Question

An object $3$ cm high is placed $12$ cm in front of a convex lens of focal length $8$ cm find the nature and size of image formed.

Answer 1

$u=-12cm$

$f=+8cm$

$ho=+3cm$

Using mirror formula,

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

$\frac{1}{v}+\frac{1}{-30}=\frac{1}{8}$

$\frac{1}{v}-\frac{1}{30}=\frac{1}{8}$

$\frac{1}{v}=\frac{1}{8}+\frac{1}{30}$

$\frac{1}{v}=\frac{3+2}{60}$

$\frac{1}{v}=\frac{5}{60}=\frac{1}{12}$

$v=+12cm$

Therefore, the image will be formed at a distance of $12cm$ behind the mirror.

$m=\frac{-v}{um}=\frac{-12}{-30m}=+0.4$

$m=\frac{h_i}{h_o}$

$0.4=\frac{h_i}{h_o}$

$0.4=\frac{h_i}{1}$

$h_i=+0.4cm$

Therefore, the image formed is diminished in size.