Question

A concave lens has been placed in an optically rarer medium. The lens will behave like

• A. Converging lens
• B. Diverging lens
• C. It will loose property of lens
• D. Concave mirror

Answer 1

The correct option is B Diverging lens

On applying lens maker's formula :

$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})......\left(i\right)$

Since, surroundings is optical rarer medium.
${\mu }_2>{\mu }_1$


For concave lens,
Sign convention: Direction of incident ray taken as +ve.
$R_1=-r_1,R_2=+r_2$

$\Rightarrow (\frac{1}{R_1}-\frac{1}{R_2})=\frac{1}{-r_1}-\frac{1}{r_2}=-(\frac{1}{r_1}+\frac{1}{r_2})$

Thus, term $(\frac{1}{R_1}-\frac{1}{R_2})$ is negative.

$(\frac{{\mu }_2}{{\mu }_1}-1)$ is positive.

Hence from Eq. (i) $\frac{1}{f}$ comes to be -ve.

$\Rightarrow$ Negative focal length means, refracted ray appear to intersect.
$\therefore$ lens is diverging in behaviour.

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: +ve focal length indicates converging}\\ \text{behaviour because the refracted rays will}\\ \text{intersect along +ve - direction due to}\\ \text{convergence by lens}\end{array}$