In Young's double slit experiment, 'I' is the minimum intensity and ' I₁ ' is the intensity at a point where the path difference is (λ)/(4) where ' λ ' is the wavelength of light used. The ratio I/I₁ is (Intensities of the two interfering waves are same) (cos 0° = 1, cos 90° = 0) (a) 0 (b) 4 (c) 3 (d) 2

Answer: (a). Equal intensities: Imin = (√(I₀) - √(I₀))² = 0, so I = 0. At λ/4: I₁ = 4I₀cos ² 45° = 2I₀ ≠ 0. I/I₁ = 0. So the answer is (a) 0.

In Young's double slit experiment, 'I' is the minimum intensity and '… | MHT CET 2024 (15 May, Shift 1)

Q46MHT CET 2024 · 15 MAY · SHIFT 1Medium

Superposition, Coherence & Intensity

In Young's double slit experiment, 'I' is the minimum intensity and ' $I_1$ ' is the intensity at a point where the path difference is $\frac{\lambda}{4}$ where ' $\lambda$ ' is the wavelength of light used. The ratio $\frac{I}{I_1}$ is (Intensities of the two interfering waves are same) $(\cos 0^\circ = 1, \cos 90^\circ = 0)$

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Answer: (a)

Equal intensities: $I_{min} = (\sqrt{I_0} - \sqrt{I_0})^2 = 0$ , so $I = 0$ .
At $\lambda/4$ : $I_1 = 4I_0\cos^2 45^\circ = 2I_0 \neq 0$ .
$\dfrac{I}{I_1} = 0$ . So the answer is (a) $0$ .

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