Interference
3511
The second bright ring is when
1:
m
=
7
4
(2(1)
1) (5.80
10
m) (0.952 m)
9.10
10
m
0.910 mm.
2
r
−
−
+
×
≈
=
×
=
So the diameter of the second bright ring is 1.82 mm.
E
VALUATE
:
The diameter of the
th
m
ring is proportional to
2
1
m
+
, so the rings get closer together as
m
increases.
This agrees with Figure 35.17b in the textbook.
35.40.
I
DENTIFY
:
As found in Problem 35.39, the radius of the
th
m
bright ring is
(2
1)
,
2
m
R
r
λ
+
≈
for
.
R
λ
"
S
ET
U
P
:
Introducing a liquid between the lens and the plate just changes the wavelength from
to
n
λ
λ
, where
n
is the refractive index of the liquid.
E
XECUTE
:
(2
1)
0.850 mm
( )
0.737 mm.
2
1.33
m
R
r
r n
n
n
λ
+
≈
=
=
=
E
VALUATE
:
The refractive index of the water is less than that of the glass plate, so the phase changes on
reflection are the same as when air is in the space.
35.41.
I
DENTIFY
:
The liquid alters the wavelength of the light and that affects the locations of the interference minima.
S
ET
U
P
:
The interference minima are located by
1
2
sin
(
)
d
m
θ
λ
=
+
.
For a liquid with refractive index
n
,
air
liq
n
λ
λ
=
.
E
XECUTE
:
1
2
(
)
sin
constant
m
d
θ
λ
+
=
=
, so
liq
air
air
liq
sin
sin
θ
θ
λ
λ
=
.
liq
air
air
air
sin
sin
/
n
θ
θ
λ
λ
=
and
air
liq
sin
sin35.20
1.730
sin
sin19.46
n
θ
θ
=
=
=
°
°
.
E
VALUATE
:
In the liquid the wavelength is shorter and
1
2
sin
(
)
m
d
λ
θ
=
+
gives a smaller
θ
than in air, for the
same
m
.
35.42.
I
DENTIFY
:
As the brass is heated, thermal expansion will cause the two slits to move farther apart.
S
ET
U
P
:
For destructive interference,
d
sin
θ
=
λ
/2. The change in separation due to thermal expansion is
dw
=
α
w
0
dT
, where
w
is the distance between the slits.
E
XECUTE
:
The first dark fringe is at
d
sin
θ
=
λ
/2
⇒
sin
θ
=
λ
/2
d.
Call
d
≡
w
for these calculations to avoid confusion with the differential. sin
θ
=
λ
/2
w
Taking differentials gives
d
(sin
θ
) =
d
(
λ
/2
w
) and cos
θ
d
θ
=
−
λ
/2
dw/w
2
.
For thermal expansion,
dw
=
α
w
0
dT
, which gives
0
2
0
0
cos
2
2
w dT
dT
d
w
w
λ α
λα
θ
θ
= −
= −
. Solving for
d
θ
gives
0
0
2
cos
dT
d
w
λα
θ
θ
= −
. Get
λ
:
w
0
sin
θ
0
=
λ
/2
→
λ
= 2
w
0
sin
θ
0
. Substituting this quantity into the equation for
d
θ
gives
0
0
0
0
0
2
sin
tan
2
cos
w
dT
d
dT
w
θ α
θ
θ α
θ
= −
= −
.
5
1
tan(32.5
)(2.0
10
K
)(115 K)
0.001465 rad
0.084
d
θ
−
−
= −
×
= −
= −
°
°
The minus sign tells us that the dark fringes move closer together.
E
VALUATE
:
We can also see that the dark fringes move closer together because sin
θ
is proportional to 1/
d
, so as
d
increases due to expansion,
θ
decreases.
35.43.
I
DENTIFY
:
Both
frequencies will interfere constructively when the path difference from both of them is an
integral number of wavelengths.
S
ET
U
P
:
Constructive interference occurs when sin
θ
=
m
λ
/
d
.
E
XECUTE
:
First find the two wavelengths.
λ
1
=
v/f
1
= (344 m/s)/(900 Hz) = 0.3822 m
λ
2
=
v/f
2
= (344 m/s)/(1200 Hz) = 0.2867 m
To interfere constructively at the same angle, the angles must be the same, and hence the sines of the angles must
be equal. Each sine is of the form sin
θ
=
m
λ
/
d
, so we can equate the sines to get
m
1
λ
1
/
d
=
m
2
λ
2/
d
m
1
(0.3822 m) =
m
2
(0.2867 m)
m
2
= 4/3
m
1
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3512
Chapter 35
Since both
m
1
and
m
2
must be integers, the allowed pairs of values of
m
1
and
m
2
are
m
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 Spring '06
 Buchler
 Physics, Light, Wavelength

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