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

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

In Young's double slit experiment, the intensity on screen at a point… | MHT CET 2025 (19 Apr, Shift 2)

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Superposition, Coherence & Intensity

In Young's double slit experiment, the intensity on screen at a point where path difference is $\frac{\lambda}{4}$ is $\frac{K}{2}$ . The intensity at a point when path difference is ' $\lambda$ ' will be

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Answer: (c)

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

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