Chapter 24
Wave Optics
Quick Quizzes
1.
(c). The fringes on the screen are equally spaced only at small angles where tan θ ≈ sin θ is
a valid approximation.
2.
(b). The space between successive bright fringes is proportional to the wavelength of the
light. Since the wavelength in water is less than that in air, the bright fringes are closer
together in the second experiment.
3.
(b). The outer edges of the central maximum occur where sin θ = ± λ a . Thus, as a, the
width of the slit, becomes smaller, the width of the central maximum will increase.
4.
The compact disc. The tracks of information on a compact disc are much closer together
than on a phonograph record. As a result, the diffraction maxima from the compact disc
will be farther apart than those from the record.
303
304
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.
(a) Two waves interfere constructively if their path difference is either zero or some
integral multiple of the wavelength; that is, if the path difference is mλ, where m is an
integer. (b) Two waves interfere destructively if their path difference is an odd multiple of
1
λ

one-half of a wavelength; that is, if the path difference equals  m +  λ = ( 2m + 1) .
2
2

10.
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.
12.
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.
14.
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.
Wave Optics
305
16.
One way to produce interference patterns is to allow light to pass through very small
openings. The opening between threads in a tautly stretched cloth like that in an umbrella
is small enough for the effects to be observed.
18.
Sound waves are longitudinal waves and cannot be polarized.
20.
The first experiment. The separation between maxima is inversely proportional to the slit
separation (see Eq. 24.5), so increasing the slit separation causes the distance between the
two maxima to decrease.
22.
The separations are greater in the second experiment when using red light having the
longer wavelength.
306
CHAPTER 24
Answers to Even Numbered Problems
1.77 µ m
2.
(a)
4.
2.61 m
6.
1.5 mm
8.
3.00 cm
1.93 µ m
(b)
1.47 µ m
(b)
δ = 3λ
(b)
81.58 nm
(b)
4.5 mm
10.
(a)
12.
1.73 km
14.
(a)
16.
193 nm
18.
233 nm
20.
290 nm
22.
8 (counting the zeroth order)
24.
6.5 × 10 2 nm
26.
99.6 nm
28.
20.0 × 10 −6 ( °C )
30.
(a)
32.
91.2 cm
34.
0.227 mm
36.
(a)
2 complete orders
(b)
10.9°
38.
(a)
13 orders
(b)
1 order
40.
7.35°
42.
469 nm and 78.1 nm
44.
632.8 nm
46.
38
123.4 nm
−1
2.3 mm
(c)
maximum
Wave Optics
48.
36.9°
50.
60.5°
52.
(a)
54.7°
(b)
63.4°
54.
(a)
I I0 = 1 2
(b)
54.7°
56.
432 nm
58.
maxima at 0°, 29.1°, and 76.3°
minima at 14.1° and 46.8°
60.
113 dark fringes
62.
(a)
(b)
0.25
68.
313 nm
70.
74 µ m
0
(c)
71.6°
307
308
CHAPTER 24
Problem Solutions
24.1
∆ybright = ym+1 − ym =
( 632.8 × 10
=
λL
−9
d
( m + 1) −
d
m ) ( 5.00 m )
0.200 × 10 −3 m
24.2
λL
m=
λL
d
= 1.58 × 10 −2 m = 1.58 cm
(a) For a bright fringe of order m, the path difference is δ = mλ , where m = 0, 1, 2,… At
the location of the third order bright fringe, m = 3 and
δ = 3 λ = 3 ( 589 nm ) = 1.77 × 10 3 nm = 1.77 µ m
1

(b) For a dark fringe, the path difference is δ =  m +  λ , where m = 0,1, 2, …
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

24.3
2
2
(a) The distance between the central maximum and the first order bright fringe is
λL
∆y = ybright
− ybright
=
, or
m =1
m=0
d
∆y =
λL
d
=
( 546.1 × 10
−9
m ) ( 1.20 m )
0.250 × 10 −3 m
= 2.62 × 10 −3 m = 2.62 mm
(b) The distance between the first and second dark bands is
∆y = ydark
24.4
m =1
− ydark
m=0
=
λL
d
= 2.62 mm as in (a) above.
λL
1
 m +  , the spacing between the first and second dark fringes is
2
d 
λL 3 1 λL
∆y =
. Thus, the required distance to the screen is
 − =
d 2 2 d
From ydark =
L=
( ∆y ) d = ( 4.00 × 10−3 m )( 0.300 × 10-3 m ) =
λ
460 × 10 −9 m
2.61 m
Wave Optics
24.5
(a) From d sin θ = mλ , the angle for the m = 1 maximum for the sound waves is
m v
m 
λ  = sin −1   sound
d 
 d  f
θ = sin −1 


