A ray of light of intensity I is incident on a parallel glass-slab at the point A and it undergoes partial reflection and refraction as shown in the figure. At each reflection 25% of the incident energy is reflected and the rest is transmitted. If the rays AB and A'B' undergo interference, then the ratio `(I_("max"))/(I_("min"))` is

Answer

`81:1`

Explanation

As given reflection coefficient `=20%thereforeAB=(I)/(5)`
`A'B'=(16I)/(125)`
If AB and A'B' interference than
`I_(max)=(sqrt((I)/(5))+sqrt((16I)/(125)))^(2)=(I)/(5)xx(81)/(25)`
`I_(min)=(sqrt((I)/(5))-sqrt((16I)/(125)))^(2)=(I)/(5)xx(1)/(25)` ltbr. `thereforeI_(max)//I_(min)=81:1`