Question

A convex lens of focal length $25cm$ is cut into two equal parts to obtain two plano-convex lenses which are then joined as shown in Fig. 1. Focal length of this combination is

• A. $25cm$
• B. $12.5cm$
• C. $50cm$
• D. $100cm$

Answer 1

The correct option is A $25cm$

It is given that for a convex lens is $25cm$ then

By uns maker formula

$\frac{1}{f}=\frac{1}{n}\left[\frac{1}{\frac{1}{e}+\frac{1}{e}}\right]$

$\Rightarrow \frac{1}{f}=\frac{1}{2n}\left[\frac{1}{R}\right]$

$\Rightarrow \frac{2}{f}=\frac{1}{n}\left[\frac{1}{e}\right]$

For $lens-1$

$\frac{1}{f_1}=\frac{1}{n}\left[\frac{1}{\frac{1}{\infty }+\frac{1}{R}}\right]=\frac{1}{n}\left[\frac{1}{R}\right]=\frac{2}{f}$

Similarly for lens$-2$

$\frac{1}{f_2}=\frac{1}{n}\left[\frac{1}{\frac{1}{R}+\frac{1}{\infty }}\right]=\frac{1}{n}\left[\frac{1}{R}\right)=\frac{2}{f}$

For combination of lens$-1$ and lens $-2$

As, $P_{eff}=P_1+P_2$ [$P\to$ Power of lens]

$\Rightarrow \frac{1}{f{e}_{ff}}=\frac{1}{f_1}+\frac{1}{f_2}$

$\Rightarrow \frac{1}{f{e}_{ff}}=\frac{2}{f}+\frac{2}{f}$

$\Rightarrow \frac{1}{f{e}_{ff}}=\frac{1}{f}$

$feff=f=25cm$

Option $A$ is correct