nguyen (jmn727) – homework 40 – Turner – (59070)
1
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001
10.0 points
Consider the setup of a single slit experiment.
The wavelength of the incident light is
λ
=
480 nm
.
The slit width and the distance between the
slit and the screen is specified in the figure.
y
1
6
.
1 m
550
μ
m
S
1
S
2
θ
viewing
screen
Find the position
y
=
y
1
of the first inten
sity minimum. Use a small angle approxima
tion sin
θ
= tan
θ .
Correct answer: 5
.
32364 mm.
Explanation:
Let :
λ
= 480 nm
,
L
= 6
.
1 m
,
and
a
= 550
μ
m
.
y
1
L
a
S
1
S
2
θ
viewing
screen
δ
≡
a
sin
θ
≈
a
bracketleftBig
y
L
bracketrightBig
For single slit diffraction, destructive in
terference occurs when,
a
2
sin
θ
=
λ
2
, or sim
ply when,
δ
≡
a
sin
θ
=
λ
.
Thus, between
the two end rays which correspond to the
first minimum, the phase angle difference is
β
1
= 2
π
and the path length difference is
δ
1
=
λ
. The small angle approximation gives
us
y
1
L
= tan
θ
1
≈
θ
1
≈
sin
θ
1
=
δ
1
a
, or
y
1
=
δ
1
a
L
=
λ L
a
=
sturt 5
.
32364 mm
.
002
10.0 points
Consider the setup of a single slit experiment.
y
6
L
a
S
1
S
2
θ
viewing
screen
×
15
Find the height
y
6
where the sixth mini
mum occurs.
Use a small angle approxima
tion sin
θ
= tan
θ .
1.
y
6
=
15
2
λ L
a
2.
y
6
= 6
λ L
a
correct
3.
y
6
=
11
2
λ L
a
4.
y
6
= 7
λ L
a
5.
y
6
= 5
λ L
a
6.
y
6
= 4
λ L
a
7.
y
6
=
17
2
λ L
a
8.
y
6
= 8
λ L
a
9.
y
6
=
9
2
λ L
a
10.
y
6
=
13
2
λ L
a
Explanation:
Let :
k
≡
2
π
λ
.
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nguyen (jmn727) – homework 40 – Turner – (59070)
2
The first minimum is at
β
= 2
π
, where
β
=
2
φ
= 2
π
, and
φ
=
π
is the phase difference of
the two rays for destructive interference.
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 Spring '08
 Turner
 Physics, Work, Light, Correct Answer, single slit, small angle approximation

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