Question

Liquid of refractive index $1.62$ is placed between the plano-convex lenses (identical) whose material is of refractive index $1.54$. Two possible arrangements $\text{P}$ and $\text{Q}$ are shown below. The system is:

• A. Divergent in $\text{P}$, convergent in $\text{Q}$
• B. Convergent in $\text{P}$, divergent in $\text{Q}$
• C. Convergent in both
• D. Divergent in both

Answer 1

The correct option is A Divergent in $\text{P}$, convergent in $\text{Q}$

If two or more lenses are in contact, the equivalent focal length is given by the formula
$$\frac{1}{f}=\sum _{i=1}^n\frac{1}{f_i}$$
The focal length of $\text{P}$ is ,
$\frac{1}{f}=\frac{1}{f_{lens}}+\frac{1}{f_{lens}}+\frac{1}{f_{liquid}}=\frac{2}{f_{lens}}+\frac{1}{f_{liquid}}$

Focal length of plano-convex lens $f_{lens}=\frac{R}{\mu -1}$

$\therefore \frac{1}{f}=\frac{2(\mu -1)}{R}+\frac{1}{f_{\text{liquid}}}$

Using lens makers' formula,

$\frac{1}{f_{\text{liquid}}}=(1.62-1)(\frac{1}{-R}-\frac{1}{R})=0.62\left(\frac{-2}{R}\right)$

$\therefore \frac{1}{f}=\frac{2(1.54-1)}{R}-\frac{1.24}{R}$

$=\frac{1.08}{R}-\frac{1.24}{R}=\frac{-0.16}{R}$

It is clear that $f$ is negative.

i.e. $\text{P}$ is divergent in nature.

Focal length of combination as shown in $\text{Q}$ is given by

$\frac{1}{f}=\frac{2}{f_{\text{lens}}}+\frac{1}{f_{\text{liquid}}}$

Here, $\frac{1}{f_{\text{liquid}}}=(1.62-1)(\frac{1}{-R}-\frac{1}{\infty })=\frac{-0.62}{R}$

$\therefore \frac{1}{f}=\frac{2(1.54-1)}{R}-\frac{0.62}{R}$

$=\frac{1.08}{R}-\frac{0.62}{R}=\frac{0.46}{R}$

It is clear that $f$ is positive.

i.e. $\text{Q}$ is convergent in nature.

Hence, option (a) is the correct answer.