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

Two identical radiators have a separation of `d=lambda//4` where `lamb

1.	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)A. `I_(0)`B. `2I_(0)`C. `3I_(0)`D. `4I_(0)`
Answer» Correct Answer - B 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)`

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