Question

A convex lens is in contact with a concave lens. The magnitude of the ratio of their focal lengths is $\frac{2}{3}$. Their equivalent focal length is $30\text{cm}.$ What are their individual focal length? (in $\text{cm}$)

• A. $-15,10$
• B. $-10,15$
• C. $75,50$
• D. $-75,50$

Answer 1

The correct option is A $-15,10$

$\text{Let,}{f}_1=f=\text{Focal length of convex lens}{f}_2=\text{Focal length of concave lens}$

Given, $\left|\begin{array}{c}\frac{f_1}{f_2}\end{array}\right|=\frac{2}{3}$

$\Rightarrow \frac{1}{f_2}=\frac{2}{3f}$

$\text{Equivalent focal length of the combination}{f}_{eq}=30\text{cm}$

$\because \frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}$

$\Rightarrow \frac{1}{30}=\frac{1}{f}-\frac{2}{3f}$

$\Rightarrow \frac{1}{30}=\frac{1}{3f}$

$\Rightarrow f=10\text{cm}=\text{Focal length of convex lens}$

$\therefore \text{Focal length of concave lens}=\frac{-3f}{2}=-15\text{cm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.