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Standard XII
Physics
Question

In double slit experiment, the distance between two slits is 0.6mm and these are illuminated with light of wavelength 4800˚A. The width of dark fringe on the screen at a distance 120cm from slits will be:
  1. 8×10−4 radian
  2. 60×10−4 radian
  3. 6×10−4 radian
  4. 16×10−4 radian

A
8×10−4 radian
B
6×10−4 radian
C
60×10−4 radian
D
16×10−4 radian
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Solution
Verified by Toppr

The correct option is A 8×10−4 radian
In Young's double slit experiment, angular width (θ) of the fringe is given by,
θ=λd
where, λ=wavelength of light used =4800˚A=4800×10−10m
d=distance between 2 slits=0.6mm=0.6×10−3m
Now,
θ=4800×10−100.6×10−3=8×10−4radian

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