 354 m s  
1
−1
  = sin 

  = 36.2°
 0.300 m  2 000 Hz  
 
(b) For 3.00-cm microwaves, the required slit spacing is
d=
( 1)( 3.00 cm )
mλ
=
= 5.08 cm
sin θ
sin36.2°
(c) The wavelength is λ =
d sin θ
; and if this is light, the frequency is
m
( 1) ( 3.00 × 108 m s )
mc
f= =
=
= 5.08 × 1014 Hz
λ d sin θ ( 1.00 × 10 -6 m ) sin 36.2°
c
24.6
The position of the first order bright fringe for wavelength λ is y1 =
λL
d
−9
( ∆λ ) L ( 700 − 400 ) × 10 m  ( 1.5 m )
Thus, ∆y1 =
=
= 1.5 × 10 −3 m = 1.5 mm
−3
0.30 × 10
d
24.7
m
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 ,
or λ =
or λ =
y
d
d
sin θ = 
2
2
2  L + y2


( 300 m ) 
2






400 m
2
(1 000 m ) + ( 400 m )2
400 m
300 m

 = 55.7 m


1 000 m
309
310
CHAPTER 24
(b) The next minimum encountered is the m = 2 minimum; and at that point,
1
5
δ = d sin θ =  m +  λ = λ

2
2


 5λ 
−1 5 ( 55.7 m )
or θ = sin −1 
 = 27.7°
 = sin 
 2d 
 2 ( 300 m ) 
y = ( 1 000 m ) tan 27.7° = 524 m
Then,
so the car must travel an additional 124 m
24.8
In a double-slit interference pattern the distance from the central maximum to the
position of the mth order bright fringe is given by
 λL 
ym = m 

 d 
where d is the distance between the splits and L is the distance to the screen. Thus, the
spacing between the first- and second-order bright fringes is
 ( 600 × 10 −9 m ) ( 2.50 m ) 
 λL 
 = 0.0300 m = 3.00 cm
∆y = y2 − y1 = [2 − 1] 
 = 1
0.050 × 10 −3 m
 d 


24.9
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,…

so
24.10
dmin =
2
δ min
2
1  λ λ 300 m

= 0 +  = =
= 75.0 m
2
4

2 4
The angular deviation from the line of the central maximum is given by
y
θ = tan −1   = tan −1 
 140 cm
L
1.80 cm 
 = 0.737°

311
Wave Optics
(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)
λ
δ = ( 1.93 × 10 −6 m ) 
 643 × 10
−9

 = 3.00 λ
m
(c) Since the path difference for this position is a whole number of wavelengths, the
waves interfere constructively and produce a maximum at this spot.
24.11
The distance between the central maximum (position of A) and the first minimum is
λL 
1
λL
y=
=
m+ 
d 
2  m=0 2 d
Thus, d =
24.12
( 3.00 m )( 150 m )
= 11.3 m
2y
2 ( 20.0 m )
λL
=
The path difference in the two waves received
at the home is δ = 2d − 30.0 km where d is defined
in the figure at the right. For minimum cloud
height and (hence minimum path difference) to
yield destructive interference, δ = λ 2 giving
1
λ
dmin =  30.0 km +  = 15.1 km , and
2
2
2
hmin = dmin
− ( 15.0 km ) =
2
24.13
Cloud
d
d
h
15.0 km
Transmitter
15.0 km
Receiver
2
2
(15.1 km ) − ( 15.0 km ) = 1.73 km
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 )
d1
occurs when θ = 25.0° , then
α = 180° − (θ + 90.0° + θ ) = 40.0° , and
d2 =
δ
1 − sin α
=
( 250 m 2 )
1 − sin 40.0°
Thus, h = d2 sin 25.0° = 148 m
q
= 350 m
a
q
d2
h
312
CHAPTER 24
24.14
With n film > nair , light reflecting from the front surface of the film (an air to film
boundary) experiences a 180° phase shift, but light reflecting from the back surface (a
film to air boundary) experiences no shift. The condition for constructive interference in
the two reflected waves is then
1

