JEE Main 2025 Apr 3 Shift 2, Physics Q6: Young’s Double Slit Experiment

JEE Main2025Apr 3, Shift 2Physics

Two monochromatic light beams have intensities in the ratio $1 : 9$ . An interference pattern is obtained by these beams. The ratio of the intensities of maximum to minimum is

A
$8 : 1$
B
$9 : 1$
C
$3 : 1$
D
$4 : 1$

Solution

With amplitudes $I_{1} : I_{2} = 1 : 3$ :

$\frac{I _{m a x}}{I _{m i n}} = \frac{( I _{1} + I _{2} ) ^{2}}{( I _{1} - I _{2} ) ^{2}} = \frac{( 1 + 3 ) ^{2}}{( 3 - 1 ) ^{2}} = \frac{16}{4} = 4$ .

$∴ I_{m a x} : I_{m i n} = 4 : 1$ .

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JEE Main 2025 Apr 3 Shift 2, Physics Q6: Young’s Double Slit Experiment

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