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In Young's double slit experiment, the r...

In Young's double slit experiment, the ratio of maximum and minimum intensities in the fringe system is `9:1` the ratio of amplitudes of coherent sources is

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The correct Answer is:
2
`(I_("max"))/(I_("min")) = ((sqrt(I_1) + sqrt(I_2))^2)/((sqrt(I_1) - sqrt(I_2))^2) = 9`
`((sqrt(I_1) + sqrt(I_2)))/((sqrt(I_1) - sqrt(I_2))) = 3`
`implies (sqrt(I_1))/(sqrt(I_2)) = (3 + 1)/(3 -1) = 4/2 = 2`.
Intensity is proportional to the square of amplitude, hence we can write the following :
`implies (sqrt(A_1^2))/(sqrt(A_2^2)) = 2`
`implies (A_1)/(A_2) = 2`.
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