Question

A pin of length 2.0 cm lies along the principal axis of a converging lens, the centre being at a distance of 11 cm from the lens. The focal length of the lens is 6 cm. Find the size of the image.

Answer 1

$\frac{1}{v_A}+\frac{1}{{\mu }_A}=\frac{1}{f}$

$\Rightarrow \frac{1}{v_A}+\frac{1}{10}=\frac{1}{6}$

$\frac{1}{v_A}=\frac{1}{6}-\frac{1}{10}=\frac{1}{15}$

$v_A=15\text{cm}$

$\frac{1}{v_B}+\frac{1}{12}=\frac{1}{6}\Rightarrow v_B=62\text{cm}$

Length of image $=v_A-b_B=15-12=3\text{cm}$