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JEE Main 2024 (6 Apr, Shift 2) · Superposition, Coherence & Intensity

Superposition, Coherence & Intensity question from JEE Main 2024 (6 Apr, Shift 2)

Q39JEE Main 2024 · 6 APR · SHIFT 2Easy

Superposition, Coherence & Intensity

Two coherent monochromatic light beams of intensities I and $4I$ are superimposed. The difference between maximum and minimum possible intensities in the resulting beam is $xI$ . The value of $x$ is __________.

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

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Answer: 8

$I_{max} = (\sqrt{I}+\sqrt{4I})^2 = (1+2)^2 I = 9I$
$I_{min} = (\sqrt{4I}-\sqrt{I})^2 = (2-1)^2 I = I$
$I_{max} - I_{min} = 8I$
So $x = 8$ .

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