PHY2061
R. D. Field
Solutions
Chapter 41
Page 1 of 10
Chapter 41 Solutions
Problem 1:
Light from a small region of a
100 Watt
incandescent bulb passes through a
yellow filter and then serves as the source for a Young’s doubleslit
interference experiment. Which of the following changes would cause the
interference pattern to be more closely spaced?
(a) Use a blue filter instead of a yellow filter
(b) Use a 10 Watt bulb
(c) Use a 500 Watt bulb
(d) Move the bulb closer to the slits
(e) Move the slits closer together
Answer:
(a) Use a blue filter instead of a yellow filter
Solution:
The angular position of the bright fringes is given by
d
m
λ
θ
=
sin
.
A more closely spaced pattern implies that for a given
m,
θ
is smaller.
Thus, we want to increase
d
or decrease
λ
and from
problem 15
we see that
λ
blue
<
λ
yellow
.
Problem 2:
The characteristic yellow light of sodium lamps arises from two
prominent
wavelengths in its spectrum, at approximately
589.0 nm
and
589.6 nm
, respectively.
The light passes through a double slit and falls on a
screen
10 m
away.
If the slits are separated by a distance of
0.01 mm
, how
far apart are the two secondorder bright fringes on the screen (
in mm
)?
Answer:
1.2
Solution:
The position of the second order bright fringe (
constructive
interference
) is given by
d
d
m
2
sin
=
=
≈
,
where I have set
m = 2
and used the small angle approximation.
We also
know that
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R. D. Field
Solutions
Chapter 41
Page 2 of 10
tan
θθ
≈=
y
L
.
The position of the second order bright fringe is thus
L
d
y
λ
2
=
,
and
mm
m
m
nm
d
L
y
y
2
.
1
10
)
10
)(
6
.
0
(
2
)
(
2
5
1
2
1
2
=
=
−
=
−
−
.
Problem 3:
A monochromatic light
placed
equal distance from two slits a
distance
d
apart produces a central
bright spot with a width of
1 cm
on a screen located a distance
L
away, as shown in the
figure
.
If
the entire apparatus is immersed
in a clear liquid with index of refraction
n
the width of the central bright
spot shrinks to
0.75 cm
.
What is the index of refraction
n
of the clear
liquid? (Note: assume L >> d)
Answer:
1.33
Solution:
We know that for doubleslit interference the position of the
first dark fringe
is given by
L
y
d
=
≈
=
≈
θ
tan
2
sin
and hence the width,
w
, of the central bright spot is
d
L
y
w
=
=
2
and
n
n
w
w
1
0
0
1
2
1
2
=
=
=
,
where I used
λ
1
=
λ
0
and
λ
2
=
λ
0
/n
where
λ
0
is the vacuum wavelength.
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 Spring '08
 FRY
 Physics, Light, Wavelength, R. D. Field, overall phase shift

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