In Young's double slit experiment the intensities at two points, for the path difference (λ)/(4) and (λ)/(3) ( λ = wavelength of light used) are I₁ and I₂ respectively. If I₀ denotes the intensity produced by each one of the individual slits then (I₁ + I₂)/(I₀) is equal to (cos 60° = 0.5, cos 45° = 1/√(2)) (a) 1 (b) 2 (c) 3 (d) 4

Answer: (c). For equal slits I = 4I₀cos ²φ/2. λ/4: φ = π/2, I₁ = 4I₀ · 1/2 = 2I₀. λ/3: φ = 2π/3, I₂ = 4I₀ · 1/4 = I₀. (I₁ + I₂)/(I₀) = 3. So the answer is (c) 3.

In Young's double slit experiment the intensities at two points, for… | MHT CET 2023 (12 May, Shift 1)

Q61MHT CET 2023 · 12 MAY · SHIFT 1Medium

Superposition, Coherence & Intensity

In Young's double slit experiment the intensities at two points, for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ( $\lambda =$ wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits then $\frac{I_1 + I_2}{I_0}$ is equal to $\left(\cos 60^\circ = 0.5, \cos 45^\circ = \frac{1}{\sqrt{2}}\right)$

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Answer: (c)

For equal slits $I = 4I_0\cos^2\dfrac{\phi}{2}$ .
$\lambda/4$ : $\phi = \pi/2$ , $I_1 = 4I_0 \cdot \dfrac{1}{2} = 2I_0$ .
$\lambda/3$ : $\phi = 2\pi/3$ , $I_2 = 4I_0 \cdot \dfrac{1}{4} = I_0$ .
$\dfrac{I_1 + I_2}{I_0} = 3$ . So the answer is (c) $3$ .

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