Question

The image of the point $P$ as seen by the eye in the figure above $P$ at $0.11\times xcm$ then the value of $x$ is ?

Answer 1

Answer: 2.00

Apparent depth of $P$ as seen by observer
$=\left(\frac{t_1}{{\mu }_1}+\frac{t_2}{{\mu }_2}+\frac{t_3}{{\mu }_3}+\frac{t_4}{{\mu }_4}+\frac{t_5}{{\mu }_5}+\frac{t_6}{{\mu }_6}\right){\mu }_7$
$=\left(\frac{1}{1}+\frac{0.2}{1.2}+\frac{1}{1}+\frac{0.3}{1.3}+\frac{1}{1}+\frac{0.4}{1.4}\right)(1)$
$=\left(3+\frac{1}{6}+\frac{3}{13}+\frac{2}{7}\right)=3.68cm$
So image of $P$ is formed at a distance $3.68cm$ below the top of slab of thickness $0.4cm$ or at a distance $(3.9-3.68)=0.22cm$ above $P$.