In Young's double slit experiment, the intensity at a point where the path difference is (λ)/(4) [ λ is wavelength of light used] is ' I '. If ' I₀ ' is the maximum intensity then I/I₀ is equal to [cos π/4 = sin π/4 = 1/√(2)] (a) 3: 2 (b) 2: 3 (c) 3: 4 (d) 1: 2

Answer: (d). Δ x = λ/4 ⇒ φ = π/2. I/I₀ = cos ²π/4 = 1/2 So the answer is (d) 1: 2.

In Young's double slit experiment, the intensity at a point where the… | MHT CET 2021 (20 Sep, Shift 2)

Q83MHT CET 2021 · 20 SEP · SHIFT 2Medium

Superposition, Coherence & Intensity

In Young's double slit experiment, the intensity at a point where the path difference is $\frac{\lambda}{4}$ [ $\lambda$ is wavelength of light used] is ' $I$ '. If ' $I_0$ ' is the maximum intensity then $\frac{I}{I_0}$ is equal to $\left[\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}\right]$

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Answer: (d)

$\Delta x = \lambda/4 \Rightarrow \phi = \pi/2$ .
$\dfrac{I}{I_0} = \cos^2\dfrac{\pi}{4} = \dfrac{1}{2}$
So the answer is (d) $1 : 2$ .

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Superposition, Coherence & Intensity question from MHT CET 2021 (20 Sep, Shift 2)

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