In Young's double slit experiment, at two points P and Q on screen, waves from slits S₁ and S₂ have a path difference of 0 and (λ)/(4) respectively. The ratio of intensities at point P to that at Q will be (cos 0° = 1, cos 45° = 1/√(2)) (a) 3: 2 (b) 2: 1 (c) √(2): 1 (d) 4: 1

Answer: (b). I = Imaxcos ²φ/2, φ = (2π)/(λ)Δ x. At P: φ = 0, IP = Imax. At Q: φ = π/2, IQ = Imax/2. So the answer is (b) 2: 1.

In Young's double slit experiment, at two points P and Q on screen… | MHT CET 2025 (25 Apr, Shift 2)

Q30MHT CET 2025 · 25 APR · SHIFT 2Medium

Superposition, Coherence & Intensity

In Young's double slit experiment, at two points $P$ and $Q$ on screen, waves from slits $S_1$ and $S_2$ have a path difference of 0 and $\frac{\lambda}{4}$ respectively. The ratio of intensities at point $P$ to that at $Q$ will be $\left(\cos 0^\circ = 1, \cos 45^\circ = \frac{1}{\sqrt{2}}\right)$

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Answer: (b)

$I = I_{max}\cos^2\dfrac{\phi}{2}$ , $\phi = \dfrac{2\pi}{\lambda}\Delta x$ .
At $P$ : $\phi = 0$ , $I_P = I_{max}$ . At $Q$ : $\phi = \pi/2$ , $I_Q = I_{max}/2$ .
So the answer is (b) $2 : 1$ .

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