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tovar (jdt436) – homework 40 – Turner – (59070)
1
This printout should have 10 questions.
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beFore answering.
001
10.0 points
Consider the setup oF a single slit experiment.
The wavelength oF the incident light is
λ
=
530 nm
.
The slit width and the distance between the
slit and the screen is specifed in the fgure.
y
1
3
.
8 m
410
μ
m
S
1
S
2
θ
viewing
screen
±ind the position
y
=
y
1
oF the frst inten
sity minimum. Use a small angle approxima
tion sin
θ
= tan
θ .
Correct answer: 4
.
9122 mm.
Explanation:
Let :
λ
= 530 nm
,
L
= 3
.
8 m
,
and
a
= 410
μ
m
.
y
1
L
a
S
1
S
2
δ
≡
a
sin
θ
≈
a
b
y
L
B
±or single slit di²raction, 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
frst minimum, the phase angle di²erence is
β
1
= 2
π
and the path length di²erence 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 4
.
9122 mm
.
002
10.0 points
Consider the setup oF a single slit experiment.
y
7
L
a
S
1
S
2
×
15
±ind the height
y
7
where the seventh mini
mum occurs. Use a small angle approximation
sin
θ
= tan
θ .
1.
y
7
= 6
λ L
a
2.
y
7
=
15
2
λ L
a
3.
y
7
= 7
λ L
a
correct
4.
y
7
= 9
λ L
a
5.
y
7
=
13
2
λ L
a
6.
y
7
= 8
λ L
a
7.
y
7
=
11
2
λ L
a
8.
y
7
= 5
λ L
a
9.
y
7
=
17
2
λ L
a
10.
y
7
=
19
2
λ L
a
Explanation:
Let :
k
≡
2
π
λ
.
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View Full Documenttovar (jdt436) – homework 40 – Turner – (59070)
2
The frst minimum is at
β
= 2
π
, where
β
=
2
φ
= 2
π
, and
φ
=
π
is the phase diFerence o±
the two rays ±or destructive inter±erence.
The
seventh
minimum occurs at
β
= 14
π
,
which corresponds to a path diFerence be
tween two end rays o±
δ
=
β
k
=
14
π
p
2
π
λ
P
= 7
λ .
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 Spring '08
 Turner
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