InterviewSolution
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Interference fringes are produced on a screen by using two light sources of intensities / and 9/. The phase difference between the beams `pi/2` is at point P and `pi`at point Q on the screen. The difference between the resultant intensities at point P and Q isA. 2IB. 4IC. 6ID. 8I |
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Answer» Correct Answer - C According to question, resultant intensity of interferring wave is given by `I_(P)=I_(1)+I_(2)+2sqrt(I_(1)I_(2)cos phi)` For `phi=(pi)/(2), I_(P)=I+9I=10I` Again at point Q, resultant intensity is given by `I_(Q)=I_(Q)=I_(1)+I_(2)+2sqrt(I_(1)I_(2)cos phi)` For `phi=pi, I_(Q)=I+9I+(-2sqrt(9(I)^(2)))` `=10I-6I=4I` Now, difference between the resultant intensity is given by `Delta I=I_(P)-I_(Q)=10I-4I=6I` |
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