JEE MAIN - Physics (2025 - 4th April Evening Shift - No. 25)
In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 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 $x \times 10^{-3} \mathrm{~cm}$, where $x$-value is _________ .
Answer
15
Explanation
Width of 20 maxima of double slit $=$ width of central maxima of single slit
$$\begin{aligned} & \frac{20 \lambda \mathrm{D}}{\mathrm{~d}}=\frac{2 \lambda \mathrm{D}}{\mathrm{a}} \\ & \frac{10}{\mathrm{~d}}=\frac{1}{\mathrm{a}} \\ & \mathrm{a}=\frac{\mathrm{d}}{10}=\frac{1.5 \times 10^{-1}}{10} \mathrm{~cm}=15 \times 10^{-3} \mathrm{~cm} \end{aligned}$$
Value of $x$ is 15
Answer is 15
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