Solutions to Exercises 12
th
Ed.
35.1.
I
DENTIFY
:
Compare the path difference to the wavelength.
S
ET
U
P
:
The separation between sources is 5.00 m, so for points between the sources the largest
possible path difference is 5.00 m.
E
XECUTE
:
(a)
For constructive interference the path difference is
m
λ
,
m
=
0,
±
1,
±
2, . . .
Thus only
the path difference of zero is possible. This occurs midway between the two sources, 2.50 m from
A.
(b)
For destructive interference the path difference is
(
m
+
1
2
)
λ
,
m
=
0,
±
1,
±
2, .. .
A path difference of
±
λ
/
2
=
3.00
m is possible but a path difference as large as
3
λ
/ 2
=
9.00
m is
not possible. For a point a distance
x
from
A
and
5.00

x
from
B
the path difference is
x

(5.00 m

x
).
x

(5.00 m

x
)
= +
3.00 m gives
x
=
4.00 m.
x

(5.00 m

x
)
= 
3.00 m gives
x
=
1.00 m
.
E
VALUATE
:
The point of constructive interference is midway between the points of destructive
interference.
35.8.I
DENTIFY
:
The value of
y
20
is much smaller than
R
and the approximate expression
y
m
=
R
m
λ
d
is
accurate.
S
ET
U
P
:
y
20
=
10.6
×
10

3
m
.
E
XECUTE
:
d
=
20
R
λ
y
20
=
(20)(1.20 m)(502
×
10

9
m)
10.6
×
10

3
m
=
1.14
×
10

3
m
=
1.14 mm
E
VALUATE
:
tan
θ
20
=
y
20
R
so
θ
20
=
0.51o
and the approximation
sin
θ
20
≈
tan
θ
20
is very accurate.
35.9.
I
DENTIFY
and
S
ET
U
P
:
The dark lines correspond to destructive interference and hence are located
by Eq.(35.5):
d
sin
θ
=
m
+
1
2
λ
so sin
θ
=
m
+
1
2
λ
d
,
m
=
0,
±
1,
±
2,
K
Solve for
θ
that locates the second and third dark lines. Use
y
=
R
tan
θ
to find the distance of
each of the dark lines from the center of the screen.
E
XECUTE
:
1st dark line is for
m
=
0
2nd dark line is for
m
=
1
and
sin
θ
1
=
3
λ
2
d
=
3(500
×
10

9
m)
2(0.450
×
10

3
m)
=
1.667
×
10

3
and
θ
1
=
1.667
×
10

3
rad
3rd dark line is for
m
=
2
and
sin
θ
2
=
5
λ
2
d
=
5(500
×
10

9
m)
2(0.450
×
10

3
m)
=
2.778
×
10

3
and
θ
2
=
2.778
×
10

3
rad
(Note that
θ
1
and
θ
2
are small so that the approximation
θ
≈
sin
θ
≈
tan
θ
is valid.) The distance of
each dark line from the center of the central bright band is given by
y
m
=
R
tan
θ
,
where
R
=
0.850 m
is the distance to the screen.
tan
θ
≈
θ
so
y
m
=
R
θ
m
y
1
=
R
θ
1
=
(0.750 m)(1.667
×
10

3
rad)
=
1.25
×
10

3
m
y
2
=
R
θ
2
=
(0.750 m)(2.778
×
10

3
rad)
=
2.08
×
10

3
m
∆
y
=
y
2

y
1
=
2.08
×
10

3
m

1.25
×
10

3
m
=
0.83 mm
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E
VALUATE
:
Since
θ
1
and
θ
2
are very small we could have used Eq.(35.6), generalized to
destructive interference:
y
m
=
R m
+
1
2
λ
/
d
.
35.10.
I
DENTIFY
:
Since the dark fringes are eqully spaced,
R
?
y
m
, the angles are small and the dark
bands are located by
y
m
+
1
2
=
R
(
m
+
1
2
)
λ
d
.
S
ET
U
P
:
The separation between adjacent dark bands is
∆
y
=
R
λ
d
.
E
XECUTE
:
∆
y
=
R
λ
d
⇒
d
=
R
λ
∆
y
=
(1.80 m) (4.50
×
10

7
m)
4.20
×
10

3
m
=
1.93
×
10

4
m
=
0.193 m.
E
VALUATE
:
When the separation between the slits decreases, the separation between dark fringes
increases.
35.11.
I
DENTIFY
and
S
ET
U
P
:
The positions of the bright fringes are given by Eq.(35.6):
y
m
=
R
(
m
λ
/
d
).
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 Fall '07
 Hoffman
 Physics, Light, Wavelength

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