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JEE Main · 4 April 2025, Shift 2 · Q46

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…

In a Young's double slit experiment, two slits are located $\displaystyle 1.5$ mm apart. The distance of screen from slits is $\displaystyle 2$ m and the wavelength of the source is $\displaystyle 400$ nm. If the $\displaystyle 20$ 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 $\displaystyle x \times 10^{-3} \mathrm{~cm}$, where $\displaystyle x$-value is $\displaystyle \_\_\_\_$.
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JEE Main 2025 Physics question, with the answer from NTA’s final answer key. Where our answers come from.