Question

An object is placed $50cm$ from a screen as shown.
A converging lens is moved such that line $MN$ is its principal axis. Sharp images are formed on the screen in two positions of lens separated by $30cm$. Find the focal length of the lens in cm.

Answer 1

Let an image be formed at the screen when the lens is at a distance of x from object. Then applying Lens equation gives,

$\frac{1}{50-x}-\frac{1}{-x}=\frac{1}{f}$

$x^2-50x+50f=0$

Solution for this quadratic equation is,

$x_1=\frac{50+D}{2}$ and $x_2=\frac{50-D}{2}$

As given, $x_1-x_2=50$

Hence $D=\sqrt{{50}^2-200f}=30$

$f=8cm$