Question

A symmetrical bi-concave thin lens is cut into two identical halves and arranged as shown.


What will be the effective focal length of the arrangement ?
[Given, $\mu =$ refracting index of material and $R=$ radius of curvature]

• A. $\frac{-2R}{(\mu -1)}$
• B. $\frac{-R}{2(\mu -1)}$
• C. $\frac{R}{(\mu -1)}$
• D. $\frac{2R}{(\mu -1)}$

Answer 1

The correct option is B $\frac{-R}{2(\mu -1)}$

Given, the radius of curvature is $R$.

Let focal length of $a=f_1$ and of $b=f_2$ respectively.

$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}\dots \dots \left(1\right)$

For part $a$,
from lens maker's formula,

$\frac{1}{f_1}=(\mu -1)(\frac{1}{\infty }-\frac{1}{R})$

$\frac{1}{f_1}=\frac{-(\mu -1)}{R}\dots \dots \left(2\right)$

Similarly, for part $b$,

$\frac{1}{f_2}=(\mu -1)(\frac{-1}{R}-\frac{1}{\infty })$

$\frac{1}{f_2}=\frac{-(\mu -1)}{R}\dots \dots \left(3\right)$

From Eq.(1), (2) and (3)

$\frac{1}{f_{eq}}=\frac{-(\mu -1)}{R}-\frac{(\mu -1)}{R}=\frac{-2(\mu -1)}{R}$

$\Rightarrow f_{eq}=\frac{-R}{2(\mu -1)}$

Hence, option $\left(b\right)$ is correct.