A narrow monochromatic beam of light of intensity I is incident on a glass plate labeled 1 as shown in figure (not drawn to scale). Another identical glass plate labeled 2 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 $\sqrt{\frac{I_{\max}}{I_{\min}}}$ of the interference pattern formed by two beams obtained after one reflection at each plate. ( $I_{\max}$ and $I_{m i n}$ represent the maximum value and minimum value of intensity respectively. Ignore the distance between plates 1 and 2 in the diagram. It is drawn for the sake of readability. Consider only the first reflection from the glass plates)

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Detailed Solution

I is the intensity of incident beam ab. The interfering waves are bc and ef, reflected from surface of I and II plate, respectively. Reflection coefficient of intensity, $r = 25 % = 0.25$

Transmission coefficient of intensity, $t = 75 % = 0.25$

The intensity of beam, $b c, I_{1} = 0.25 I = \frac{1}{4} I$

The intensity of beam, $b d = 0.75 I$

The intensity of beam, $d e = 0.25 \times 0.75 I$

The intensity of beam ef, $I_{2} = 0.75 \times 0.25 \times 0.75 I = \frac{9}{64} I$

Ratio of maximum and minimum intensities. $\frac{\sqrt{I_{\max}}}{\sqrt{I_{\min}}} = \frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}} = 7$