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.

• A. $\frac{40}{3}cm$
  
• B. $\frac{20}{3}cm$
• C. $\frac{60}{3}cm$
• D. $\frac{80}{3}cm$

Answer 1

The correct option is D

$\frac{80}{3}cm$

$u+v=40$

$\Rightarrow v=40-u$

Now,

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

$\Rightarrow \frac{1}{40-u}-\frac{1}{u}=\frac{1}{f}\dots \dots \left(1\right)$

Also;

$\frac{v}{u}=\frac{1}{2}\Rightarrow \frac{u_0-u}{u}=\frac{1}{2}$

$\Rightarrow 80-2u=u$

$\Rightarrow 3u=80$

$\Rightarrow u=\frac{80}{3}$

Putting this value in eqn (1) we get,

$\frac{1}{\frac{u_0}{3}}-\frac{1}{\frac{80}{3}}=\frac{1}{f}$

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

$\Rightarrow f=\frac{80}{3}cm$