Question

A plano-convex lens has a thickness of $4cm$. When placed on a horizontal table, with the curved surface in contact with it, the apparent depth of the bottom-most point of the lens is found to be $3cm$. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of the plane face of the lens is found to be $25/8cm$. Find the focal length of the lens.

Answer 1

Use lens's maker formula

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

$\frac{1}{25/8}-\frac{4/3}{4}=\frac{1-4/3}{R}$

$\frac{8}{25}-\frac{1}{3}=\frac{-1}{3}\times \frac{1}{R}$

$\frac{24-25}{25\times 3}=\frac{-1}{3R}$

$\frac{-1}{75}=\frac{-1}{3R}$

$R=25cm$

${\mu }_2=1$
Apparent depth $=\frac{h}{\mu }$
$3cm=\frac{4cm}{\mu }$
$\mu =4/3$

Using lens maker formula

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\frac{1}{f}=\frac{(4-1)}{3}\left(\frac{1}{25}\right)=\frac{1}{3\times 25}=\frac{1}{75cm}$
$f=75cm$