2n filmt =  m +  λ
2

with
For minimum thickness, m = 0 , giving
24.15
m = 0, 1, 2, …
tmin =
λ
4 n film
(a)
With λ = 656.3 nm and n film = 1.330 , tmin =
656.3 nm
= 123.4 nm
4 ( 1.330 )
(b)
When λ = 434.0 nm and n film = 1.330 , tmin =
434.0 nm
= 81.58 nm
4 ( 1.330 )
Light reflecting from the upper surface undergoes phase reversal while that reflecting
from the lower surface does not. The condition for constructive interference in the
reflected light is then
2t −
λn
1λ 
1 λ

, m = 0, 1, 2, …
= mλn , or t =  m +  n =  m + 
2
2 2 
2  2 n film

For minimum thickness, m = 0 giving
t=
24.16
λ
4 n film
=
500 nm
= 91.9 nm
4 ( 1.36 )
With nglass > nair and nliquid < nglass , light reflecting from the air-glass boundary experiences
a 180° phase shift, but light reflecting from the glass-liquid boundary experiences no
shift. Thus, the condition for destructive interference in the two reflected waves is
2nglass t = mλ
where
m = 0, 1, 2, …
For minimum (non-zero) thickness, m = 1 giving
t=
λ
2nglass
=
580 nm
= 193 nm
2 ( 1.50 )
Wave Optics
24.17
313
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 =
24.18
2ncoating t
= 2 ( 1.55 )( 177.4 nm ) = 550 nm
1
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 = mλn = m
λ
n film
 λ
, or t = m 
 2 n film


 where m = 0, 1, 2,…


Assuming that m = 1 , the thickness of the oil slick is
t = (1)
24.19
λ
2 n film
=
600 nm
= 233 nm
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
1
λ

where m = 0, 1, 2,…
2 t =  m +  λn , or t = ( 2m + 1)
4 n film
2

If the film is as thin as possible, then m = 0 and the needed thickness is
t=
λ
4 n film
=
3.00 cm
= 0.500 cm
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!
314
CHAPTER 24
24.20
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
n film
condition is
24.21
=
λ
1.00
= λ , the minimum non-zero plate separation satisfying this
d = t = ( 1)
λ
2
=
580 nm
= 290 nm
2
(a) 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 n film or
t=
λ
2 n film
=
656.3 nm
= 238 nm
2 ( 1.378 )
(b) The filter will expand. As t increases in 2n film t = λ , so does λ increase
(c) Destructive interference for reflected light happens also for λ in 2 t = 2 λ n film , or
λ = n filmt = ( 1.378 )( 238 nm ) = 328 nm (near ultraviolet)
Wave Optics
24.22
315
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 
2 t = mλn = m 
 = m λ , where m = 0, 1, 2,…
 nair 
At the thickest part of the film ( t = 2.00 µ m ) , the order number is
m=
2t
λ
=
2 ( 2.00 × 10 −6 m )
546.1 × 10 −9 m
= 7.32
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.
24.23
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 
2 t = mλn = m 
 = m λ , where m = 0, 1, 2,…
 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 = 2r , where r is the radius of
the wire. Thus,
−9
t 29 λ 29 ( 600 × 10 m )
=
= 4.35 µ m
r= =
2
4
4
Incident
light
n
t3
t2
t1
O
316
CHAPTER 24
h=R–t
24.24
R
Glass
air
no phase reversal
phase reversal
r
t{
Glass
From the geometry shown in the figure, R2 = ( R − t ) + r 2 , or
2
t = R − R2 − r 2
= 3.0 m −
2
( 3.0 m ) − ( 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


