In a Young's double slit experiment the intensity at a point where the path difference is (λ)/(6) ( λ being the wavelength of light used) is I. If I₀ denotes the maximum intensity, I/I₀ is equal to (a) 3/4 (b) 1/√(2) (c) √(3)/2 (d) 1/2

Answer: (a). Phase difference φ = (2π)/(λ)Δ x = (2π)/(λ)·(λ)/(6) = π/3 I = I₀cos ²φ/2 = I₀cos ²π/6 = 3/4I₀ So the answer is (a) 3/4.

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Superposition, Coherence & Intensity

In a Young's double slit experiment the intensity at a point where the path difference is $\frac{\lambda}{6}$ ( $\lambda$ being the wavelength of light used) is $I$ . If $I_0$ denotes the maximum intensity, $\frac{I}{I_0}$ is equal to

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Answer: (a)

Phase difference $\phi = \frac{2\pi}{\lambda}\Delta x = \frac{2\pi}{\lambda}\cdot\frac{\lambda}{6} = \frac{\pi}{3}$
$I = I_0\cos^2\frac{\phi}{2} = I_0\cos^2\frac{\pi}{6} = \frac{3}{4}I_0$
So the answer is (a) $\frac{3}{4}$ .

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