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KCET 2020 Physics Solved Paper
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© examsnet.com
Question : 16
Total: 60
In Young’s Double Slit Experiment, the distance between the slits and the screen is 1.2 m and the distance between the two slits is 2.4 mm. If a thin transparent mica sheet of thickness 1 𝜇𝑚 𝑎𝑛𝑑 𝑅.𝐼. 1.5 is introduced between one of the interfering beams, the shift in the position of central bright fringe is
0.25 mm
2 mm
0.5 mm
0.125 mm
Validate
Solution:
Path difference due to insertion of mica sheet
Δ
x
=
(
µ
−
1
)
t
Let the shift in the fringe pattern be
′
y
′
Also, path difference
Δ
x
=
y
×
d
D
, so comparing both
(
µ
−
1
)
t
=
y
×
d
D
y
=
(
µ
−
1
)
t
×
D
d
where
µ
=
1.5
,
D
=
2.4
and
d
=
1.2
putting the values, we get
y
=
0.25
m
m
.
© examsnet.com
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