Question

A narrow monochromatic beam of light of intensity $I$ is incident on a glass plate as shown in the 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. The ratio $\sqrt{\frac{I_{\max }}{I_{\min }}}$ for the interference pattern formed by two beams obtained after one reflection at each plate is.

Answer 1

$I$ is the intensity of incident beam $ab$.
Let the intensities of interfering waves $bc$ and $ef$ be $I_1$ and $I_2$ respectively.
Intensity reflection coefficient will be $r=25\%=0.25$
Intensity transmission coefficient will be
$t=75\%=0.75$
The intensity of beam $bc,I_1=0.25I=\frac{1}{4}I$
The intensity of beam $bd=0.75I.$
The intensity of beam $de=0.25\times 0.75I$
$\therefore I_2=0.75\times 0.25\times 0.75I=\frac{9}{64}I$
Ratio of maximum and mininum amplitudes is given by
$\frac{\sqrt{I_{max}}}{\sqrt{I_{min}}}=\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}=7$