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. `I_(0)`
B. `2I_(0)`
C. `3I_(0)`
D. `4I_(0)`

Answer

B

Explanation

The intensity at a point on screen is given by
`I=4I_(0)cos^(2)(phi//2)`
When `phi` is the phase difference. In this problem `phi` arises
(i) due to initial phase differnce of `(pi)/(4)` and (ii) due to path difference for the observation point situated at `theta=30^(@)`.
Thus,
`phi=(pi)/(4)+(2pi)/(lambda)(d sin theta)=(pi)/(4)+(2pi)/(lambda) xx (lambda)/(4) (sin30^(@))=(pi)/(4)+(pi)/(4)=(pi)/(2)`
Hence, `(pi)/(2)=(pi)/(4) and I=4I_(0) cos^(2)((pi)/(4))=2I_(0)`