Question

An equi-convex lens $(\mu =\frac{3}{2})$ has radius of curvature of magnitude $20\text{cm}$. Which one of the following options best describe the image formed of an object of height $2\text{cm}$ placed $30\text{cm}$ from the lens?

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

Answer 1

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

Given,
Magnitude of radius of curvature, $R=20\text{cm}$
Height of object, $h_0=2\text{cm}$
Object distance , $u=-30\text{cm}$

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

We know that, for the equi-convex lens, $R_1=-R_2$
So, $R_1=20\text{cm}$ and $R_2=-20\text{cm}$
Substituting the values,

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

$\Rightarrow \frac{1}{f}=\left(\frac{1}{2}\right)\times \frac{2}{20}$

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

Applying lens formula, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

We have,

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

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

$\Rightarrow \frac{1}{v}=\frac{10}{600}$

$\therefore v=60\text{cm}$

Now applying magnification formula,

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

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

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

Since, the magnification is negative, we can conclude that the image is inverted.

Hence, option $\left(c\right)$ is correct answer.