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A parallel beam of light of intensity I is incident on a glass plate. `25%` of light is reflected in any reflection by upper surface and `50%` of light is reflected by any reflection from lower surface. Rest is refracted The ratio of maximum to minimum intensity in interference region of reflected rays is

A

`((1/2+sqrt(3/8))/(1/2-sqrt(3/8)))^(2)`

B

`((1/4+sqrt(3/8))/(1/2-sqrt(3/8)))^(2)`

C

`5/8`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
D

`I_(1)=I/4 and I_(2)=(9I)/32`
`:. I_(1)/I_(2)=8/9`
or `sqrt(I_(1)/I_(2))=(2sqrt(2))/3`
`:. I_(max)/I_(min)=((sqrtI_(1)//I_(2)+1)/(sqrtI_(1)//I_(2)-1))^(2)=((2sqrt(2)+3)/(2sqrt(2)-3))^(2)`
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