nguyen (jmn727) – oldhomework 40 – Turner – (59070)
1
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001
(part 1 oF 3) 10.0 points
A standard setup oF the double slit experiment
as shown in the sketch. The distance between
the two slits is
d
, the wave length oF the
incident wave is
λ
, the distance between the
slits and the screen is
L
.
A point on the screen is specifed by its
ycoordinate, or the corresponding angle
θ
.
y
L
d
S
1
S
2
θ
viewing
screen
θ
δ
At the Fourth minimum on the screen what
is
φ
(the phase angle di±erence oF the two
rays From slit
S
1
and slit
S
2
) and
δ
(the corre
sponding path di±erence)?
1.
φ
= 7
π
and
δ
= 7
λ
2.
φ
= 6
π
and
δ
= 6
λ
3.
φ
= 4
π
and
δ
= 2
λ
4.
φ
= 5
π
and
δ
= 5
λ
5.
φ
= 5
π
and
δ
=
5
2
λ
6.
φ
= 8
π
and
δ
= 4
λ
7.
φ
= 6
π
and
δ
= 3
λ
8.
φ
= 7
π
and
δ
=
7
2
λ
correct
9.
φ
= 8
π
and
δ
= 8
λ
10.
φ
= 4
π
and
δ
= 4
λ
Explanation:
In general, the phase angle di±erence For
minima is given by
φ
= (2
n
+ 1)
π,
with
n
= 0
,
1
,
2
···
.
The Fourth minimum corresponds to
n
= 3,
so
φ
= 7
π
and
δ
=
p
λ
2
π
P
φ
=
λ
2
π
(7
π
) =
7
2
λ .
002
(part 2 oF 3) 10.0 points
What is the vertical distance
y
For the frst
maximum (which is adjacent to the central
maximum)? Use the small angle approxima
tion.
1.
y
=
2
λ L
d
2.
y
=
λ d
2
L
3.
y
=
2
λ d
L
4.
y
=
λ L
2
d
5.
y
=
d L
λ
6.
y
=
d L
2
λ
7.
y
=
λ L
d
correct
8.
y
=
λ
9.
y
=
λ d
L
10.
y
=
2
d L
λ
Explanation:
The frst maximum occurs when the path
di±erence is
λ
, so
θ
=
λ
d
=
y
L
y
=
λ L
d
.
003
(part 3 oF 3) 10.0 points
²ind the minimum positive
θ
value such that
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View Full Documentnguyen (jmn727) – oldhomework 40 – Turner – (59070)
2
I
I
0
=
1
4
, where
I
0
and
I
are the intensities
of light at 0
◦
and at
θ
, respectively. Use the
small angle approximation.
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 Spring '08
 Turner
 Physics, Work, Wavelength, Doubleslit experiment, Thomas Young, phase angle difference

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