Question

An object is placed $15cm$ in front of a convex lens of focal length $10cm$ . Find the nature and position of the image formed. Where should a concave mirror of radius of curvature $20$ cm be placed so that the final image is formed at the position of the object itself?

Answer 1

Given: For convex lens $u=-15cm$ $f=10cm$

To find: $v$

Solution: From lens formula

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

$\frac{1}{v}+\frac{1}{15}=\frac{1}{10}$

$\frac{1}{v}=\frac{1}{30}$

$v=30cm$

And magnification is $=\frac{30}{-15}=-2$

So for image is inverted and magnified and when a mirror is placed at a distance d behind lens then u for mirror is $u^1=-30+\alpha$

$=\alpha -30$

and $v=-30-14+\alpha$

From mirror formula

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

$\frac{1}{\alpha -45}+\frac{1}{\alpha -30}=\frac{-1}{10}$

$\frac{\alpha -30+\alpha -45}{(\alpha -45)(\alpha -30)}=\frac{-1}{10}$

$\frac{2\alpha -75}{(\alpha -45)(\alpha -30)}=\frac{-1}{10}$

$20\alpha -750=-{\alpha }^2+75\alpha -1350$

${\alpha }^2=\frac{+}{-}\sqrt{(55{)}^2-4/100}2$

$\alpha =\frac{55+-25}{2}=40cm$ or $15cm$

Hence $d=15cm$(as 40 cm not possible)