2 t =  m +  λn =  m +
2



1 λ 
= m+

2  nair 
1
 λ , where m = 0, 1, 2,…
2
At the 50 th bright fringe, m = 49 , and the wavelength is found to be
2 ( 1.6 × 10 −5 m )
2t
λ=
=
= 6.5 × 10 −7 m = 6.5 × 10 2 nm
m+1 2
49.5
24.25
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 = mλn = m
λ
nair
= mλ , where m = 0, 1, 2,…
At the 19th dark ring (in addition to the dark center spot), the order number is m = 19 ,
and the thickness of the film is
mλ 19 ( 500 × 10
=
t=
2
2
−9
m)
= 4.75 × 10 −6 m = 4.75 µ m
Wave Optics
24.26
317
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

2t =  m +

1

 λn =  m +
2

1 λ
, where m = 0, 1, 2,…

2  n film
For minimum thickness, m = 0 , and the thickness is
t = ( 2 m + 1)
24.27
λ
4 n film
( 550 × 10
= ( 1)
−9
4 ( 1.38 )
m)
= 9.96 × 10 −8 m = 99.6 nm
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 = mλn = m
λ
n film
, where m = 1, 2, 3, …
With t = 1.00 × 10 −5 cm = 100 nm , and n film = 1.38 , the wavelengths intensified in the
reflected light are
λ=
2 n film t
m
=
2 ( 1.38 )( 100 nm )
, with m = 1, 2, 3, …
m
Thus, λ = 276 nm, 138 nm, 92.0 nm …
and none of these wavelengths are in the visible spectrum
318
CHAPTER 24
24.28
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
2
(
− mf +
1
2
) λ2
air
= ∆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 )
λair
2
= ( 200 )
( 500 × 10
2
−9
m)
= 5.00 × 10 −5 m
From ∆L = L0α ( ∆T ) , the coefficient of linear expansion of the rod is
α=
24.29
∆L
5.00 × 10 −5 m
−1
=
= 20.0 × 10 −6 ( °C )
L0 ( ∆T ) ( 0.100 m ) ( 25.0 °C )
The distance on the screen from the center to either edge of the central maximum is
λ
y = L tan θ ≈ L sin θ = L  
a
 632.8 × 10 −9 m 
−3
= ( 1.00 m ) 
 = 2.11 × 10 m=2.11 mm
−3
 0.300 × 10 m 
The full width of the central maximum on the screen is then
2 y = 4.22 mm
Wave Optics
24.30
319
(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 
−3
= ( 1.5 m ) 
 = 2.25 × 10 m = 2.3 mm
−3
 0.40 × 10 m 
(b) The width of the central maximum is 2 y1 = 2 ( 2.25 mm ) = 4.5 mm
24.31
(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
L = y1   = ( 0.85 × 10 −3 m ) 
 = 1.1 m
-9
λ
 587.5 × 10 m 
(b) 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 θ = λ a or
λ
5.00 cm 
θ = sin −1   = sin −1 
 = 7.98°
a
 36.0 cm 
a = 36.0 cm = 0.360 m
L = 6.50 m
24.32
Central
Maximum
Then, y1 = L tan θ = ( 6.50 m ) tan 7.98° = 0.912 m = 91.2 cm
q
y1 First Order
Minimum
320
CHAPTER 24
24.33
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
  λ 
ym = L tan θ m ≈ L sin θ m = m  L   
  a 

 500 × 10 −9 m  
= m ( 1.20 m ) 
  = m ( 1.20 mm )
-3
 0.500 × 10 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
24.34
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, y3 − y1 = ( 3 − 1)  L ( λ a )  = 2  L ( λ a ) 
and
24.35
2 ( 0.500 m ) ( 680 × 10 −9 m )
2Lλ
a=
=
= 2.27 × 10 −4 m = 0.227 mm
−3
y3 − y1
3.00 × 10 m
The grating spacing is d =
1
1
cm =
m and d sin θ = mλ
3 660
3.66 × 10 5
(a) The wavelength observed in the first-order spectrum is λ = d sin θ , or
 10 4 nm 
1 m   109 nm 
sin
θ
=



 sin θ
5
 3.66 × 10   1 m 
 3.66 
λ = 
This yields:
at 10.1°, λ = 479 nm ;
and
at 14.8°, λ = 698 nm
at 13.7°, λ = 647 nm ;
Wave Optics
321
(b) 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 [ 2sin θ1 ]
24.36
This yields:
for λ = 479 nm , θ 2 = 20.5° ;
and
for λ = 698 nm , θ 2 = 30.7°
for λ = 647 nm , θ 2 = 28.3° ;
(a) The longest wavelength in the visible spectrum is 700 nm, and the grating spacing is
1 mm
d=
= 1.67 × 10 −3 mm = 1.67 × 10 −6 m
600
Thus, mmax =
d sin 90.0°
λred
(1.67 × 10
=
−6
m ) sin 90.0°
700 × 10 −9 m
= 2.38
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
 700 × 10 −9 m 
m
λ 
λ

