Question

If a concave lens of focal length $20\text{cm}$ is kept at a distance of $5\text{cm}$ from a convex lens of focal length $10\text{cm}$. The equivalent focal length and nature of this combination is :

• A. $\frac{10}{3}\text{cm},$ concave
• B. $\frac{10}{3}\text{cm},$ convex
• C. $\frac{40}{3}\text{cm},$ convex
• D. $\frac{40}{3}\text{cm},$ concave

Answer 1

The correct option is C $\frac{40}{3}\text{cm},$ convex

Given :
Focal length of concave lens $\left(f_1\right)=-20\text{cm}$
(According to the sign convention)

Focal length of convex lens is $\left(f_2\right)=10\text{cm}$

Distance between the lenses $\left(d\right)=5\text{cm}$

The equivalent focal length of the combination is given by :

$\frac{1}{f_E}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1{f}_2}$

From the given data :

$\frac{1}{f_E}=\frac{1}{-20}+\frac{1}{10}-\frac{5}{-20\times 10}$

$\Rightarrow \frac{1}{f_E}=\frac{3}{40}$

Or, $f_E=\frac{40}{3}\text{cm}$

Hence, option $\left(c\right)$ is the correct answer.