1.

In YDSE, the amplitude of intensity variation fo the two sources is found to be `5%` of the average intensity. The ratio of the intensities of two interfering sources isA. 2564B. 1089C. 1681D. 869

Answer» Correct Answer - c
`(I_(max))/(I_(min)) = ((sqrt I_(1) + sqrt I_(2))^(2))/((sqrt I_(1) - sqrt I_(2))^(2))`
`= ((1 + sqrt((I_(2))/(I_(1))))/(1 - sqrt((I_(2))/(I_(1)))))^(2)`
For given information,
`I_(max) = (105)/(100) I`
`I_(min) = ((95)/(100)) I`
where `I + I_(1) + I_(2)` is average intensity
`:. (I_(max))/(I_(min)) = (105)/(95) = ((1 + sqrt((I_(2))/(I)))/(1 - sqrt((I_(2))/(I_(1)))))^(2)`
`implies sqrt((I_(2))/(I_(1))) = 0.0244 = gt (I_(2))/(I_(1)) = 0.0059`
`implies (I_(1))/(I_(2)) = 1681`.


Discussion

No Comment Found

Related InterviewSolutions