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Q.

A parallel beam of monochromatic light of wavelength 6000 A0   is incident normally on a narrow slit of width of  0.002 mm . The light is focussed by a convex lens on a screen placed on a focal plane. The first minimum is formed for the angle of diffraction equal to

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a

00

b

150

c

sin−1(0.3)

d

600

answer is C.

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Detailed Solution

Wavelength λ=6000 A0=6000×10−10 m Slit width a=0.002 mm=0.002×10−3 mAngle of diffraction θ=?    asinθ=nλ⇒sinθ=nλa=1×6×10−72×10−6=0.3          ∴  θ=sin−1(0.3)
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