θ = sin −1 ( 0.486 m ).
For m = 0, 1, and 2 , this yields maxima at 0°, 29.1°, and 76.3° .
No solutions exist for m ≥ 3 since that would imply sin θ > 1.

λ 
Minima occur where d sin θ = ( m + 1 2 ) λ or θ = sin −1 ( 2 m + 1)
for m = 0, 1, 2, K.
2
d 

With λ d = 0.486 , this becomes θ = sin −1 ( 2 m + 1) ( 0.243) .
For m = 0 and 1, we find minima at 14.1° and 46.8° .
No solutions exist for m ≥ 2 since that would imply sin θ > 1.
24.67
The source and its image, located 1.00 cm below the mirror, act as a pair of coherent sources.
This situation may be treated as double-slit interference, with the slits separated by 2.00 cm, if
it is remembered that the light undergoes a phase reversal upon reflection from the mirror. The
existence of this phase change causes the conditions for constructive and destructive interference
to be reversed. Therefore, dark bands (destructive interference) occur where
y = m ( λ L d ) for m = 0, 1, 2, …
The m = 0 dark band occurs at y = 0 (that is, at mirror level). The first dark band above the
mirror corresponds to m = 1 and is located at
)
−9
 λ L  ( 500 × 10 m (100 m )
=
= 2 .50 × 10 −3 m = 2 .50 mm
y = (1) 
 d 
2 .00 × 10 −2 m
56157_24_ch24_p335-360.indd 357
3/20/08 1:38:32 AM
358
24.68
Chapter 24
Assuming the glass plates have refractive indices greater than that of both air and water, there
will be a phase reversal at the reflection from the lower surface of the film but no reversal from
reflection at the top of the film. Therefore, the condition for a dark fringe is
2 t = mλn = m ( λ nfilm ) for m = 0, 1, 2, …
If the highest order dark band observed is m = 84 (a total of 85 dark bands counting the m = 0
order at the edge of contact), the maximum thickness of the wedge is
t max =
mmax  λ  84  λ 
=
 = 42 λ

