Question

Focal length of a convex lens $(\mu = 3/2)$ is $24 \,cm$ in air. When it is immersed in water $(\mu = 4/3)$, its focal length will be

• A. $24\,cm$
• B. $48\,cm$
• C. $36\,cm$
• D. $96\,cm$

Answer 1

Answer: (D)

According to lens maker's formula
$\frac{1}{f}=\left(\mu-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ with $\mu=\frac{\mu_{L}}{\mu_{M}}$
$\therefore \frac{1}{fair}=\left(\frac{\left(3 /2\right)}{1}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \ldots\left(i\right)$
and $\frac{1}{f_{water}}=\left(\frac{\left(3 /2\right)}{\left(4/ 3\right)}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\ldots\left(ii\right)$
Dividing $\left(i\right)$ by $\left(ii\right)$, we get
$\frac{f_{water}}{f_{air}}=\frac{\left[\left(3/ 2\right)-1\right]}{\left[\frac{\left(3 /2\right)}{4 /3}-1\right]}=\frac{\left(1 /2\right)}{\left(1 /8\right)}=4$
$f_{water}=4 f_{air}=4\times24\, cm=96\,cm$