Question

A pin of length 2.00 cm is placed perpendicular to the principal axis of a converging lens. An inverted image of size 1.00 cm is formed at a distance of 40.0 cm from the pin. Find the focal length of the lens and its distance from the pin.

Answer 1

Given $h_0=2\text{cm}$, $h_1=-2$

$\frac{h_i}{h_0}=\frac{v}{u}$, $\frac{-2}{2}=\frac{-v}{u}$

$u=2v$

Also gives,

v + u = 40 cm, 3v = 40 cm

$v=13.3=\frac{40}{3}\text{cm}$, $u=\frac{80}{3}$

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{3}{40}-\frac{3}{(-80)}=\frac{1}{f}$

$\Rightarrow \frac{1}{f}=\frac{3}{40}[1+\frac{1}{2}]=\frac{3}{40}\times \frac{3}{2}$

$f=\frac{80}{9}=8.9\text{cm}$

$f=8.9\text{cm}$, $u=26.66\text{cm}$