Unformatted text preview: flected from the
top surface of the plastic is onehalf of a wavelength out of phase with the light reflected 1444 INTERFERENCE AND THE WAVE NATURE OF LIGHT from the bottom surface of the glass. This halfwavelength difference means destructive
interference occurs, resulting in the dark fringe at A.
At the second dark fringe (the first is at A), the downandback distance traveled by the light
in the air wedge is one wavelength in the air film. At the third dark fringe, the downandback distance traveled by the light in the air wedge is two wavelengths. Therefore, at the
seventh dark fringe at B, the downandback distance traveled by the light in the air wedge
is six wavelengths. Since the downandback distance is six wavelengths, the thickness of
the air wedge at B is onehalf of this distance, or three wavelengths.
SOLUTION If the thickness of the air wedge is t, the condition for destructive interference
is
1λ
= 1 λair , 3 λair , 5 λair , K
12t3 +
2
2 air
2
24
2
123
14444 244444
3
Extra distance
traveled by wave
in the air Onehalf wavelength
net phase change
due to reflection Condition for
destructive interference ( ) = m + 1 λair
2 m = 0, 1, 2, 3, K The case m = 0 corresponds to the first dark fringe at A. Therefore, the dark fringe at B
corresponds to m = 6. Solving this equation for the thickness of the air wedge at B gives ( 2t = m + 1
2 ) λair − 1 λair = mλair
2 t = 1 mλair = 1 ( 6 ) ( 520 nm ) = 1560 nm
2
2 19. SSM WWW REASONING To solve this problem, we must express the condition for
constructive interference in terms of the film thickness t and the wavelength λ film of the
light in the soap film. We must also take into account any phase changes that occur upon
reflection.
SOLUTION For the reflection at the top film surface, the light travels from air, where the
refractive index is smaller (n = 1.00), toward the film, where the refractive index is larger
(n = 1.33). Associated with this reflection there is a phase change that is equivalent to onehalf of a wavelength. For the reflection at the bottom film surface, the light travels from the
film, where the refractive index is larger (n = 1.33), toward air, where the refractive index is
smaller (n = 1.00). Associated with this reflection, there is no phase change. As a result of
these two reflections, there is a net phase change that is equivalent to onehalf of a
wavelength. To obtain the condition for constructive interference, this net phase change
must be added to the phase change that arises because of the film thickness t, which is
traversed twice by the light that penetrates it. For constructive interference we find that
1
2 t + 2 λ film = λ film , 2 λ film , 3 λ film , ... Chapter 27 Problems or ch 1
2 t = m + 2 λ film , 1445 where m = 0, 1, 2, ... Equation 27.3 indicates that λ film = λ vacuum / n . Using this expression and the fact that
m = 0 for the minimum thickness t, we find that the condition for constructive interference
becomes ch c 1
1
2 t = m + 2 λ film = 0 + 2 or
t= λ vacuum
4n = FJ
hλ n I
GK
H
vacuum 611 nm
= 115 nm
4 1.33 bg 20. REASONING When the light strikes the film of oil from above, the wave reflected from
the top surface of the film undergoes a phase shift that is equivalent to onehalf of a
wavelength, since the light travels from the smaller refractive index of air toward the larger
refractive index of oil. On the other hand, there is no phase shift when the light reflects from
the bottom surface of the film, since the light travels from the larger refractive index of oil
toward the smaller refractive index of water. Thus, the net phase change due to reflection
from the two surfaces is equivalent to onehalf of a wavelength in the film. This
halfwavelength must be combined with the extra distance 2t traveled by the wave reflected
from the bottom surface, where t is the film thickness. Thus, the condition for destructive
interference is + 12t3
2
Extra distance
traveled by wave
in the film 1λ 12 24
4 film
3 = 1λ
, 3λ , 5λ , K
2 film 2 film 2 film
144444
244444
3 Halfwavelength
net phase change
due to reflection Condition for
destructive interference Note that the lefthand side of this equation is greater than
must also be greater than
3λ
.
2 film Therefore, 1λ
.
2 film we 1λ
.
2 film Thus, the righthand side The smallest value that is greater than have that 2t + 1 λfilm =
2 3λ
,
2 film or 1λ
2 film is the term 2t = λfilm . Since λfilm = λvacuum / nfilm = ( 640.0 nm ) / nfilm (see Equation 27.3), the condition for destructive
interference becomes 2t = λfilm = 640.0 nm
nfilm The condition for constructive interference is (1) 1446 INTERFERENCE AND THE WAVE NATURE OF LIGHT
1 λ′
′
= mλfilm
1 24
43
12 24
4 film
3 + 12t3
2
Extra distance
traveled by wave
in the film Halfwavelength
net phase change
due to reflection m = 1, 2, 3,K Condition for
constructive
interference where λf′ilm is the wavelength that produces constructive interference in the film. Solving
this relation for 2t gives
λ′
′
2t = m − 1 λfil...
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 Spring '13
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 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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