Question

A thin equiconvex lens $(\mu =\frac{3}{2})$ of focal length $10cm$ is cut and separated and a material of refractive index $3$ is filled between them. What is the focal length of the combination?

• A. $-10cm$
• B. $\frac{-10}{3}cm$
• C. $\frac{-10}{4}cm$
• D. None of these

Answer 1

The correct option is B $\frac{-10}{3}cm$

The system can be seen as a combination of three lenses (assuming all three to be thin lenses i.e width negligible)


and $P_{net}=P_1+P_2+P_3$
$\Rightarrow \frac{1}{f_{net}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}$

When a biconvex lens is spliced longitudinally, the focal length of each of the resulting half is doubled.

i.e $f_1=f_3=20cm$

For lens 2 (biconcave),

Using lens makers' formula,

$\frac{1}{f_2}=\frac{{\mu }_2-{\mu }_1}{{\mu }_1}[\frac{1}{R_1}-\frac{1}{R_2}]$

$=\frac{3-1}{1}[\frac{-1}{R}-\frac{1}{R}]$
$=2\times \frac{-2}{R}$

The radius of curvature $R$ will be the same as that of the initial biconvex lens.

For the initial biconvex lens,

$\frac{1}{f}=\frac{\mu -1}{1}[\frac{1}{R}-\frac{-1}{R}]$

$\Rightarrow \frac{1}{10}=\frac{3/2-1}{1}\left[\frac{2}{R}\right]$

$\Rightarrow R=10cm$

Substituting this, we get,

$\frac{1}{f_2}=\frac{-4}{10}$

Therefore, $\frac{1}{f_{net}}=\frac{1}{20}+\frac{-4}{10}+\frac{1}{20}$
$=\frac{-6}{20}$
$\Rightarrow f_{net}=\frac{-10}{3}cm$