−1  400 × 10
∆θ = sin −1  red  − sin −1  violet  = sin −1 
−
sin



−6
−6
 d 
 d 
 1.67 × 10 m 
 1.67 × 10 m 
or ∆θ = 10.9°
24.37
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 ) 
 − sin −1 

∆θ = sin −1 
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.
322
CHAPTER 24
24.38
(a) If d =
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
mmax =
(b) If d =
λ
=
( 6.67 × 10
−6
500 × 10
m ) ( 1)
−9
m
= 13.3 or 13 orders
1 cm
= 6.67 × 10 −5 cm = 6.67 × 10 −7 m , then
15 000
mmax =
24.39
d sin 90°
d sin 90°
λ
( 6.67 × 10
=
−7
m ) ( 1)
500 × 10 −9 m
= 1.33 or 1 order
1 cm
= 2.00 × 10 −4 cm = 2.00 × 10 −6 m , and d sin θ = mλ gives
5 000
the 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


 = 28.7°
θ1 = sin
= 37.6° and θ 2 = sin 
−6
−6
 2.00 × 10 m 
 2.00 × 10 m 
−1
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 = L tan θ . 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
323
Wave Optics
24.40
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
 = 7.35°
θ = sin 
 = sin 
−6
 d 
 5.00 × 10 m 
−1
24.41
The grating spacing is d =
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
 2 ( 700 × 10 −9 m ) 
 2λred 
−1
 = 22.65°
θ r = sin 
 = sin 
3.636 × 10 −6 m 
 d 


−1
and
 2λviolet
 d
θ v = sin −1 
 2 ( 400 × 10 −9 m ) 

−1

 = 12.71°
sin
=

 3.636 × 10 −6 m 



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
Grating
Screen
Dy
the screen is ∆y = L ( tan θ r − tan θ v ) .
Since it is given that ∆y = 1.75 cm , the distance
from the grating to the screen must be
L=
or
∆y
1.75 cm
=
tan θ r − tan θ v tan ( 22.65° ) − tan ( 12.71° )
L = 9.13 cm
qv
L
qr
yv
yr
324
CHAPTER 24
24.42
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
d
maximum to 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
Using sin θ =
∆y = ym+1 − ym = λ L d .
The longer wavelength in the light is found to be
λlong =
( ∆y ) d = ( 8.44 × 10−3 m )( 8.33 × 10−6 m ) =
0.150 m
L
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
3 λshort L  0 + 1  λlong L
1
or λshort =   ( 469 nm ) = 78.1 nm
=

d
 2  d
6
24.43
The grating spacing is d =
1 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 ( 2λ d ) .
(a) When the grating is surrounded by air, the wavelength is λair = λ nair ≈ λ and
 2 ( 541 × 10 −9 m ) 
 2 λair 
−1
 = 25.6°
θ a = sin 
 = sin 
−6
 d 
 2.50 × 10 m 
−1
(b) If the grating is immersed in water,
then λn = λwater =
λ
nwater
 2 λwater
 d
θ b = sin −1 
=
λ
1.333
, yielding


2 ( 541 × 10 −9 m )

