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Physics · 2026

JEE Main · 2 April 2026, Shift 2 · Q46

In a Young's double slit experiment, the intensity at some point on the screen is found to be 3/4 times of the maximum of the interference pattern.…

In a Young's double slit experiment, the intensity at some point on the screen is found to be $\displaystyle \frac{3}{4}$ times of the maximum of the interference pattern. The path difference between the interfering waves at this point is $\displaystyle \frac{\lambda}{x}$ where $\displaystyle \lambda$ is wavelength of the incident light. The value of $\displaystyle x$ is $\displaystyle \_\_\_\_$.
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JEE Main 2026 Physics question, with the answer from NTA’s final answer key. Where our answers come from.