Question

At what distance eye 'E' will observe the fourth image (after four refractions from plane surfaces) of object 'O' from itself?

Answer 1

The fourth image will be formed when the incident ray emerging from object 'O' suffers the refraction at 4 interfaces.
Here, 2 different mediums are present : ${\mu }_1=1.5$ and ${\mu }_2=2$

$\Rightarrow$ shift produced in the direction of incident ray from 'O' will be,

$S_{total}=S_1+S_2$

$\Rightarrow S_{total}=t_1(1-\frac{1}{{\mu }_1})+t_2(1-\frac{1}{{\mu }_2})$

or
$S_{total}=10(1-\frac{1}{3/2})+10(1-\frac{1}{2})$

$\Rightarrow S_{total}=\frac{10}{3}+5=\frac{25}{3}\text{cm}$

Initially the distance between object (O) & eye (E) is,

$d=10\times 5=50\text{cm}$

$\Rightarrow$ Fourth image will form at a distance d' from 'E',

$\therefore d^{\prime }=d-S_{total}=50-\frac{25}{3}=\frac{125}{3}\text{cm}$

Hence, option (c) is the correct answer.

Why this question? Caution: Here observer is in air and two different media are also separated by $10\text{cm}$ of air, thus shift will not be produced by $10\text{cm}$ layer of air.