−1

 = 19.0°
sin
=

−6

 ( 2.50 × 10 m ) ( 1.333 ) 
mλn 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 θ b , or the angles of parts (a) and (b) satisfy
(c) From sin θ =
Snell’s law.
Wave Optics
24.44
When light of wavelength λ passes through a single slit of width a, the first minimum is
observed at angle θ where
sin θ =
λ
λ
or θ = sin −1  
a
a
This will have no solution if a < λ , so the maximum slit width if no minima are to be
seen is a = λ = 632 .8 nm .
24.45
(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  ( 1.00 ) sin 48.0° 
 = sin 
 = 42.0°
1.11


 n2 
θ 2 = sin −1 
Note that when θ1 = θ p , θ 2 = 90.0° − θ p as it should.
24.46
Unpolarized light incident on a polarizer contains electric field vectors at all angles to
the transmission axis of the polarizer. Malus’s law then gives the intensity of the
transmitted light as I = I 0 ( cos 2 θ )av . Since the average value of cos 2 θ is 1 2 , the
intensity of the light passed by the first polarizer is I1 = I 0 2 , where I 0 is the incident
intensity.
Then, from Malus’s law, the intensity passed by the second polarizer is
I
3
 I  3 
I 2 = I1 cos 2 ( 30.0° ) =  0    , or 2 =
I0
8
 2  4 
24.47
325
The more general expression for Brewster’s angle is (see Problem 51)
tan θ p = n2 n1
n 
 1.52 
(a) When n1 = 1.00 and n2 = 1.52 , θ p = tan −1  2  = tan −1 
 = 56.7°
 1.00 
 n1 
(b)
n
When n1 = 1.333 and n2 = 1.52 , θ p = tan −1  2
 n1

−1  1.52 
 = tan 
 = 48.8°
 1.333 

326
CHAPTER 24
24.48
The polarizing angle for light in air striking a water surface is
 n2
 n1
θ p = tan −1 

−1  1.333 
 = tan 
 = 53.1°
 1.00 

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°
24.49
n 
 1.65 
The polarizing angle is θ p = tan −1  2  = tan −1 
 = 58.8°
 1.00 
 n1 
When the light is incident at the polarizing angle, the angle of refraction is
θ r = 90.0° − θ p = 90.0° − 58.8° = 31.2°
24.50
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
 n2 
−1  1.77 
 = tan 
 = 60.5°
 1.00 
 n1 
θ p = tan −1 
24.51
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 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 cosθ p or tan θ p = n2 n1
Wave Optics
24.52
24.53
I = I 0 cos 2 θ

I
 I0
⇒
θ = cos −1 
(a)
I
1
=
⇒
I 0 3.00
θ = cos −1 
(b)
I
1
=
⇒
I 0 5.00
θ = cos −1 
(c)
I
1
=
I 0 10.0
θ = cos −1 
⇒




1 
 = 54.7°
 3.00 

1 
 = 63.4°
 5.00 

1 
 = 71.6°
 10.0 
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
24.54
327
(a) Using Malus’s law, the intensity of the transmitted light is found to be
(
I = I 0 cos 2 ( 45° ) = I 0 1
)
2
2 , or I I 0 = 1 2
(b) From Malus’s law, I I 0 = cos 2 θ . Thus, if I I 0 = 1 3 we obtain
(
cos 2 θ = 1 3 or θ = cos −1 1
)
3 = 54.7°
328
CHAPTER 24
24.55
(a) 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
λn =
1
λ
546.1 nm
= 413.7 nm
1.320
=
n1
λn =
and
2
λ
n2
=
546.1 nm
= 409.7 nm
1.333
(b) The number of cycles of vibration each component completes while passing
through the specimen are
N1 =
t
=
λn
1
and
1.000 × 10 -6 m
= 2.417
413.7 × 10 -9 m
N2 =
t
λn
2
=
1.000 × 10 -6 m
= 2.441
409.7 × 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°
24.56
Bright lines occur at angles given by sin θ = m ( λ d ) . Thus, if the m = 4 bright line of
wavelength λ1 and the m = 5 bright line of wavelength λ2 occur at the same angle, we
have
λ 
λ 
4
4
sin θ = 5  2  = 4  1  , or λ2 =   λ1 =   ( 540 nm ) = 432 nm
5
5
 d 
 d
24.57
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° , the slit spacing is
 60.0 min 
θ = ( 18.0 min ) 

d = m+

−9
1 λ
 3  ( 546 × 10 m )
= 
= 1.56 × 10 −4 m = 0.156 mm

2  sin θ  2  sin ( 0.300° )
Wave Optics
24.58
The wavelength is λ =
329
vsound 340 m s
=
= 0.170 m
2 000 Hz
f
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 ( 2m + 1)  for m = 0, 1, 2, …
2d

