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?

• 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 $\frac{-R}{2(\mu -1)}$

Let the radius of curvature of each surface be $R$.

Let the focal length of $\text{a}$ and $\text{b}$ be $f_1$ and $f_2$ respectively.

Let the focal lenght of the new arrangement be $f_{new}$ which can be written as,

$\frac{1}{{\left(f_e\right)}_{new}}=\frac{1}{f_1}+\frac{1}{f_2}...\left(1\right)$

Applying lens maker's formula formula we get for part $\text{a}$ in new arrangement,

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

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

Applying lens maker's formula formula, we get for part $\text{b}$ in new arrangement,

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

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

Adding $\left(2\right)$ and $\left(3\right)$,

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

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

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

Hence, option (b) is the correct answer.