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In interference, I(max)/I(min) = alpha, ...

In interference, `I_(max)/I_(min) = alpha,` find.
`(a) A_(max)/A_(min) (b) A_1/A_2 (c) I_1/I_2`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
(a) `A_(max)/A_(min)=sqrt(I_(max)/I_(min)) = sqrt alpha`
(b) `A_(max)/A_(min) = sqrt alpha = ((A_1+A_2)/(A_1-A_2)) = ((A_1//A_2)+1)/((A_1//A_2)-1)`
Solving this equation, we get
`A_1/A_2 = (sqrt alpha+1)/(sqrtalpha-1)`
(c) `I_1/I_2 = (A_1/A_2)^2 ((sqrt alpha+1)/(sqrtalpha)-1)^2`.
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