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 π / 4 . Then the intensity on

Question

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 π / 4 . Then the intensity on the screen at a distant point situated at an angle θ = 3 0 ∘ from the radiators is (here I 0 is the intensity at that point due to one radiator alone)

TestHub · Accepted Answer

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 $θ = 3 0^{\circ}$ . Thus

$ϕ = \frac{π}{4} + \frac{2 π}{λ} (d sin θ) = \frac{π}{4} + \frac{2 π}{λ} \cdot \frac{λ}{4} (sin 3 0^{\circ}) = \frac{π}{4} + \frac{π}{4} = \frac{π}{2}$

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

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