In a Young's double slit experiment, the intensity at some point on… | JEE Main 2026 (2 Apr, Shift 2)

Q18JEE Main 2026 · 2 APR · SHIFT 2Medium

Superposition, Coherence & Intensity

In a Young's double slit experiment, the intensity at some point on the screen is found to be $\frac{3}{4}$ times of the maximum of the interference pattern. The path difference between the interfering waves at this point is $\frac{\lambda}{x}$ where $\lambda$ is wavelength of the incident light. The value of $x$ is _______.

Numerical answer type. Work it out, then open the answer.

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Answer: 6

$I = I_{max}\cos^2\frac{\phi}{2} = \frac{3}{4}I_{max} \Rightarrow \cos\frac{\phi}{2} = \frac{\sqrt{3}}{2}$
$\frac{\phi}{2} = \frac{\pi}{6} \Rightarrow \phi = \frac{\pi}{3}$
Path difference $\Delta x = \frac{\lambda}{2\pi}\phi = \frac{\lambda}{6}$
So $x = 6$ .

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