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 of minimum and maximum intensities of the resultant beam obtained after one reflection at each plate is

• A. $19:1$
• B. $1:49$
• C. $1:16$
• D. $16:1$

Answer 1

The correct option is B $1:49$

At each surface, reflection and transmission takes place. Intensity of reflected and transmitted beam at first plate is $\frac{I}{4}\text{and}\frac{3I}{4}$ respectively.


Now, $25\%$ of $I$ is reflected and $75\%$ is transmitted. Intensity of reflected and transmitted beam at second plate is $\frac{3I}{16}\text{and}\frac{9I}{16}$ respectively.

The reflected beam from the second plate are incident at the first plate again. The reflected, and transmitted intensities of this beam are $\frac{3I}{64}$ and $\frac{9I}{64}$ respectively.

So, here the two interfering waves are having the intensities $\frac{I}{4}\text{and}\frac{9I}{64}$.

$I_{max}=(\sqrt{I_1}+\sqrt{I_2}{)}^2=I{[\frac{1}{2}+\frac{3}{8}]}^2=\frac{49I}{64}$

$I_{min}=(\sqrt{I_1}-\sqrt{I_2}{)}^2=I{[\frac{1}{2}-\frac{3}{8}]}^2=\frac{I}{64}$

So, $\frac{I_{min}}{I_{max}}=\frac{\frac{I}{64}}{\frac{49I}{69}}=\frac{1}{49}=1:49$

Hence, option $\left(\text{B}\right)$ is correct.