Interference fringes are produced on the screen by using two light sources of intensities I and 9I. The phase difference between the beams is π/2 at point P and π at point Q on the screen. The difference between the resultant intensities at points P and Q is (cos 90° = 0, cos π = -1) (a) 2 I (b) 4 I (c) 6 I (d) 8 I

Answer: (c). I = I₁ + I₂ + 2√(I₁ I₂)cos φ, with 2√(9I²) = 6I. At P ( φ = π/2): IP = 10I. At Q ( φ = π): IQ = 10I - 6I = 4I. IP - IQ = 6I. So the answer is (c) 6I.

Interference fringes are produced on the screen by using two light… | MHT CET 2025 (21 Apr, Shift 1)

Q33MHT CET 2025 · 21 APR · SHIFT 1Medium

Superposition, Coherence & Intensity

Interference fringes are produced on the screen by using two light sources of intensities I and 9I. The phase difference between the beams is $\pi/2$ at point P and $\pi$ at point Q on the screen. The difference between the resultant intensities at points P and Q is $(\cos 90^\circ = 0, \cos\pi = -1)$

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Answer: (c)

$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$ , with $2\sqrt{9I^2} = 6I$ .
At $P$ ( $\phi = \pi/2$ ): $I_P = 10I$ . At $Q$ ( $\phi = \pi$ ): $I_Q = 10I - 6I = 4I$ .
$I_P - I_Q = 6I$ . So the answer is (c) $6I$ .

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