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JEE Main Physics Class 12 Wave Optics Part 2 Questions
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© examsnet.com
Question : 23
Total: 58
A beam of light consisting of two wavelengths
7000
Å
and
5500
Å
is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is
2.5
mm
and the distance between the place of slits and the screen is
150
cm
. The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is
n
×
10
−
5
m
. The value of
n
is ________
[6-Apr-2023 shift 2]
Your Answer:
Validate
Solution:
👈: Video Solution
Let
n
1
maxima of
7000
Å
coincides with
n
2
maxima of
5500
Å
therefore
n
1
β
1
=
n
2
β
2
or
n
1
n
2
=
λ
2
λ
1
=
5500
7000
=
11
14
therefore
11
th
maxima of
7000
Å
will coincide with
14
th
maximum of
5500
Å
To find the least distance of this
y
=
n
1
β
1
or
y
=
n
1
λ
1
D
d
=
11
×
7000
×
10
−
10
×
150
×
10
−
2
2.5
×
10
−
3
=
11
×
7
×
5
2.5
×
10
−
5
m
or
y
=
462
×
10
−
5
m
therefore
n
=
462
© examsnet.com
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