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Halliday Resnick Walker
FUNDAMENTALS OF PHYSICSSIXTH EDITION
Selected Solutions
Chapter 36
36.2136.2936.3736.45
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21. Consider the two waves, one from each slit, that produce the
seventh bright fringe in the absence of themica. They are in phase
at the slits and travel different distances to the seventh bright
fringe, wherethey have a phase difference of 2 m = 14 . Now a piece
of mica with thickness x is placed in front of one of the slits,
and an additional phase difference between the waves develops.
Specically, their phasesat the slits differ by
2x m
2x
=
2x
(n 1)
where m is the wavelength in the mica and n is the index of
refraction of the mica. The relationship m = /n is used to
substitute for m . Since the waves are now in phase at the
screen,
2x
(n 1) = 14
orx =
7n 1
=7(550 10 9 m)
1.58 1= 6 .64 10 6 m .
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29. We take the electric eld of one wave, at the screen, to
be
E 1 = E 0 sin(t)
and the electric eld of the other to be
E 2 = 2 E 0 sin(t + ) ,
where the phase difference is given by
=2d
sin .
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E 0
2E 0
E
t
Here d is the center-to-center slit separation and is the
wavelength. The resultant wave can be writtenE = E 1 + E 2 = E
sin(t + ), where is a phase constant. The phasor diagram is shown
above. Theresultant amplitude E is given by the trigonometric law
of cosines:
E 2 = E 20 + (2 E 0 )2 4E 20 cos(180
) = E 20 (5 + 4cos ) .
The intensity is given by I = I 0 (5 + 4 cos ), where I 0 is the
intensity that would be produced bythe rst wave if the second were
not present. Since cos = 2cos 2 (/ 2) 1, this may also be writtenI
= I 0 1 + 8cos 2 (/ 2) .
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37. For complete destructive interference, we want the waves
reected from the front and back of the coatingto differ in phase by
an odd multiple of rad. Each wave is incident on a medium of higher
index of refraction from a medium of lower index, so both suffer
phase changes of rad on reection. If L isthe thickness of the
coating, the wave reected from the back surface travels a distance
2 L farther thanthe wave reected from the front. The phase
difference is 2 L(2/ c ), where c is the wavelength in thecoating.
If n is the index of refraction of the coating, c = /n , where is
the wavelength in vacuum,and the phase difference is 2 nL (2/ ). We
solve
2nL2
= (2 m + 1)
for L. Here m is an integer. The result is
L =(2m + 1)
4n.
To nd the least thickness for which destructive interference
occurs, we take m = 0. Then,
L =
4n=
600 10 9 m4(1.25)
= 1 .2 10 7 m .
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45. Assume the wedge-shaped lm is in air, so the wave reected
from one surface undergoes a phase changeof rad while the wave
reected from the other surface does not. At a place where the lm
thickness isL , the condition for fully constructive interference
is 2 nL = ( m + 12 ) , where n is the index of refractionof the lm,
is the wavelength in vacuum, and m is an integer. The ends of the
lm are bright.Suppose the end where the lm is narrow has thickness
L 1 and the bright fringe there correspondsto m = m 1 . Suppose the
end where the lm is thick has thickness L 2 and the bright fringe
therecorresponds to m = m 2 . Since there are ten bright fringes, m
2 = m 1 + 9. Subtract 2 nL 1 = ( m 1 + 12 )from 2nL 2 = ( m 1 + 9 +
12 ) to obtain 2 n L = 9 , where L = L 2 L 1 is the change in the
lmthickness over its length. Thus,
L =92n
=9(630 10 9 m)
2(1.50)= 1 .89 10 6 m .
