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

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

In Young's double slit experiment using monochromatic light of… | MHT CET 2024 (3 May, Shift 2)

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Superposition, Coherence & Intensity

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

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Answer: (b)

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

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