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Two beams of ligth having intensities I and 4I interface to produce a fringe pattern on a screen. The phase difference between the beams is `(pi)/(2)` at point A and `pi` at point B. Then the difference between the resultant intensities at A and B isA. 2 IB. 4 IC. 5 ID. 7 I |
Answer» Correct Answer - b We know that the resultant amplitude of two interfering waves is given by `R^(2) = a^(2) + B^(2) + 2a b cos phi`, where R is the amplitude of resultant wave, a is the amplitude of one wave, b is the amplitude of second wave, and `phi` as the phase difference between the two wave at a point. Also `I prop ("amplitude")^(2)` `:. I prop I_(1) + I_(2) + 2sqrt I_(1) sqrt I_(2) cos phi` Applying Eq. (i) when phase diffenence is `pi // 2` `I_(pi // 2) prop I + 4 I` `implies I_(pi // 2) prop 5 I` Again, applying Eq. (i) when phase difference is `pi` `I_(pi) prop I + 4 I + 2 sqrt I sqrt( 4I) cos pi` `:. I_(pi) prop I implies I_(pi//2) - I_(pi) prop 4 I` Correct option is (b) |
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