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Question

A single slit is illuminated with a parallel beam of light of wavelength 6000 A and a diffraction pattern is obtained on a screen 1.2 m away from the slit. Distance between position of zero intensity on both side of central maxima is found to be 3.5 mm, then width of slit is

A
0.4 mm
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B
0.8 mm
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C
1.2 mm
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D
0.7 mm
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Solution

The correct option is A 0.4 mm
At minima,
bsinθ=λ, and sinθ=tanθ=D2x
b×D2x=λb=2λxD
b=2×6000×1010×1.23.5×103=0.41 mm

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