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A parallel beam of light of wavelength $5000 Å$ is incident normally on a single slit of width 0.001mm . The light is focused by a convex lens on a screen placed in focal plane. The first minimum is formed for the angle of diffraction equal to (in degrees)

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answer is 2.

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Detailed Solution

$θ = \sin^{- 1} (\frac{m λ}{d}); dsin θ = m λ; θ = \sin^{- 1} (\frac{(1) (5000 \times 10^{- 10})}{0.001 \times 10^{- 3}}); θ = 30^{\circ}$

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