Question

A biconvex lens has a radius of curvature of magnitude $20cm$. Which one of the following options best describe the image formed of an object of height $2cm$ placed $30cm$ from the lens?

• A. Virtual, upright, height $=1cm$
• B. Virtual, upright, height $=0.5cm$
• C. Real, inverted, height $=4cm$
• D. Real, inverted, height $=1cm$

Answer 1

The correct option is C Real, inverted, height $=4cm$

Given,

Radius of curvature, $R=20cm$

Height of object, $h_0=2$

Object distance, $u=30cm$

We have, $\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

$=(\frac{3}{2}-1)[\frac{1}{20}-(-\frac{1}{20})]$

$\Rightarrow \frac{1}{f}=(\frac{3}{2}-1)\times \frac{2}{20}$

$\therefore f=20cm$

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

$\Rightarrow \frac{1}{20}=\frac{1}{v}+\frac{1}{30}$

$\frac{1}{v}=\frac{1}{20}-\frac{1}{30}$

$=\frac{10}{600}$

$v=60cm$

$m=\frac{h_i}{h_0}=\frac{v}{u}$

$\Rightarrow h_i=\frac{v}{u}\times h_0$

$=\frac{60}{30}\times 2=-4cm$

so, image is inverted.