In Young's double slit experiment, the wavelength of light used is λ. The intensity on the screen at a point for path difference ' λ ' is 'X'. The intensity at the point for path difference ((λ)/(6)) is ( cos 180° = -1, cos 30° = √(3)/2) (a) X/6 (b) X/2 (c) 3X/4 (d) 4X/3

Answer: (c). Path difference λ gives φ = 2π, a bright fringe, so X = Imax. For λ/6: φ = π/3, I = Xcos ² 30° = 3X/4. So the answer is (c) 3X/4.

In Young's double slit experiment, the wavelength of light used is λ… | MHT CET 2026 (19 Apr, Shift 1)

Q11MHT CET 2026 · 19 APR · SHIFT 1Medium

Superposition, Coherence & Intensity

In Young's double slit experiment, the wavelength of light used is $\lambda$ . The intensity on the screen at a point for path difference ' $\lambda$ ' is 'X'. The intensity at the point for path difference $\left(\frac{\lambda}{6}\right)$ is ( $\cos 180^\circ = -1,\ \cos 30^\circ = \frac{\sqrt{3}}{2}$ )

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Answer: (c)

Path difference $\lambda$ gives $\phi = 2\pi$ , a bright fringe, so $X = I_{max}$ .
For $\lambda/6$ : $\phi = \pi/3$ , $I = X\cos^2 30^\circ = \dfrac{3X}{4}$ .
So the answer is (c) $\dfrac{3X}{4}$ .

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