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 received on the screen. The linear magnification of the image is 2.5. The lens is now moved 30 cm nearer to the screen and a sharp image is again formed on the screen. The focal length of the lens is:

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

Answer 1

The correct option is B $14.3cm$

Given $\frac{v}{u}=2.5$
$v=2.5u$
again $v-u=30$
$v=30+u$
$2.5u=30+u$
$1.5u=30$
$u=20$
Now, $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}=\frac{u+v}{uv}$
$f=\frac{uv}{u+v}=\frac{2.5u^2}{3.5u}$
$=\frac{5}{7}u=\frac{5\times 20}{7}=\frac{100}{7}=14.3cm$