Question

A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm from each other. A point object placed 40 cm in front of the convex lens. Find the position of the image formed by this combination.

Answer 1

For convex lens,

Focal length, $f=+20cm$

Object distance, $u=-40cm$

Using lens formula, we have

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{20}=\frac{1}{v}-\frac{1}{-40}$

$\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{40}=\frac{2-1}{40}=\frac{1}{40}$

$\Rightarrow v=+40cm$

Now, for convex mirror, we have

Focal length, $f=\frac{R}{2}=\frac{20}{2}=+10cm$

Object distance, $u=+(40-15)cm=+25cm$

Using mirror formula, we have

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\frac{1}{10}=\frac{1}{v}+\frac{1}{25}$

$\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{25}=\frac{5-2}{50}=\frac{3}{50}$

$\Rightarrow v=+\frac{50}{3}cm$

$\therefore$ The final image is formed at a distance of $\frac{50}{3}cm$ from the convex mirror or at a distance of $(\frac{50}{3}+15)cm=\frac{95}{3}cm$ from the convex lens.

The final image is a virtual image.