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JEE Main 08 Apr 2018 Code A Solved Paper
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Question : 23
Total: 90
The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1µm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.)
25 µm.
50 µm.
75 µm.
100 µm.
Validate
Solution:
d
sin
30
=
λ
λ
=
0.5
×
10
−
6
m
β
=
λ
D
d
=
0.5
×
10
−
6
×
50
1
=
25
µ
m
© examsnet.com
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