Two waves of equal frequencies have their amplitudes in the
ratio of 3:5. They are superimposed on each other. Calculate the ratio of
maximum and minimum intensities of the resultant wave.

Answer

A

Explanation

Given, `A_1/A_2 = 3/5 `
` :. sqrt(I_1/I_2) = 3/5` (as `I prop A^2`)
Maximum intensity is obtained, where
` cos phi = 1`
and `I_(max) = (sqrtI_1 + sqrtI_2)^2`

Minimum intensity is found, where
` cos phi = -1`
and `I_(min) = (sqrtI_1 - sqrtI_2)^2`
Hence, `I_(max)/I_(min) = ((sqrt I_1+sqrtI_2)/(sqrt I_1 -sqrt I_2))^2 = ((sqrt (I_1/I_2)+1)/(sqrt((I_1/I_2))-1))`
` = ((3/5 +1)/(3/5-1))^2 = 64/4 = 16/1`.