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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)/(sqrt alpha-1)`
` (c) I_1/I_2 = (A_1/A_2)^2 = (((sqrt(alpha))+1)/((sqrt alpha)-1)) ^2 ` .
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