Question

A narrow monochromatic beam of light of intensity $I$ 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 the remaining. Find the ratio $\sqrt{\frac{I_{max}}{I_{min}}}$ the interference pattern formed by two beams obtained after one reflection at each plate.

Answer 1

If the incident intensity is $I$,

Intensity of ray bc=$\frac{I}{4}$

And intensity of ray bd=$\frac{3I}{4}$

Due to reflection from glass plate 2, intensity of ray de=$\frac{3I}{4}\times \frac{1}{4}=\frac{3I}{16}$

Due to refraction from glass plate 1 again,
Intensity of ray ef=$\frac{3I}{16}\times \frac{3}{4}=\frac{9I}{64}$

Rays bc and ef form the interference pattern.

For rays of intensity $I_1,I_2$ interfering,

$I_{max}=(\sqrt{I_1}+\sqrt{I_2}{)}^2$

$I_{min}=(\sqrt{I_1}-\sqrt{I_2}{)}^2$

Here, $I_1=\frac{I}{4}$, $I_2=\frac{9I}{64}$

Hence $\sqrt{\frac{I_{max}}{I_{min}}}=7$