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

Options

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. `(8)/(5)`

Answer

A

Explanation

The intensity of light reflected from upper surface is
`I_(0)=I_(0)xx25%=I_(0)xx(25)/(100)=(I_(0))/(4)`
The intensity of transmitted light from upper surface is
`I=I_(0)-(I_(0))/(4)=(3I_(0))/(4)`
`therefore` The intensity of reflected light from lower surface is
`I_(2)=(3I_(0))/(4)xx(50)/(100)=(3I_(0))/(8)`
`therefore (I_("max"))/(I_("min")) =((sqrt(I_(1))+sqrt(I_(2)))^(2))/((sqrt(I_(1))-sqrt(I_(2)))^(2))=(((1)/(2)+sqrt((3)/(8)))^(2))/(((1)/(2)-sqrt((3)/(8)))^(2))`