Q 12-10-065JEE MainJEE Main 2024 (29 Jan, Shift 2)Easy
In a single slit diffraction pattern, a light of wavelength $6000\ \text{Å}$ is used. The distance between the first and third minima in the diffraction pattern is found to be $3\ \text{mm}$ when the screen is placed $50\ \text{cm}$ away from the slit. The width of the slit is ______ $\times10^{-4}\ \text{m}$.
Numerical value type. Enter your answer.
Answer: 2
Minima are at $y_n = \dfrac{n\lambda D}{a}$, so
$$y_3 - y_1 = \frac{2\lambda D}{a} \;\Rightarrow\; a = \frac{2\times6\times10^{-7}\times0.5}{3\times10^{-3}} = 2\times10^{-4}\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics