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 `lambda//4`. Then the intensity on the screen at a distant point situated at an angle `theta=30^@` from the radiators is (here `I_0` is intensity at that point due to one radiator alone)

Options

A. (a) `I_0`
B. (b) `2I_0`
C. (c) `3I_0`
D. (d) `4I_0`

Answer

B

Explanation

The intensity at a point on screen is given by
`I=4I_0cos^2(varphi//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^@`. Thus
`varphi=pi/4+(2pi)/(lambda)(dsintheta)=pi/4+(2pi)/(lambda)*lambda/4(sin 30^@)`
`=pi/4+pi/4=pi/2`
Thus `varphi/2=pi/2` and `I=4I_0cos^2(pi//4)=2I_0`