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A parallel beam of monochromatic light of wavelength 5000 $\overset{o}{A}$ is incident normally on a single narrow slit of width 0.001 mm. The light is focused by a convex lens on a screen placed on the focal plane. The first minimum will be formed for the angle of diffraction equal to :

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0°

15°

30°

60°

answer is C.

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Detailed Solution

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

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