Q 12-10-052JEE MainJEE Main 2024 (1 Feb, Shift 2)Easy
A microwave of wavelength $2.0\ \text{cm}$ falls normally on a slit of width $4.0\ \text{cm}$. The angular spread of the central maximum of the diffraction pattern obtained on a screen $1.5\ \text{m}$ away from the slit will be
Answer: (C) $60^\circ$
First minimum: $a\sin\theta = \lambda \Rightarrow \sin\theta = \dfrac{2}{4} = \dfrac12 \Rightarrow \theta = 30^\circ$.
The central maximum spreads from $-30^\circ$ to $+30^\circ$, so its angular spread is $60^\circ$.
Solution by Sreeraj P, M.Sc Physics