Light of wavelength λ = 5000 Å falls normally on a narrow slit. A screen placed at a distance of 1 m from the slit and perpendicular to the direction of light. The first minima of the diffraction pattern is situated at 5 mm from the centre of central maximum. The width of the slit is
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a
0.1 mm
b
1.0 mm
c
0.5 mm
d
0.2 mm
answer is A.
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Detailed Solution
Position of nth minima xn=nλDd⇒5×10−3=1×5000×10−10×1d⇒d=10−4m=0.1mm