JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Wave Optics

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A microwave of wavelength 2.0 cm falls normally on a slit of width 4.0 cm . The angular spread of the central maxima of the diffraction pattern obtained on a screen 1.5 m away from the slit, will be:

(A) $3 0^{\circ}$
(B) $1 5^{\circ}$
(C) $6 0^{\circ}$ ✓ Correct
(D) $4 5^{\circ}$

Answer: (C)

Solution

For diffraction of light of wavelength

λ

through a single slit of width a, the first minimum occurs where

a sin θ = λ

. Substituting

λ = 2.0

cm and

a = 4.0

cm:

sin θ = \frac{2.0}{4.0} = \frac{1}{2}

, so

θ = 3 0^{\circ}

. The angular spread of the central maximum, measured between the first minima on either side of the axis, is

2 θ = 6 0^{\circ}

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option

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