In an interference experiment similar to Young’s double slit experiment, the slits S₁and S₂ are illuminated with coherent microwave sources, each of frequency $10^{6} Hz$ . The source are synchronised to have zero phase difference. The slits are separated by a distance d = 150 m. The intensity $I (θ)$ is measured as a function of $θ$ , where $θ$ is defined as shown. If I₀ is the maximum intensity, then $I (θ) for 0 \leq θ \leq 90^{\circ}$ is given by

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$I (θ) = \frac{I_{0}}{2} for θ = 30^{\circ}$

$I (θ)$ is constant for all

$I (θ) = \frac{I_{0}}{4} for θ = 90^{\circ}$

$I (θ) = I_{0} for θ = 0^{\circ}$

answer is A, C.

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Detailed Solution

For microwaves $λ = \frac{c}{f} = \frac{3 \times 10^{8}}{10^{6}} = 300 m$
Path difference $δ = dsin θ$
Phase difference $δ = \frac{2 π}{λ} = Δ = \frac{2 π}{λ} (d \sin θ) = \frac{2 π}{300} (150 \sin θ) = π \sin θ$
intensity $I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos δ$
When $I_{1} ∣= I_{2}, δ = π \sin θ$
I will be maximum when $\cos^{2} (\frac{π \sin θ}{2})$ is maximum =1
$I_{max} = 4 I_{1} = I_{0}$ (given)
(A) when $θ = 0, I = I_{0} \sin^{2} 0 = I_{0}$
(B) when $θ = 30^{\circ}, I = I_{0} \cos^{2} (\frac{π \sin 30^{\circ}}{2}) = I_{0} \cos^{2} \frac{π}{4} = \frac{I_{0}}{2}$
(C) when $θ = 90^{\circ}, I = I_{0} \cos^{2} (\frac{π \sin 90^{\circ}}{2}) = I_{0} \cos^{2} \frac{π}{2} = 0$
i.e., (a) and (c) are correct