Question

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

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

Answer 1

The correct option is C Real, inverted, height $=4\text{cm}$

Given, Radius of curvature, $R=20\text{cm}$

height, $h_1=2\text{cm}$

Object distance, $u=-30\text{cm}$

Using,

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

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

$\therefore f=20\text{cm}$

Now from lens formula,

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

$\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{30}=\frac{10}{600}$

$\therefore v=60$

As magnification, $m=-\frac{v}{u}=\frac{h_i}{h_0}$

$\Rightarrow -\frac{60}{30}=\frac{h_i}{2}$

$\therefore h_i=-4\text{cm}$

Hence, $\left(C\right)$ is the correct answer.