24.3
(a) The distance between the central maximum and the first order bright fringe
λL
− ybright
=
is Δy = ybright
, or (Note, it is safer to not use the
m =1
m =0
d
approximation used here that tanθ=sinθ)
Δy =
λL
d
( 546.1× 10
=
−9
) (1.20 m ) = 2.62× 10
m
−3
−3
0.250× 10 m
m = 2.62 m m
(b) The distance between the first and second dark bands is
Δy = ydark m =1 − ydark m =0 =
24.7
λL
d
= 2.62 m m
as in (a) above.
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 = dsinθ = 2λ ,
or λ =
or λ =
24.21
y
d
d⎛
sinθ = ⎜
2
2 ⎜ L2 + y2
⎝
⎡
300 m
⎞
⎟
⎟
⎠
1 000 m
⎤
⎥ = 55.7 m
2
2 ⎥
⎢
⎣⎢ ( 1000 m ) + ( 400 m ) ⎦⎥
( 300 m ) ⎢
2
400 m
400 m
(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
2n 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
2t= 2λ n film , or
λ = n film t= ( 1.378)( 238 nm ) = 328 nm
24.22
(near ultraviolet)
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 ⎞
2t= m λn = 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=
−6
2t 2( 2.00 × 10 m )
=
= 7.32
546.1× 10−9 m
λ
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.
24.27
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
2t= m λn = 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=
2n film t 2( 1.38)( 100 nm )
=
, with m = 1,2,3,…
m
m
Thus, λ = 276 nm ,138 nm ,92.0 nm …
and none ofthese w avelengths are in the visible spectrum
24.36
(a) The longest wavelength in the visible spectrum is 700 nm, and the grating
1m m
spacing is d =
= 1.67 × 10−3 m m = 1.67 × 10−6 m
600
Thus, m m ax =
dsin90.0°
λred
(1.67 × 10
=
−6
m ) sin 90.0°
700 × 10−9 m
= 2.38
so 2 com plete orders will be observed.
(b) From λ = dsinθ , 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
Δθ = sin −1 ⎢ red ⎥ − sin −1 ⎢ violet ⎥ = sin −1 ⎢
−
si
n
⎥
⎢ 1.67 × 10−6 m ⎥
−6
×
d
d
1.
67
10
m
⎣
⎦
⎣
⎦
⎣
⎦
⎣
⎦
or Δθ = 10.9°
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
2000
Then, from dsinθ = m λ , the location of the first order for the red light is
⎡ ( 1) ( 640× 10−9 m
mλ ⎞
−1
⎢
si
n
=
⎟
−6
⎝ d ⎠
⎢⎣ 5.00× 10 m
θ = sin−1 ⎜⎛
) ⎤⎥ =
⎥⎦
7.35°
24.42
The grating spacing is d =
1 cm
= 8.33× 10−4 cm = 8.33× 10−6 m
1200
mλ
and the small angle approximation, the distance from the
d
central 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
Using sinθ =
maxima is Δy = ym +1 − ym = λ L d .
The longer wavelength in the light is found to be
λlong
−3
−6
Δy) d ( 8.44 × 10 m )( 8.33× 10 m )
(
=
=
=
L
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
3λshortL ⎛ 0 + 1 ⎞ λlongL
⎛ 1⎞
=⎜
or λshort = ⎜ ⎟ ( 469 nm ) = 78.1 nm
⎟
d
2
d
⎝
⎠
⎝ 6⎠
24.58
The wavelength is λ =
vsound 340 m s
=
= 0.170 m
2000 H z
f
Maxima occur where dsinθ = m λ , 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 ( 0.486m )
For m = 0,1, and 2 , this yields m axim a at0°,29.1°,and 76.3°
No solutions exist for m ≥ 3 since that would imply sin θ > 1
⎡
λ ⎤
Minima occur where dsin θ = ( m + 1 2) λ or θ = sin −1 ⎢( 2m + 1) ⎥ for
2d ⎦
⎣
m = 0,1,2,K
With λ d = 0.486 , this becomes θ = sin −1 ⎡⎣( 2m + 1)( 0.243) ⎤⎦
For m = 0 and 1, we find m inim a at14.1° and 46.8°
No solutions exist for m ≥ 2 since that would imply sin θ > 1