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.
Answer & explanation
Correct answer: option 3
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:
$μ=\frac{μ_1}{μ_m}=\frac{2}{1.5}=\frac{4}{3}$
$⇒\frac{1}{f}=(μ-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$
$=(\frac{4}{3}-1)\left[\frac{1}{10}-\left(\frac{1}{-10}\right)\right]$
$=\left(\frac{1}{3}\right)×\frac{2}{10}$
$⇒f=15cm$