In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is λ is x units, λ being the wavelength of light used. The intensity at a point where the path difference is (λ)/(4) will be (cos 2π = 1, cos π/2 = 0) (a) x/4 (b) x/2 (c) x (d) zero

Answer: (b). Path difference λ is a bright fringe, so x = Imax. For λ/4: φ = π/2, I = xcos ² 45° = x/2. So the answer is (b) x/2.

In Young's double slit experiment, the intensity of light at a point… | MHT CET 2024 (15 May, Shift 2)

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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 x units, $\lambda$ being the wavelength of light used. The intensity at a point where the path difference is $\frac{\lambda}{4}$ will be $\left(\cos 2\pi = 1, \cos\frac{\pi}{2} = 0\right)$

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Answer: (b)

Path difference $\lambda$ is a bright fringe, so $x = I_{max}$ .
For $\lambda/4$ : $\phi = \pi/2$ , $I = x\cos^2 45^\circ = \dfrac{x}{2}$ .
So the answer is (b) $\dfrac{x}{2}$ .

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