PHYS1200
PHYSICS II
SPRING 2005
HOMEWORK SOLUTIONS
CLASS 21.
36.06, 36.17, 36.28, 36.65; Q38.04, Q38.06
36.06.
a)
The minima of single slit diffraction are determined by
λ
θ
m
a
=
sin
.
Using the
small angle approximation, we have
λ
m
D
y
a
m
=
. Here
m
y
is the distance from the
center of central maximum to the
m
th minimum. So we get
(
29
m
10
35
.
0
)
m
40
.
0
)(
m
10
550
(
4
4
so
,
5
3
9
1
5
1
5


×
×
=

=

=

y
y
D
a
y
y
D
a
λ
λ
λ
a
= 2.5 ×10
3
m = 2.5 mm
b)
m
10
5
.
2
m
10
550
3
9
1


×
×
=
≈
a
λ
θ
θ
1
= 2.2 ×10
4
rad = 0.013°
36.17.
a)
θ
λ
R
d
=
=
×
×


122
122
550
10
50
10
9
3
.
.
.
m
m
θ
R
= 1.34 ×10
4
rad = 7.69 ×10
3
deg = 27.7 arc seconds
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
b)
Let
s
be the distance between the headlights, and
D
the distance from the headlights to
the eye. Then,
4

10
1.34
m
.4
1
so
,
×
=
=
=
R
R
s
D
D
s
θ
θ
D
= 1.0 ×10
4
m = 10 km
36.28.
a)
The angular position of the first minimum of the diffraction pattern is given by,
d
λ
θ
22
.
1
=
. That means that longer wavelengths spread out farther. Therefore, the
outer ring is produced by the longest visible wavelength. It is red.
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 Spring '06
 Stoler
 Energy, Work, Diffraction, Light, Wavelength, single slit diffraction

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