PHYS 4D
Solution to HW 7
February 21, 2011
Problem Giancoli 352
(I) Monochromatic light falls on a slit that is 2
.
60
×
10
−
3
mm
wide.
If the angle
between the first dark fringes on either side of the central maximum is 32
.
0
◦
(dark fringe to dark fringe), what is the
wavelength of the light used?
Solution:
The angle from the central maximum to the first dark fringe is equal to half the width of the central
maximum. Using the angle and Eq. 351, we calculate the wavelength used
θ
1
=
1
2
∆
θ
=
1
2
32
.
0
◦
= 16
.
0
◦
,
sin
θ
1
=
λ
D
⇒
λ
= 717
nm.
Problem Giancoli 354
(II) Consider microwaves which are incident perpendicular to a metal plate which has
a 1
.
6
cm
slit in it. Discuss the angles at which there are diffraction minima for wavelengths of (a) 0
.
50
cm
, (b) 1
.
0
cm
,
and (c) 3
.
0
cm
.
Solution:
(a) Using Eq. 352,
m
= 1
,
2
,
3
, ...
to calculate the possible diffraction minima, when the wavelength
is 0
.
50
cm
,
D
sin
θ
m
=
mλ
⇒
θ
m
= sin
−
1
(
mλ
D
)
.
⇒
θ
1
= 18
.
2
◦
, θ
2
= 38
.
7
◦
, θ
3
= 69
.
6
◦
, θ
4
=
no solution.
θ
1
, θ
2
and
θ
3
give three diffraction minima.
(b) Using the new wavelength and repeat the above process, we find
θ
1
= 38
.
7
◦
, θ
2
=
no solution.
θ
1
gives the only diffraction minima.
(c) Using the new wavelength and repeat the above process, we find
θ
1
=
no solution.
No diffraction minima.
Problem Giancoli 356
(II) Monochromatic light of wavelength 633
nm
falls on a slit. If the angle between the
first bright fringes on either side of the central maximum is 35
◦
, estimate the slit width. .
Solution:
The angle from the central maximum to the first bright maximum is half the angle between the ﬃrst
bright maxima on either side of the central maximum. The angle to the first maximum is about halfway between
the angle to the first and second minima. We use Eq. 352, setting
m
=
3
2
, to calculate the slit width,
D
.
θ
1
=
1
2
∆
θ
=
1
2
35
.
0
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 Winter '08
 staff
 Physics, Light, coherent, 2L, 3 mm, 633nm, 618nm

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