In Young's double slit experiment, the intensity on screen at a point, where path difference is (λ)/(4) is K/4. The intensity at a point when path difference is ' λ ' will be [cos π/2 = 0, cos 2π = 1] (a) 4 K (b) 2 K (c) K (d) K/2

Answer: (d). At Δ x = λ/4, φ = π/2, so I = Imaxcos ² 45° = Imax2 = K/4. So Imax = K/2. At Δ x = λ, φ = 2π, a maximum: I = K/2. So the answer is (d) K/2.

In Young's double slit experiment, the intensity on screen at a… | MHT CET 2025 (5 May, Shift 2)

Q29MHT CET 2025 · 5 MAY · SHIFT 2Medium

Superposition, Coherence & Intensity

In Young's double slit experiment, the intensity on screen at a point, where path difference is $\frac{\lambda}{4}$ is $\frac{K}{4}$ . The intensity at a point when path difference is ' $\lambda$ ' will be $\left[\cos\frac{\pi}{2} = 0, \cos 2\pi = 1\right]$

Show answer and solution

Answer: (d)

At $\Delta x = \lambda/4$ , $\phi = \pi/2$ , so $I = I_{max}\cos^2 45^\circ = \dfrac{I_{max}}{2} = \dfrac{K}{4}$ .
So $I_{max} = \dfrac{K}{2}$ .
At $\Delta x = \lambda$ , $\phi = 2\pi$ , a maximum: $I = \dfrac{K}{2}$ .
So the answer is (d) $\dfrac{K}{2}$ .

Report a mistake

Superposition, Coherence & Intensity question from MHT CET 2025 (5 May, Shift 2)

More questions on Superposition, Coherence & Intensity