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KCET 2014 Physics Solved Paper
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© examsnet.com
Question : 24
Total: 60
A fringe width of a certain interference pattern is β = 0.002 cm. What is the distance of
5
th
dark fringe from center ?
1
×
10
−
2
c
m
11
×
10
−
2
c
m
1.1
×
10
−
2
c
m
3.28
×
10
6
c
m
None of the above
Validate
Solution:
Given, fringe width β = 0.002 cm; distance of
n
th
fringe is given as
x
n
=
(
2
n
+
1
)
λ
D
2
d
We know fringe width is
β
=
λ
D
2
For 5th dark fringe,
n
=
4
α
n
=
(
2
×
4
+
1
)
×
0.002
2
=
9
×
0.001
=
0.009
c
m
© examsnet.com
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