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PHYS 4D
Solution to HW 6
February 10, 2011
Problem Giancoli 344
(II) Monochromatic light falls on two very narrow slits 0
.
048
mm
apart. Successive
fringes on a screen 6
.
00
m
away are 8
.
5
cm
apart near the center of the pattern. Determine the wavelength and
frequency of the light.
Solution:
For constructive interference, the path diﬀerence is a multiple of the wavelength, as given by Eq.
342a. The location on the screen is given by
x
=
l
tan
θ
, as seen in Fig. 347(c). For small angles, we have
sin
θ
≈
tan
θ
≈
x/l
. Adjacent fringes will have ∆
m
= 1.
d
x
l
=
d
sin
θ
=
mλ
⇒
x
=
λml
d
.
x
1
=
λml
d
, x
2
=
λ
(
m
+ 1)
l
d
⇒
∆
x
=
x
2
−
x
1
=
λl
d
λ
=
d
∆
x
l
=
(0
.
048
mm
)
×
(8
.
5
cm
)
6
.
00
m
= 6
.
8
×
10
−
7
m
f
=
c
λ
=
3
.
0
×
10
8
m/s
6
.
8
×
10
−
7
m
= 4
.
4
×
10
14
Hz
Problem Giancoli 348
(II) Light of wavelength 680
nm
falls on two slits and produces an interference pattern
in which the thirdorder bright fringe is 38
mm
from the central fringe on a screen 2
.
6
m
away. What is the separation
of the two slits?
Solution:
d
x
l
=
mλ
⇒
d
=
λml
x
=
(680
nm
)3(2
.
6
m
)
38
mm
= 1
.
4
×
10
−
4
m.
Problem Giancoli 3412
(II) In a doubleslit experiment it is found that blue light of wavelength 480
nm
gives a
secondorder maximum at a certain location on the screen. What wavelength of visible light would have a minimum
at the same location?
Solution:
We equate the expression from Eq. 342a for the second order blue light to Eq. 342b, since the slit
separation and angle must be the same for the two conditions to be met at the same location.
d
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 Winter '08
 staff
 Physics, Light

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