In a Young's double slit experiment, the intensities at two points, for the path difference (λ)/(4) and (λ)/(3) ( λ being the 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) = (cos 45° = 1/√(2), cos 60 = 1/2) (a) 2 (b) 3 (c) 4 (d) 5

Answer: (b). For equal slits I = 4I₀cos ²φ/2. Δ x = λ/4: φ = π/2, I₁ = 4I₀cos ² 45° = 2I₀. Δ x = λ/3: φ = 2π/3, I₂ = 4I₀cos ² 60° = I₀. (I₁ + I₂)/(I₀) = 3. So the answer is (b) 3.

In a Young's double slit experiment, the intensities at two points… | MHT CET 2026 (11 Apr, Shift 1)

Q14MHT CET 2026 · 11 APR · SHIFT 1Medium

Superposition, Coherence & Intensity

In a Young's double slit experiment, the intensities at two points, for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ( $\lambda$ being the 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 $\frac{I_1 + I_2}{I_0} =$

$\left(\cos 45^\circ = \frac{1}{\sqrt{2}}, \cos 60 = \frac{1}{2}\right)$

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Answer: (b)

For equal slits $I = 4I_0\cos^2\dfrac{\phi}{2}$ .
$\Delta x = \lambda/4$ : $\phi = \pi/2$ , $I_1 = 4I_0\cos^2 45^\circ = 2I_0$ .
$\Delta x = \lambda/3$ : $\phi = 2\pi/3$ , $I_2 = 4I_0\cos^2 60^\circ = I_0$ .
$\dfrac{I_1 + I_2}{I_0} = 3$ . So the answer is (b) $3$ .

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