Q 12-10-070JEE MainJEE Main 2024 (30 Jan, Shift 1)Medium
The diffraction pattern of a light of wavelength $400\ \text{nm}$ diffracting from a slit of width $0.2\ \text{mm}$ is focused on the focal plane of a convex lens of focal length $100\ \text{cm}$. The width of the 1st secondary maxima will be:
Answer: (A) $2\ \text{mm}$
A secondary maximum lies between two adjacent minima, so its width is
$$\beta = \frac{\lambda f}{a} = \frac{400\times10^{-9}\times1}{0.2\times10^{-3}} = 2\times10^{-3}\ \text{m} = 2\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics