Question

A liquid of refractive index $1.33$ is placed between two identical plano-convex lenses, with refractive index $1.50$. Two possible arrangements, $\text{P}$ and $\text{Q}$ are shown. 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 C convergent in both

Given,
refractive index of lens, ${\mu }_1=1.5$
refractive index of liquid, ${\mu }_2=1.33$


For the arrangement $\text{P}$:
Let, the focal length of the combination is $f_P$. So, we can write,

$\frac{1}{f_P}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}.....\left(1\right)$

From Lens maker's formula for the plano-convex lens,
$\frac{1}{f_1}=({\mu }_1-1)(\frac{1}{R_1}-\frac{1}{R_2})...\left(2\right)$

Here, $R_1=R$ and for plane surface, $R_2=\infty$.

Substituting the values in $\left(2\right)$, we have
$\frac{1}{f_1}=(1.5-1)(\frac{1}{R}-\frac{1}{\infty })$

$\Rightarrow \frac{1}{f_1}=\frac{0.5}{R}$

Since both the plano-convex combination are identical, we can say that,

$\frac{1}{f_2}=\frac{0.5}{R}$

When the intervening medium is filled with liquid, then focal length of the concave lens formed by the liquid, by lens makers formula,

$\frac{1}{f_3}=(1.33-1)(\frac{1}{-R}-\frac{1}{R})$

$\Rightarrow \frac{1}{f_3}=\frac{0.66}{-R}$

Now, substituting the values of $f_1$, $f_2$ and $f_3$ in equation $\left(1\right)$, we get

$\frac{1}{f_P}=\frac{0.5}{R}+\frac{0.5}{R}+\frac{0.66}{-R}$

$\therefore \frac{1}{f_P}=\frac{0.34}{R}$

$\therefore f_P>0$

Similarly, for the arrangement $\text{Q}$:
We can write,
$\frac{1}{f_Q}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3^{\prime }}.......\left(2\right)$

Liquid medium forms a plano - concave lens. So,

$\frac{1}{f_3^{\prime }}=(1.33-1)(\frac{1}{-R}-\frac{1}{\infty })$

$\Rightarrow \frac{1}{f_3^{\prime }}=(1.33-1)\left(\frac{1}{-R}\right)$

Now, substituting the values of $f_1$, $f_2$ and $f_3^{\prime }$ in equation $\left(2\right)$, we get

$\frac{1}{f_Q}=\frac{1.5-1}{R}+\frac{1.33-1}{-R}+\frac{1.5-1}{R}$

$\Rightarrow \frac{1}{f_Q}=\frac{0.67}{R}$

$\therefore f_Q>0$

Since, both the focal lengths $f_P$ and $f_Q$ are positive, so combination behaves as convergent lens.

Hence, option (c) is the correct answer.