Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point A and π at point B. Then the difference between the resultant intensities at A and B is (a) 4I (b) 5I (c) 2I (d) 3I

Answer: (a). IR = I₁ + I₂ + 2√(I₁ I₂)cos φ, with 2√(4I²) = 4I. At A ( φ = π/2): IA = 5I. At B ( φ = π): IB = 5I - 4I = I. IA - IB = 4I. So the answer is (a) 4I.

Two beams of light having intensities I and 4I interfere to produce a… | MHT CET 2021 (24 Sep, Shift 1)

Q80MHT CET 2021 · 24 SEP · SHIFT 1Medium

Superposition, Coherence & Intensity

Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\pi/2$ at point $A$ and $\pi$ at point $B$ . Then the difference between the resultant intensities at $A$ and $B$ is

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Answer: (a)

$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$ , with $2\sqrt{4I^2} = 4I$ .
At $A$ ( $\phi = \pi/2$ ): $I_A = 5I$ . At $B$ ( $\phi = \pi$ ): $I_B = 5I - 4I = I$ .
$I_A - I_B = 4I$ . So the answer is (a) $4I$ .

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