Diffraction
369
EVALUATE:
Note that
/
3.0
da
=
.
Figure 36.27
36.28.
IDENTIFY:
The maxima are located by
sin
dm
θλ
=
.
SET UP:
The order corresponds to the values of
m
.
EXECUTE:
Firstorder:
1
sin
d
=
. Fourthorder:
4
sin
4
d
=
.
4
41
4
1
sin
4
, sin
4sin
4sin8.94 and
38.4
sin
d
d
θθ
θ
==
=
°
=
°
.
EVALUATE:
We did not have to solve for
d
.
36.29.
IDENTIFY
and
SET UP:
The bright bands are at angles
given by
sin
.
=
Solve for
d
and then solve for
for the specified order.
EXECUTE: (a)
78.4
=°
for
3
m
=
and
681 nm,
λ
=
so
4
/sin
2.086 10 cm
λθ
−
×
The number of slits per cm is 1/
4790 slits/cm
d
=
(b)
1st order:
1,
m
=
so
96
sin
/
(681 10 m)/(2.086 10 m)
d
−−
==×
×
and
19.1
2nd order:
2,
m
=
so sin
2 /
d
=
and
40.8
(c)
For
4, sin
4 /
md
is greater than 1.00, so there is no 4thorder bright band.
EVALUATE:
The angular position of the bright bands for a particular wavelength increases as the order increases.
36.30.
IDENTIFY:
The bright spots are located by
sin
=
.
SET UP:
Thirdorder means
3
m
=
and secondorder means
2
m
=
.
EXECUTE:
constant
sin
m
d
, so
rr
vv
rv
sin
sin
mm
λλ
=
.
vr
24
0
0
n
m
sin
sin
(sin65.0 )
0.345
37
0
0
n
m
m
m
⎛⎞
⎛
⎞
=
⎜⎟
⎜
⎟
⎝⎠
⎝
⎠
°
and
v
20.2
=
°
.
EVALUATE:
The thirdorder line for a particular
occurs at a larger angle than the secondorder line.
In a given
order, the line for violet light (400 nm) occurs at a smaller angle than the line for red light (700 nm).
36.31.
IDENTIFY
and
SET UP:
Calculate
d
for the grating. Use Eq.(36.13) to calculate
for the longest wavelength in
the visible spectrum and verify that
is small. Then use Eq.(36.3) to relate the linear separation of lines on the
screen to the difference in wavelength.
EXECUTE: (a)
5
1
cm
1.111 10 m
900
d
−
×
For
2
700 nm,
/
6.3 10 .
d
−
×
The firstorder lines are located at sin
/ ;
d
=
sin
is small enough for
sin
≈
to be an excellent approximation.
(b)
/
,
yxd
=
where
2.50 m.
x
=
The distance on the screen between 1st order bright bands for two different wavelengths is
()
/
,
yx
d
Δ= Δ
so
53
(
)/
(1.111 10 m)(3.00 10 m)/(2.50 m)
13.3 nm
dyx
=
×
×
=
EVALUATE:
The smaller
d
is (greater number of lines per cm) the smaller the
Δ
that can be measured.
36.32.
IDENTIFY:
The maxima are located by
sin
=
.
SET UP:
6
51
1
350 slits mm
2.86 10
m
3.50 10 m
d
−
−
⇒
×
×
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Chapter 36
EXECUTE:
7
400
6
4.00 10
m
1:
arcsin
arcsin
8.05
2.86 10
m
m
d
λ
θ
−
−
⎛⎞
×
==
=
=
⎜⎟
×
⎝⎠
D
.
7
700
6
7.00 10
m
arcsin
arcsin
14.18
2.86 10
m
d
−
−
×
=
×
D
.
1
14.18
8.05
6.13
Δ=
°
−
°
=
°
.
7
400
6
33
(
4
.
0
0
1
0
m
)
3:
arcsin
arcsin
24.8
2.86 10
m
m
d
−
−
×
=
=
°
×
.
7
700
6
(
7
.
0
0
1
0
m
)
arcsin
arcsin
47.3
2.86 10
m
d
−
−
×
=
°
×
.
1
47.3
24.8
22.5
°
−
°
=
°
.
EVALUATE:
Δ
is larger in third order.
36.33.
IDENTIFY:
The maxima are located by
sin
dm
θλ
=
.
SET UP:
6
1.60 10 m
d
−
=×
EXECUTE:
7
6
[6.328 10 m]
arcsin
arcsin
arcsin([0.396] )
mm
m
d
−
−
×
=
×
.
For
1
1,
23.3
m
°
.
For
2
m
=
,
2
52.3
=
°
.
There are no other maxima.
EVALUATE:
The reflective surface produces the same interference pattern as a grating with slit separation
d
.
36.34.
IDENTIFY:
The maxima are located by
sin
=
.
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 Spring '06
 Buchler
 Physics, Diffraction, Wavelength, Sin, Δθ

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