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361
D
IFFRACTION
36.1.
IDENTIFY:
Use
tan
yx
θ
=
to calculate the angular position
of the first minimum. The minima are located by
Eq.(36.2):
sin
,
m
a
λ
=
1,
2,
m
=±
±
…
First minimum means
1
m
=
and
1
sin
/
a
θλ
=
and
1
sin .
a
λθ
=
Use this
equation to calculate
.
SET UP:
The central maximum is sketched in Figure 36.1.
EXECUTE:
11
tan
=
1
1
tan
y
x
==
3
3
1.35 10 m
0.675 10
2.00 m
−
−
×
=×
3
1
0.675 10 rad
−
Figure 36.1
33
1
sin
(0.750 10 m)sin(0.675 10 rad)
506 nm
a
−−
==×
×
=
EVALUATE:
1
is small so the approximation used to obtain Eq.(36.3) is valid and this equation could have been
used.
36.2.
IDENTIFY:
The angle is small, so
m
m
a
=
.
SET UP:
1
10.2 mm
y
=
EXECUTE:
7
5
1
3
1
(0.600 m)(5.46 10
m)
3.21 10 m.
10.2 10 m
xx
ya
ay
λλ
−
−
−
×
=
⇒
×
×
EVALUATE:
The diffraction pattern is observed at a distance of 60.0 cm from the slit.
36.3.
IDENTIFY:
The dark fringes are located at angles
that satisfy
sin
,
1,
2, .
...
m
m
a
±
±
SET UP:
The largest value of
sin
is 1.00.
EXECUTE: (a)
Solve for
m
that corresponds to sin
1
=
:
3
9
0.0666 10 m
113.8
585 10 m
a
m
−
−
×
=
×
. The largest value
m
can have is 113.
1
m
,
2
±
, …,
113
±
gives 226 dark fringes.
(b)
For
113
m
,
9
3
sin
113
0.9926
−
−
⎛⎞
×
⎜⎟
×
⎝⎠
and
83.0
°
.
EVALUATE:
When the slit width
a
is decreased, there are fewer dark fringes. When
a
<
there are no dark
fringes and the central maximum completely fills the screen.
36.4.
IDENTIFY
and
SET UP:
/
a
is very small, so the approximate expression
m
m
yR
a
=
is accurate. The distance
between the two dark fringes on either side of the central maximum is
1
2
y
.
EXECUTE:
9
3
1
3
(633 10 m)(3.50 m)
2.95 10 m
2.95 mm
0.750 10 m
R
y
a
−
−
−
×
=
×
=
×
.
1
25
.
9
0
m
m
y
=
.
EVALUATE:
When
a
is decreased, the width
1
2
y
of the central maximum increases.
36.5.
IDENTIFY:
The minima are located by
sin
m
a
=
SET UP:
12.0 cm
a
=
.
40.0 cm
x
=
.
36
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Chapter 36
EXECUTE:
The angle to the first minimum is
θ
= arcsin
a
λ
⎛⎞
⎜⎟
⎝⎠
= arcsin
9.00 cm
48.6 .
12.00 cm
=°
So the distance from the central maximum to the first minimum is just
1
tan
yx
==
(40.0 cm)tan(48.6 )
45.4 cm.
°=±
EVALUATE:
2/
a
is greater than 1, so only the
1
m
=
minimum is seen.
36.6.
IDENTIFY:
The angle that locates the first diffraction minimum on one side of the central maximum is given by
sin
a
=
. The time between crests is the period
T.
1
f
T
=
and
v
f
=
.
SET UP:
The time between crests is the period, so
1.0 h
T
=
.
EXECUTE: (a)
1
11
1.0 h
1.0 h
f
T
−
=
.
1
800 km/h
800 km
1.0 h
v
f
−
=
.
(b)
AfricaAntarctica:
800 km
sin
4500 km
=
and
10.2
=
°
.
AustraliaAntarctica:
800 km
sin
3700 km
=
and
12.5
=
°
.
EVALUATE:
Diffraction effects are observed when the wavelength is about the same order of magnitude as the
dimensions of the opening through which the wave passes.
36.7.
IDENTIFY:
We can model the hole in the concrete barrier as a single slit that will produce a singleslit diffraction
pattern of the water waves on the shore.
SET UP:
For singleslit diffraction, the angles at which destructive interference occurs are given by sin
m
=
m
/
a
,
where
m
= 1, 2, 3, ….
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 Spring '11
 Shaefer

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