Q 12-10-037JEE MainJEE Main 2026 (2 Apr, Shift 1)Easy
In single slit diffraction pattern, the wavelength of light used is $628$ nm and slit width is $0.2$ mm, the angular width of central maximum is $\alpha\times10^{-2}$ degrees. The value of $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 36
Angular width of the central maximum $=\dfrac{2\lambda}{a}=\dfrac{2\times628\times10^{-9}}{0.2\times10^{-3}}=6.28\times10^{-3}$ rad.
In degrees: $6.28\times10^{-3}\times\dfrac{180}{\pi}=0.36^\circ=36\times10^{-2}$ degrees, so $\alpha=36$.
Solution by Sreeraj P, M.Sc Physics