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In a Young’s double slit experiment, the slit separation is 1 mm and the screen is 1 m from the slit. For a monochromatic light of wavelength 500 nm, the distance of 3rd minima from the central maxima is

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a

0.50 mm

b

1.25 mm

c

1.50 mm

d

1.75 mm

answer is B.

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Detailed Solution

Distance of nth minima from central bright fringexn=(2n−1)λD2d For n=3 i.e. 3rd minima x3=(2×3−1)×500×10−9×12×1×10−3 =5×500×10−62=1.25×10−3m=1.25mm.

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