Question

A biconvex lens has a radius of curvature of magnitude $20\text{cm}$ and the refractive index of the lens is $1.5$. Which of the following option 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 $=1\text{cm}$
• C. Virtual, upright, height $=0.5\text{cm}$
• D. Real, inverted, height $=4\text{cm}$

Answer 1

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

Sign convention: Direction along incident ray is taken as +ve.
For the given biconvex lens,

Distance of object from lens, $u=-30\text{cm}$
$R_1=+20\text{cm},R_2=-20\text{cm}$
Hence, by using lens maker's formula
$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\Rightarrow \frac{1}{f}=(\frac{{\mu }_g}{1}-1)(\frac{1}{20}-\frac{1}{(-20)})$
or, $\frac{1}{f}=(1.5-1)\left(\frac{2}{20}\right)$
$\therefore f=+20\text{cm}$
We know,
$m=\frac{f}{f+u}$
$\Rightarrow m=\frac{20}{20-30}=-2$
Also, $m=\frac{\text{Height of image}\left(h_i\right)}{\text{Height of object}\left(h_o\right)}$
$\Rightarrow \frac{h_i}{(+2\text{cm})}=-2$
$\therefore h_i=-4\text{cm}$ (real and inverted)

$$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: Negative sign of height of image}\text{represents that image is real, inverted.}\text{Thus image has formed below principle}\text{axis.}\\ .\end{array}$$