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Two identical radiators have a separation of $d = λ / 4$ where $λ$ is the wavelength of the waves emitted by either source. The initial phase difference between the sources is $λ / 4$ . Then the intensity on the screen at a distant point situated at an angle $θ = 30^{\circ}$ from the radiators is (here I₀ is intensity at that point due to one radiator alone)

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I₀

4I₀

2I₀

3I₀

answer is B.

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Detailed Solution

The intensity at a point on screen is given by

$I = 4 I_{0} \cos^{2} (ϕ / 2)$

where $ϕ$ is the phase difference. In this problem $ϕ$ arises (i) due to initial phase difference of $π / 4$ and (ii) due to path difference for the observation point situated at $θ = 30^{\circ} .$ Thus

$ϕ = \frac{π}{4} + \frac{2 π}{λ} (dsinθ) = \frac{π}{4} + \frac{2 π}{λ} \cdot \frac{λ}{4} (\sin 30^{\circ})$

$= \frac{π}{4} + \frac{π}{4} = \frac{π}{2}$

Thus $\frac{ϕ}{2} = \frac{π}{4} and I = 4 I_{0} \cos^{2} (π / 4) = 2 I_{0}$

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