CHAPTER
27
INTERFERENCE AND
THE WAVE NATURE OF LIGHT
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1.
(e)
The first intensity minimum occurs when the difference in path lengths is
1
2
λ
, the
second when the difference is
3
2
λ
, and the third when the difference is
5
2
λ
.
2.
(a)
The angle
θ
that specifies the
m
th
bright fringe is given by
sin
m
d
λ
θ
=
(Equation 27.1).
If
sin
, then
m
d
λ
θ
θ
θ
≈
=
. Thus, if
λ
and
d
are both doubled, the angle does not change.
3.
(b)
According to the discussion in Section 27.2, the difference in path lengths of the light
waves increases by one wavelength as one moves from one bright fringe to the next one
farther out.
4.
d
= 1.1
×
10

5
m
5.
(d)
The wavelength
λ
water
of the light in water is related to the wavelength
λ
vacuum
in a
vacuum by
water
vacuum
water
/
n
λ
λ
=
(Equation 27.3). Since the index of refraction of water
n
water
is greater than one, the wavelength decreases when the apparatus is placed in water.
The angle
θ
that specifies the
m
th
bright fringe is given by
sin
m
d
λ
θ
=
(Equation 27.1).
When the wavelength decreases, the angle also decreases.
6.
(d)
The downandback distance traveled by the light wave in the film is 2400 nm. The
wavelength of the light within the film is (see Equation 27.3)
λ
film
=
λ
vacuum
/
n
film
=
(
29
600 nm
/1.5 = 400 nm
. Thus, the downandback distance is equivalent to
film
film
(2400 nm)
= 6
400 nm
λ
λ
÷
.
7.
(c)
In drawings 1 and 2, light travels through a material with a smaller refractive index
toward a material with a larger refractive index, so the reflection at the boundary occurs
with a phase change that is equivalent to onehalf a wavelength in the material. In drawings
3 and 4, light travels through a material with a larger refractive index toward a material with
a smaller refractive index, so there is no phase change upon reflection at the boundary.
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149
INTERFERENCE AND THE WAVE NATURE OF LIGHT
8.
(e)
In film 4, the reflection at the top surface occurs with a phase change that is equivalent
to onehalf a wavelength in the film, since light travels from a material with a smaller
refractive index (
n
air
= 1.0) toward a material with a larger refractive index (
n
film
= 1.5). The
reflection at the bottom surface occurs without a phase change, since light travels from a
material with a larger refractive index (
n
film
= 1.5) toward a material with a smaller
refractive index (
n
air
= 1.0). The downandback distance traveled by the light is onehalf of
a wavelength. Therefore, the net phase change is one wavelength, which leads to
constructive interference.
9.
(b)
A bright fringe occurs at a point where the thickness of the air wedge has a certain
value. As the thickness of the air wedge becomes smaller, the fringe moves to the right
where the air wedge is thicker.
10.
t
= 2.6
×
10

6
m
11. (c)
The amount of diffraction depends on the angles that locate the dark fringes on either
side of the central bright fringe (see Section 27.5). These angles (one for each value of the
integer
m
) are related to the wavelength
λ
of the light and the width
W
of the slit by
(
29
sin
/
m
W
θ
λ
=
, (Equation 27.4). When
λ
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 Fall '11
 rollino
 Physics, Light, Wavelength, refractive index, tan θ

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