Question

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

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

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 $\mu =\frac{{\mu }_L}{{\mu }_M}$
$\therefore \frac{1}{fair}=\left(\frac{\left(3/2\right)}{1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\dots \left(i\right)$
and $\frac{1}{f_{water}}=\left(\frac{\left(3/2\right)}{\left(4/3\right)}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\dots \left(ii\right)$
Dividing $\left(i\right)$ by $\left(ii\right)$, we get
$\frac{f_{water}}{f_{air}}=\frac{\left[\left(3/2\right)-1\right]}{\left[\frac{\left(3/2\right)}{4/3}-1\right]}=\frac{\left(1/2\right)}{\left(1/8\right)}=4$
$f_{water}=4f_{air}=4\times 24cm=96cm$