With λ d = 0.486 , this becomes θ = sin −1 ( 2m + 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
24.59
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 )
y = ( 1) 
=
= 2.50 × 10 −3 m = 2.50 mm

−2
2.00 × 10 m
 d 
330
CHAPTER 24
24.60
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 n film 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
tmax =
mmax
2
 λ

 n film

 84  λ 
= 42 λ
= 
 2  1.00 

When the film consists of water, the highest order dark fringe appearing will be
 n film
mmax = 2 tmax 
 λ

 1.333 
 = 2 ( 42 λ ) 
 = 112
 λ 

Counting the zeroth order, a total of 113 dark fringes are now observed.
24.61
With nair <noil < nwater , there is a 180° phase shift in light reflecting from each surface of the oil
film. In such a case, the conditions for constructive and destructive interference are reversed
from those valid when a phase shift occurs at only one surface. Thus, in this case, the
condition for constructive interference is
2noil t = mλ
where
m = 0, 1, 2 …
If noil = 1.25 and λ = 525 nm , thicknesses of the oil film producing constructive interference are
 525 nm 
 λ 
t = m
 = m ( 210 nm ) or any positive integral multiple of 210 nm
 = m
 2noil 
 2 ( 1.25 ) 
24.62
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 )
Wave Optics
331
(a) 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
24.63
Source
In the figure at the right, observe that the path difference
between the direct and the indirect paths is
Receiver
d¤2
δ = 2x − d = 2 h 2 + ( d 2 ) − d
d¤2
x
2
h
x
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) The wavelengths that will interfere constructively are λ =
δ
m+1 2
. The longest of
these is for the m = 0 case and has a value of
λ = 2δ = 4 h 2 + ( d 2 ) − 2d
2
=4
2
2
( 50.0 m ) + ( 300 m ) − 2 ( 600 m ) = 16.6 m
(b) 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 h 2 + ( d 2 ) − d = 2 ( 50.0 m ) + ( 300 m ) − 600 m = 8.28 m
2
2
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 = mλn = m 
 n film


 m = 0, 1, 2,…


h=R–t
24.64
2
Glass, n
film, nfilm
t{
Glass, n
R
r
332
CHAPTER 24
From the figure, note that R2 = r 2 + ( R − t ) = r 2 + R 2 − 2Rt + t 2 which reduces to
2
r 2 = 2Rt − t 2 . Since t will be very small in comparison to either r or R, we may neglect the
term t 2 , leaving r ≈ 2Rt .
For a dark fringe, t =
mλ
so the radii of the dark rings will be
2n film
 mλ
r ≈ 2R 
 2n film

24.65

=


mλ R
for m = 0, 1, 2,…
n film
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
Receiver
R
qp
Antenna
A
qp
5.00 m
90.0 m
n

θ p = tan  water  = tan −1 ( 1.33 ) = 53.1°
 nair 
−1
S
air, n = 1.00
water, n = 1.33
x
qp q
p
T
y
B
From the triangle RST is 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
24.66
(a) From Snell’s law, n1 sin θ = n2 sin φ with θ and φ
defined as shown in the figure at the right.
However, φ = 180° − (θ + β ) so Snell’s law becomes
n1 sin θ = n2 sin 180° − (θ + β )  .
q
q
n1
b
n2
f
Reflected
Ray
Refracted
Ray
Wave Optics
333
Apply the given identity, realizing that
sin  − (θ + β )  = − sin (θ + β )
cos  − (θ + β )  = cos (θ + β ) , to obtain
and
n1 sin θ = n2 sin ( 180° ) cos (θ + β ) − cos ( 180° ) sin (θ + β )  = n2 sin (θ + β )
Apply the identity once again to get
n1 sin θ = ( n2 cos β ) sin θ + ( n2 sin β ) cosθ ,
or ( n1 − n2 cos β ) sin θ = ( n2 sin β ) cosθ
Since sin θ cosθ = tan θ , this simplifies to tan θ =
n2 sin β
n1 − n2 cos β
(b) If β = 90°, n1 = 1, and n2 = n , the above result reduces to
tan θ = n , which is Brewster’s law.
24.67
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
−9
3λ L 3 ( 500 × 10 m ) ( 1.40 m )
a=
=
= 3.50 × 10 −4 m = 0.350 mm
2y
2 ( 3.00 × 10 −3 m )
334
CHAPTER 24
24.68
Phase reversals occur in the reflections at both surfaces of the oil layer, so there is zero
net phase difference due to reflections. The condition for constructive interference is
then
 λbright
2 t = mλn = m 
 n film





