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JEE Main Physics Class 12 Wave Optics Part 2 Questions
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© examsnet.com
Question : 34
Total: 58
In Young's double slit experiment, two slits
S
1
and
S
2
are '
d
' distance apart and the separation from slits to screen is D (as shown in figure). Now if two transparent slabs of equal thickness
0.1
mm
but refractive index
1.51
and
1.55
are introduced in the path of beam
(
λ
=
4000
Å
)
from
S
1
and
S
2
respectively. The central bright fringe spot will shift by _______ number of fringes.
[30-Jan-2023 Shift 1]
Your Answer:
Validate
Solution:
Path difference at
P
be
∆
x
∆
x
=
(
µ
2
−
µ
1
)
t
=
(
1.55
−
1.51
)
0.1
mm
=
0.04
×
10
−
4
∆
x
=
4
×
10
−
6
=
4
µ
m
y
=
∆
xD
d
=
4
×
10
−
6
D
d
y is the distance of central maxima from geometric center
fringe width
(β)
=
λ
D
d
=
4
×
10
−
6
m
D
d
=
4
µ
m
D
d
Central bright fringe spot will shift by '
x
'
Number of shift
=
y
β
=
4
×
10
−
6
D
∕
d
4
×
10
−
7
D
∕
d
=
10
Ans
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