Q 12-10-067JEE MainJEE Main 2024 (8 Apr, Shift 2)Medium
Two slits are $1\ \text{mm}$ apart and the screen is located $1\ \text{m}$ away from the slits. A light of wavelength $500\ \text{nm}$ is used. The width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern is ______ $\times10^{-4}\ \text{m}$.
Numerical value type. Enter your answer.
Answer: 2
Width of the central diffraction maximum: $\dfrac{2\lambda D}{a}$. Fringe width of the double slit pattern: $\dfrac{\lambda D}{d}$.
$$\frac{2\lambda D}{a} = 10\frac{\lambda D}{d} \;\Rightarrow\; a = \frac{d}{5} = 0.2\ \text{mm} = 2\times10^{-4}\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics