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

Answer: (b). For equal slits I = Imaxcos ²φ/2 with φ = (2π)/(λ)Δ x. At P: Δ x = 0, φ = 0, IP = Imax. At Q: Δ x = λ/4, φ = π/2, IQ = Imaxcos ² 45° = Imax2. So the answer is (b) 2: 1.

At two points P and Q on the screen in Young's double slit… | MHT CET 2026 (15 May, Shift 2)

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Superposition, Coherence & Intensity

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

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Answer: (b)

For equal slits $I = I_{max}\cos^2\dfrac{\phi}{2}$ with $\phi = \dfrac{2\pi}{\lambda}\Delta x$ .
At $P$ : $\Delta x = 0$ , $\phi = 0$ , $I_P = I_{max}$ .
At $Q$ : $\Delta x = \lambda/4$ , $\phi = \pi/2$ , $I_Q = I_{max}\cos^2 45^\circ = \dfrac{I_{max}}{2}$ .
So the answer is (b) $2 : 1$ .

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