(1)
and the condition for destructive interference is

2t =  m +

1

 λn =  m +
2

1   λdark

2   n film




(2)
Solving for the order number, m, in equation (1),
and substituting into equation (2) gives the film thickness as
t=
24.69
λdark
500 nm
=
= 313 nm
4 n film ( 1 − λdark λbright ) 4 ( 1.20 ) ( 1 − 500 nm 750 nm )
(a) Assuming that n > n1 in the figure at the
right, Ray 1 undergoes a phase reversal as it
reflects at point A, but Ray 2 has no reversal
as it reflects at B. Therefore, the condition for
the two rays to interfere constructively is
that the difference in their optical path lengths
be an odd number of half-wavelengths,
or δ = ( m + 1 2 ) λ for m = 0, 1, 2,…
n1
q1
D
E q1
A
n2 = n
n1
Ray 1
q2
t
q2 q2
B
The difference in the optical path lengths is
δ = n ( AB + BC ) − n1 AD = 2n AB − n1 AD
But, AB =
 n

AE
− n1 sin θ1 
and AD = 2 AEsin θ1 , so δ = 2 AE 
sin θ 2
 sin θ 2

Now, observe that
AE = t tan θ 2 , and n1 sin θ1 = n sin θ 2 (from Snell’s law)
Ray 2
C
Wave Optics
Thus,
335


1
2nt
1 − sin 2 θ 2 ) = 2nt cosθ 2
− sin θ 2  =
(
 sin θ 2
 cosθ 2
δ = 2 nt tan θ 2 
The condition for constructive interference is then
2nt cosθ 2 = ( m + 1 2 ) λ
(b) When θ1 = 30.0° , then
n1 sin θ1 
−1  ( 1.00 ) sin 30.0° 
 = 21.2°
 = sin 
n
1.38




θ 2 = sin −1 
For minimum thickness, m = 0 which gives a thickness of
t=
24.70
( 0 + 1 2) λ =
2n cosθ 2
590 nm
= 115 nm
4 ( 1.38 ) cos 21.2°
The indirect ray suffers a phase reversal as it
reflects from the mirror, but there is no reversal
for the direct ray. Therefore, the condition for
constructive interference, with the two sources
separated by distance 2d, is
1
δ = ( 2d ) sin θ =  m +  λ for m = 0, 1, 2,…

2
s
d
y
q
mirror
d
s¢
L
The location of these maxima on the screen is
given by
ym = L tan θ ≈ L sin θ =
λL 
1
m + 
2d 
2
For the first bright fringe, m = 0 , giving
y=
−9
1  λ L ( 620 × 10 m ) ( 1.20 m )
+
=
=
= 7.4 × 10 −5 m = 74 µ m
0


-3
2d 
2  4d
4 ( 2.5 × 10 m )
λL 
336
CHAPTER 24
24.71
The refractive index, n, of the wedge material
is greater than that of the surrounding air.
Thus, when illuminated from above, light
reflecting from the upper surface of the wedge
experiences a 180° phase shift while light
reflecting from the bottom surface experiences
no shift. The condition for constructive
interference of light reflecting from the thin
film of transparent material is then
t
x l
1

2nt =  m +  λ where m = 0, 1, 2, …
2

and the condition for destructive interference is
2nt = mλ where m = 0, 1, 2, …
The thickness, t, at distance x from the edge of the wedge is found using the similar
triangles shown in the above sketch. Observe that
t h
=
x
or
h
t = x 
 
The condition for constructive interference or a bright fringe then becomes
1
h 
2nx   =  m +  λ
2
  
or
x=
λ 
1
m+ 
2 hn 
2
where
m = 0, 1, 2, …
and the condition for destructive interference or a dark fringe becomes
h
2nx   = mλ
 
or
x=
λ m
2 hn
where
m = 0, 1, 2, …
h