A spherical air lens of radii $R_1=R_2=10 cm$ is cut from a glass $(μ=1.5)$ cylinder as shown in figure. Its focal length is $f_1$. If a liquid of refractive index 2 is filled in the space then the focal length of liquid lens becomes $f_2$. Calculate $f_1$ and $f_2$. Choose the correct options from the following.

$f_1=-15 cm, f_2= +15cm$

The correct answer is Option (3) → $f_1=-15 cm, f_2= +15cm$

Using len's maker formula,

$\frac{1}{f}=(μ-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$⇒\frac{1}{f}=(\frac{1}{1.5}-1)\left[\frac{1}{10}\left(\frac{1}{-10}\right)\right]$

$⇒\frac{1}{f}=\left(-\frac{1}{3}\right)\left[\frac{2}{10}\right]$

$⇒f=-\frac{30}{2}=-15cm$

Case 2:

$μ=\frac{μ_1}{μ_m}=\frac{2}{1.5}=\frac{4}{3}$

$⇒\frac{1}{f}=(μ-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$

$=(\frac{4}{3}-1)\left[\frac{1}{10}-\left(\frac{1}{-10}\right)\right]$

$=\left(\frac{1}{3}\right)×\frac{2}{10}$

$⇒f=15cm$