Two light beams of intensities 4I and 9I interfere on a screen. The phase difference between these beams on the screen at point A is zero and at point B is π. The difference of resultant intensities, at the point A and B, will be _________ I. Numerical answer type. Work it out, then open the answer.

Answer: 24. At A ( φ = 0): IA = (√(4I) + √(9I))² = (2+3)² I = 25I At B ( φ = π): IB = (√(9I) - √(4I))² = (3-2)² I = I IA - IB = 24I, so the answer is 24.

Two light beams of intensities 4I and 9I interfere on a screen. The… | JEE Main 2022 (29 Jul, Shift 1)

Q71JEE Main 2022 · 29 JUL · SHIFT 1Medium

Superposition, Coherence & Intensity

Two light beams of intensities 4I and 9I interfere on a screen. The phase difference between these beams on the screen at point A is zero and at point B is $\pi$ . The difference of resultant intensities, at the point A and B, will be _________ I.

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

Show answer and solution

Answer: 24

At A ( $\phi = 0$ ): $I_A = (\sqrt{4I} + \sqrt{9I})^2 = (2+3)^2 I = 25I$
At B ( $\phi = \pi$ ): $I_B = (\sqrt{9I} - \sqrt{4I})^2 = (3-2)^2 I = I$
$I_A - I_B = 24I$ , so the answer is $24$ .

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