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

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

In Young's double slit experiment, the wavelength of light used is '… | MHT CET 2023 (9 May, Shift 2)

Q63MHT CET 2023 · 9 MAY · SHIFT 2Medium

Superposition, Coherence & Intensity

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

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Answer: (b)

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

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Superposition, Coherence & Intensity question from MHT CET 2023 (9 May, Shift 2)

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