2  nfilm 
2  1.00 
When the film consists of water, the highest order dark fringe appearing will be
n 
 1.333 
= 112
mmax = 2 t max  film  = 2 ( 42 λ ) 
 λ 
 λ 
Counting the m = 0 order, a total of 113 dark fringes are now observed.
24.69
In the figure at the right, observe that the path difference
between the direct and the indirect paths is
δ = 2 x − d = 2 h2 + (d 2) − d
2
With a phase reversal (equivalent to a half-wavelength
shift) occurring on the reflection at the ground, the
condition for constructive interference is δ = ( m + 1 2 ) λ , and
the condition for destructive interference is δ = m λ . In both cases,
the possible values of the order number are m = 0, 1, 2, … .
(a)
δ
The wavelengths that will interfere constructively are λ =
. The longest of these is
m +1 2
for the m = 0 case and has a value of
λ = 2δ = 4 h 2 + ( d 2 ) − 2 d
2
=4
(b)
( 50.0 m )2 + ( 300 m )2
The wavelengths that will interfere destructively are λ = δ m, and the largest finite one of
these is for the m = 1 case. That wavelength is
λ = δ = 2 h2 + (d 2) − d = 2
2
24.70
− 2 ( 600 m ) = 16.6 m
( 50.0 m )2 + ( 300 m )2
− 600 m = 8.28 m
From Malus’s law, the intensity of the light transmitted by the first polarizer is I1 = I i cos 2 θ1.
The plane of polarization of this light is parallel to the axis of the first plate and is incident on the second plate. Malus’s law gives the intensity transmitted by the second plate as
I 2 = I1 cos 2 (θ 2 − θ1 ) = I i cos 2 θ1 cos 2 (θ 2 − θ1 ) . This light is polarized parallel to the axis of the
second plate and is incident upon the third plate. A final application of Malus’s law gives the
transmitted intensity as
I f = I 2 cos 2 (θ 3 − θ 2 ) = I i cos 2 θ1 cos 2 (θ 2 − θ1 ) cos 2 (θ 3 − θ 2 )
continued on next page
56157_24_ch24_p335-360.indd 358
3/20/08 1:38:33 AM
Wave Optics
(a)
359
If θ1 = 45°, θ 2 = 90°, and θ 3 = 0° , then
I f I i = cos 2 45° cos 2 ( 90° − 45°) cos 2 ( 0° − 90°) = 0
(b)
If θ1 = 0°, θ 2 = 45°, and θ 3 = 90°, then
I f I i = cos 2 0° cos 2 ( 45° − 0°) cos 2 ( 90° − 45°) = 0.25
24.71
If the signal from the antenna to the
receiver station is to be completely
polarized by reflection from the water,
the angle of incidence where it strikes the
water must equal the polarizing angle from
Brewster’s law. This is given by
n 
θ p = tan −1  water  = tan −1 (1.33) = 53.1°
 nair 
From the triangle RST in the sketch, the
horizontal distance from the point of refection, T, to shore is given by
x = ( 90.0 m ) tan θ p = ( 90.0 m ) (1.33) = 120 m
and from triangle ABT, the horizontal distance from the antenna to this point is
y = ( 5.00 m ) tan θ p = ( 5.00 m ) (1.33) = 6.65 m
The total horizontal distance from ship to shore is then x + y = 120 m + 6.65 m = 127 m .
24.72
There will be a phase reversal associated with
the reflection at one surface of the film but no
reversal at the other surface of the film.
Therefore, the condition for a dark fringe
(destructive interference) is
 λ 
2 t = mλ n = m 
m = 0, 1, 2, …
 nfilm 
From the figure, note that R 2 = r 2 + ( R − t ) = r 2 + R 2 − 2 Rt + t 2, which reduces to r 2 = 2 Rt − t 2.
Since t will be very small in comparison to either r or R, we may neglect the term t 2 , leaving
r ≈ 2 Rt .
2
For a dark fringe, t =
mλ
so the radii of the dark rings will be
2 nfilm
 mλ 
=
r ≈ 2R 
 2 nfilm 
56157_24_ch24_p335-360.indd 359
mλ R
nfilm
for
m = 0, 1, 2, …
3/20/08 1:38:34 AM
360
24.73
Chapter 24
In the single slit diffraction pattern, destructive interference (or minima) occur where
sin θ = m ( λ a ) for m = 0, ± 1, ± 2, … . The screen locations, measured from the center of the
central maximum, of these minima are at
ym = L tan θ m ≈ L sin θ m = m ( λ L a )
If we assume the first-order maximum is halfway between the first- and second-order minima,
then its location is
y=
y1 + y2 (1 + 2 ) ( λ L a ) 3λ L
=
=
2
2
2a
and the slit width is
a=
24.74
)
−9
3λ L 3 ( 500 × 10 m (1.40 m )
=
= 3.50 × 10 −4 m = 0.350 mm
2y
2 ( 3.00 × 10 −3 m
)
As light emerging from the glass reflects from the top of the air layer, there is no phase reversal
produced. However, the light reflecting from the end of the metal rod at the bottom of the air
layer does experience phase reversal. Thus, the condition for constructive interference in the
reflected light is 2 t = ( m + 12 ) λair .
As the metal rod expands, the thickness of the air layer decreases. The increase in the length of
the rod is given by
∆L = ∆t = ( mi +
1
2
) λair − ( m f + 12 ) λair
2
2
= ∆m
λair
2
The order number changes by one each time the film changes from bright to dark and back to
bright. Thus, during the expansion, the measured change in the length of the rod is
∆L = ( 200 )
( 500 × 10
λair
= ( 200 )
2
2
−9
m
) = 5.00 × 10
−5
m
From ∆L = L0α ( ∆T ) , the coefficient of linear expansion of the rod is
α=
56157_24_ch24_p335-360.indd 360
∆L
5.00 × 10 −5 m
−1
=
= 20.0 × 10 −6 ( °C )
L0 ( ∆T ) ( 0.100 m ) ( 25.0°C )
3/20/08 1:38:35 AM