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. The ratio of the minimum and the maximum intensities in the interference pattern formed by the two beams obtained after one reflection at each plate is $1:n$ . The value of $n$ will be

• A. $21$
• B. $49$
• C. $24$
• D. $64$

Answer 1

The correct option is B $49$

The diagram shown below represents the intensity of reflected and refracted light.



Hence, the two interfering waves $1\text{and}2$, have intensities, $\frac{I_0}{4}\text{and}\frac{9I_0}{64}$

So, $I_1=\frac{I_0}{4}\text{and}{I}_2=\frac{9I_0}{64}$

Now, $\Rightarrow \frac{I_{min}}{I_{max}}=\frac{{(\sqrt{\frac{I_0}{4}}-\sqrt{\frac{9I_0}{64}})}^2}{{(\sqrt{\frac{I_0}{4}}+\sqrt{\frac{9I_0}{64}})}^2}$

$=\frac{{(\frac{1}{2}-\frac{3}{8})}^2}{{(\frac{1}{2}+\frac{3}{8})}^2}=\frac{1}{49}$

Hence, $\left(B\right)$ is the correct answer.