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Question

In Young's double slit experiment, fringes of width β are produced on the screen kept at a distance of 1m from the slit. When the screen is moved away by 5×102m, fringe width changes by 3×105m. The separation between the slits is 1×103m. The wavelength of light used is __________nm.

A
500
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B
600
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C
700
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D
400
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Solution

The correct option is B 600
Since, we know that in Young's double slit experiment.
β=Dλd ...(i)
β=Dλd ...(ii)
Subtract equation (ii) from equation (i), we get
ββ=λ(DD)d
Substituting the given values, we get
3×105=λ×5×1021×103
λ=3×1085×102=0.6×106m
λ=600×109m
λ=600nm

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