Question

A thin lens of glass $(\mu =1.5)$ of focal length $\pm 10cm$ is immersed in water $(\mu =1.33)$. The new focal length is

• A. $20cm$
• B. $40cm$
• C. $48cm$
• D. $12cm$

Answer 1

Answer: (B)

When lens is immersed in water,
From Lens maker's formula,
$\frac{1}{f_2}=\left({}^w{\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\Rightarrow \frac{1}{f_2}=\left(\frac{1.5}{1.33}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)...(i)$
When lens is in air, $\frac{1}{f_1}=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)...(iii)$
On dividing Eq. (ii) by Eq. (i), we get
$f_2=\left(\frac{0.5\times 1.33}{0.17}\right)\times f_1$
$=4f_1=4\times 10=40cm$