Interference fringes are produced on a 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 π/2 = 0, cos π = -1) (a) 2I (b) 4I (c) 6I (d) 8I

Answer: (c). I = I₁ + I₂ + 2√(I₁ I₂)cos φ, with 2√(I · 9I) = 6I. At P ( φ = π/2): IP = I + 9I + 0 = 10I. At Q ( φ = π): IQ = 10I - 6I = 4I. IP - IQ = 6I. So the answer is (c) 6I.

Interference fringes are produced on a screen by using two light… | MHT CET 2026 (14 May, Shift 2)

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Superposition, Coherence & Intensity

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

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Answer: (c)

$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$ , with $2\sqrt{I \cdot 9I} = 6I$ .
At $P$ ( $\phi = \pi/2$ ): $I_P = I + 9I + 0 = 10I$ .
At $Q$ ( $\phi = \pi$ ): $I_Q = 10I - 6I = 4I$ .
$I_P - I_Q = 6I$ . So the answer is (c) $6I$ .

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