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Solutions to WebAssign HW #5  Giancoli Ch. 352, 3, 8, 25, 31, 36, 53, 54, and 80;
(due 2/27)
2.
The angle from the central maximum to the first dark fringe is equal to half the width of the
central maximum.
Using this angle and Eq. 351, we calculate the wavelength used.
( )
()
11
1
22
34
32
16
sin
sin
2.60 10 mm sin 16
7.17 10 mm
717 nm
D
D
θθ
λ
θλ
θ
−−
=∆=
°= °
=→=
=
×
°
= ×
=
3.
The angle to the first maximum is about halfway between the angles to the first and second
minima.
We use Eq. 352 to calculate the angular distance to the first and second minima.
Then we average these to values to determine the approximate location of the first
maximum.
Finally, using trigonometry, we set the linear distance equal to the distance to the
screen multiplied by the tangent of the angle.
(
)
1
99
12
66
1
sin
sin
1 580 10 m
2 580 10 m
sin
8.678
sin
17.774
3.8 10 m
8.678
=
13.23
tan
10.0 m tan 13.23
2.35 m
mm
m
Dm
D
y
−
⎛⎞
=→
=
⎜⎟
⎝⎠
⎛
⎞
××
==
°
=
=
°
⎜
⎟
⎝
⎠
+
°+
°
°
°
=
l
8.
(
a
)
There will be no diffraction minima if the angle for the first minimum is greater
than 90 .
°
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 Spring '08
 LIN
 Quantum Physics

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