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JEE Main 2025 (3 Apr, Shift 1) · Superposition, Coherence & Intensity

Superposition, Coherence & Intensity question from JEE Main 2025 (3 Apr, Shift 1)

Q24JEE Main 2025 · 3 APR · SHIFT 1Easy

Superposition, Coherence & Intensity

Two coherent monochromatic light beams of intensities 4I and 9I are superimposed. The difference between the maximum and minimum intensities in the resulting interference pattern is $xI$ . The value of $x$ is___________.

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

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Answer: 24

$I_{max} = (\sqrt{4I}+\sqrt{9I})^2 = (2+3)^2 I = 25I$
$I_{min} = (\sqrt{9I}-\sqrt{4I})^2 = (3-2)^2 I = I$
$I_{max} - I_{min} = 24I$
So $x = 24$ .

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