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Question

A long narrow horizontal slit lies 1 mm above a plane mirror. The interference pattern produced by the slit and its image is viewed on a screen at a distance 1m from the slit. The wavelength of light is 700 nm. Then the distance of the first maxima above the mirror is equal to (d<<D);
1116141_cf911e12f20a4a32826b48aeec9c7915.png

A
0.30mm
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B
0.35mm
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C
60mm
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D
7.5mm
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Solution

The correct option is B 0.35mm
Given that,

D=1m

λ=700nm=700×109m

Since,

a=2mm, d=2a=2mm=2×103m

Fringe width =λDd=700×109m×1m2×103m=0.35mm

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