Question

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

• A. $1.2cm$
• B. $14.3cm$
• C. $14.6cm$
• D. $14.9cm$

Answer 1

Answer: (B)

$14.3cm$

Referring to figure, ilet 0 and $I$ represent objec and the screen respectively and $L_1$ and $L_2$ the and the screen respectively and $L_1$ two conjugate positions of the lens, then For lens at $L_1$, we have $u=-x$

$v=30+x$

Linear magnification, $m=\frac{-v}{u}=2.5$

$\text{Thus,}\left|\frac{-(30+k)}{c}\right|=|2.5|$

$\begin{array}{cc}+30+x& =2.5x\\ \frac{30}{1.5}& =x\\ \therefore x& =20cm\end{array}$

$u=-20cm,v=20+30=50cm$

Now, $\frac{1}{v}-\frac{1}{4}=\frac{1}{f}$

$\Rightarrow \frac{1}{s0}-\frac{1}{(-20)}=\frac{1}{f}$

$\Rightarrow \frac{20+50}{20\times 50}=\frac{1}{f}$

$\Rightarrow f\simeq 14.3cm$

Hence option (B) is correct