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Physics Question on single slit diffraction

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

A

0∘^{\circ}

B

15∘^{\circ}

C

30∘^{\circ}

D

50∘^{\circ}

Answer

30∘^{\circ}

Explanation

Solution

: For first minimum, asinθ=nλ=Iλa sin \theta = n \lambda = I \lambda sinθ=λa=5000×10−100.001×10−3=0.5sin \theta = \frac{\lambda}{a} = \frac{ 5000 \times {10}^{-10}}{0.001 \times {10}^{-3}} = 0.5 θ=30∘.\theta = 30 ^ \circ .