JEE Main 2025 Apr 3 Shift 1, Physics Q23: Young’s Double Slit Experiment

JEE Main2025Apr 3, Shift 1Physics

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

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Solution

For coherent superposition,

$I_{m a x} = (I_{1} + I_{2})^{2} = (4 I + 9 I)^{2} = (2 I + 3 I)^{2} = 25 I,$

$I_{m i n} = (I_{1} - I_{2})^{2} = (3 I - 2 I)^{2} = I .$

Difference: $I_{m a x} - I_{m i n} = 24 I .$ So $x = 24.$

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JEE Main 2025 Apr 3 Shift 1, Physics Q23: Young’s Double Slit Experiment

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