In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is λ is ' I '. The intensity at a point where the path difference is λ/6 is [cos π/6 = √(3)/2][λ = wavelength of light ][cos π = -1] (a) I (b) 3I/4 (c) I/2 (d) I/4

Answer: (b). Path difference λ is a bright fringe, so I = Imax. For λ/6: φ = π/3, intensity = Icos ²π/6 = 3I/4. So the answer is (b) 3I/4.

In Young's double slit experiment, the intensity of light at a point… | MHT CET 2025 (23 Apr, Shift 2)

Q31MHT CET 2025 · 23 APR · SHIFT 2Medium

Superposition, Coherence & Intensity

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is ' $I$ '. The intensity at a point where the path difference is $\lambda/6$ is $\left[\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\right][\lambda =$ wavelength of light $][\cos\pi = -1]$

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Answer: (b)

Path difference $\lambda$ is a bright fringe, so $I = I_{max}$ .
For $\lambda/6$ : $\phi = \pi/3$ , intensity $= I\cos^2\dfrac{\pi}{6} = \dfrac{3I}{4}$ .
So the answer is (b) $\dfrac{3I}{4}$ .

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