A narrow monochromatic beam of light of intensity 1 is incident on a glass plate as shown in figure Another identical glass plate is kept close to the first one and parallel to it. Each glass plate reflects `25%` of the light incident on it and transmits intensities in the interference pattern formed by two beams obtained after one reflection at each plate.

Explanation

Let `I` be the intensity of beam no. `1` incident on first glass plate. Fig. It is partially reflected and partially transmitted. As each plate reflects `25%` of light falling on it and transmits `75%`. therefore, as is clear from Fig.
`I_(2) = (25)/(100)I = (I)/(4)` `I_(3) = (75)/(100)I = (3)/(4)I`
`I_(4) = (25)/(100)I_(3) = (1)/(4) xx (3)/(4) = (3)/(16)I` `I_(5) = (75)/(100)I_(4) = (3)/(4) xx (3)/(16)I = (9)/(64)I`
If `a` and `b` are amplitude of beam `I_(2)` and `I_(5)`.
when `I_(2) = (I)/(4) = Ka^(2)` or `a = sqrt((I)/(4 K)) = (1)/(2) sqrt((I)/(K))`
`I_(5) = (9)/(64)I = Kb^(2)` or `b = sqrt((9 I)/(64 K)) = (3)/(8) sqrt((1)/(K))`
`(I_(min))/(I_(max)) = ((a - b)^(2))/(a + b)^(2) = (((1)/(2) - (3)/(8))^(2) I//K)/(((1)/(2) + (3)/(8))^(2) I//K) = ((1)/(8))^(2)/((7)/(8))^(2) = (1)/(49)`