JEE Main 2025 — Wave Optics Question with Solution
JEE Main 2025 (4 Apr Shift 2)
Question
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 centre maximum of the single slit diffraction pattern, then the width of each slit is , where x -value is ________
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Show full solutionCorrect answer: 15
Correct answer
15
Step-by-step explanation
Width of 20 maxima of double slit width of central maxima of single slit
$\begin{aligned}
& \frac{20 \lambda \mathrm{D}}{\mathrm{~d}}=\frac{2 \lambda \mathrm{D}}{\mathrm{a}} \\ & \frac{10}{\mathrm{~d}}=\frac{1}{\mathrm{a}} \\ & \mathrm{a}=\frac{\mathrm{d}}{10}=\frac{1.5 \times 10^{-1}}{10} \mathrm{~cm}=15 \times 10^{-3} \mathrm{~cm}
\end{aligned}x$ is 15
Answer is 15
$\begin{aligned}
& \frac{20 \lambda \mathrm{D}}{\mathrm{~d}}=\frac{2 \lambda \mathrm{D}}{\mathrm{a}} \\ & \frac{10}{\mathrm{~d}}=\frac{1}{\mathrm{a}} \\ & \mathrm{a}=\frac{\mathrm{d}}{10}=\frac{1.5 \times 10^{-1}}{10} \mathrm{~cm}=15 \times 10^{-3} \mathrm{~cm}
\end{aligned}x$ is 15
Answer is 15
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This is a previous-year question from JEE Main 2025, covering the Wave Optics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.