Q 12-10-164JEE MainJEE Main 2025 (4 Apr, Shift 2)Medium
In a Young's double slit experiment, two slits are located $1.5\ \text{mm}$ apart. The distance of the screen from the slits is $2\ \text{m}$ and the wavelength of the source is $400\ \text{nm}$. If the 20 maxima of the double slit pattern are contained within the central maximum of the single slit diffraction pattern, then the width of each slit is $x\times10^{-3}\ \text{cm}$, where the value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 15
The width of 20 interference fringes equals the width of the central diffraction maximum:
$$20\cdot\frac{\lambda D}{d} = \frac{2\lambda D}{a} \Rightarrow a = \frac{d}{10} = \frac{0.15\ \text{cm}}{10} = 15\times10^{-3}\ \text{cm}$$
So $x = 15$.
Solution by Sreeraj P, M.Sc Physics