Question

An object and a screen are mounted on an optical bench and a converging lens is placed between then so that a sharp image is obtained on the screen. The linear magnification of the image is $25$. The lens is now moved $30cm$ towards the screen and a sharp image is again formed on the screen. Find the focal length of the lens.

• A. $14.0cm$
• B. $14.3cm$
• C. $14.6cm$
• D. $14.9cm$

Answer 1

The correct option is B $14.3cm$

$\begin{array}{cc}& \text{Cose - I}\\ & \frac{1}{f}=\frac{1}{v}-\frac{1}{-u}\\ & \text{magnification}=25=\frac{v}{u}\Rightarrow [v=25u]-\left(1\right)\\ & \frac{1}{f}=\frac{1}{25u}+\frac{1}{u}\Rightarrow \frac{u}{f}=\frac{26}{25}\Rightarrow u=\frac{26}{25}\\ & \mathrm{case}-2\end{array}$

$\begin{array}{c}\frac{1}{f}=\frac{1}{v-30}-\frac{1}{-(u+30)}\\ \frac{1}{f}=\frac{1}{25u-30}+\frac{1}{u+30}(B_y\text{(1)})\\ \text{substitute (2)}\\ \begin{array}{c}\frac{1}{f}=\frac{1}{25\left(\frac{26}{25}\right)f-30}+\frac{1}{\frac{26f}{25}+30}\\ \frac{1}{f}=\frac{1}{26f-30}+\frac{25}{26f+750}\\ \text{solving we get}f=14.3cm\end{array}\end{array}$

So option (B) is correct