Thank you for reporting, we will resolve it shortly
Q.
A thin lens of glass $(\mu=1.5)$ of focal length $\pm 10 \,cm$ is immersed in water $(\mu=1.33)$. The new focal length is
Ray Optics and Optical Instruments
Solution:
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
$f_{2}=\left(\frac{0.5 \times 1.33}{0.17}\right) \times f_{1}$
$=4 f_{1}=4 \times 10=40\, cm$