Two identical radiators have a separation of d = /4 where is the … - Vidyadip

MCQ

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

A
${I_o}$
✓
$2{I_o}$
C
$3{I_o}$
D
$4{I_o}$

✓

Answer

Correct option: B.

$2{I_o}$

b
(b)The intensity at a point on screen is given by
$I = 4{I_0}{\cos ^2}(\phi /2)$
where $\phi$ is the phase difference. In this problem $\phi$ arises $(i)$ due to initial phase difference of $\pi/4$ and $(ii)$ due to path difference for the observation point situated at $\theta = {30^o}.$ Thus
$\phi = \frac{\pi }{4} + \frac{{2\pi }}{\lambda }\,(d\sin \theta ) = \frac{\pi }{4} + \frac{{2\pi }}{\lambda }.\frac{\lambda }{4}(\sin 30^\circ ) = \frac{\pi }{4} + \frac{\pi }{4} = \frac{\pi }{2}$
Thus $\frac{\phi }{2} = \frac{\pi }{4}$ and $I = 4{I_0}{\cos ^2}(\pi /4) = 2{I_0}$

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