JEE Main 2025 (3 Apr, Shift 2) Physics Q43: Wave Optics

Question

Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :

Accepted Answer

Option 2: $\left(3 + 2\sqrt{2}\right) : \left(3 - 2\sqrt{2}\right)$

Answer

3 : 1

Answer

$\left(3 + 2\sqrt{2}\right) : \left(3 - 2\sqrt{2}\right)$

Answer

$\left(2\sqrt{2} + 1\right) : \left(2\sqrt{2} - 1\right)$

Answer

9 : 1

JEE Main 3 April 2025, Shift 2, Physics Q43

Same idea in other shifts