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In an interference experiment the ratio ...

In an interference experiment the ratio of amplitudes of coherent waves is `a_1/a_2=1/3` The ratio of maximum and minimum intensities of fringes is …………..

A

18

B

4

C

2

D

9

Text Solution

Verified by Experts

The correct Answer is:
B
`(I_("max"))/(I_("min")) = ([ sqrt(I_(1)) + sqrt(I_(2))]^(2))/((sqrt(I_(1)) - sqrt(I_(2)))^(2)) = ((A_(1) + 3A_(1))/(A_(1) - 3A_(1)))^(2) = ((4)/(2))^(2) = (4)/(1)`
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