InterviewSolution
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InterviewSolution
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Two identical sources each of intensity `I_(0)` have a separation `d = lambda // 8`, where `lambda` is the wavelength of the waves emitted by either source. The phase difference of the sources is `pi // 4` The intensity distribution `I(theta)` in the radiation field as a function of `theta` Which specifies the direction from the sources to the distant observation point P is given byA. `I(theta) = I_(0)cos^(2)theta`B. `I(theta)=(I)_(0)/(4)cos^(2)((pi theta)/(8))`C. `I(theta)=4I_(0)cos^(2)((pi)/(8)(sin theta+1))`D. `I(theta)= I_(0) sin^(2)theta` |
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Answer» Correct Answer - C `Delta x = d sin theta = (lambda)/(8) sin theta` `Delta phi = (2pi)/(lambda)xx(lambda)/(8) sin theta + Delta phi_(0)` (`Delta phi_(0)` is initial phase difference between the slits) `Delta phi = (pi)/(4) sin theta + (pi)/(4)` `Delta phi = (pi)/(4) (1+sin theta)` Resultant intensity `I = 4I_(0) cos^(2) ((Delta phi)/(2))` `I = 4I_(0) cos^(2) {(pi)/(8) (1+sin theta)}` |
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