Question

Consider the situation shown in figure. A plane mirror is fixed at a height $h$ above the bottom of a beaker containing water $\left(\text{Referactive index =}\mu \right)$ up to a height $d$. The position of the image of the bottom of the beaker formed by the mirror will be:

Answer 1

The rays from the bottom of the beaker bends away from the normal during refraction at air-water interface.
Thus, for plane mirror the object (bottom of the beaker) will appear shifted above by $\Delta d$.
Here, $d_{real}=d$
So, $\mu =\frac{d_{real}}{d_{app}}$
$\therefore d_{app}=\frac{d}{\mu }$
or, $\Delta d=d-d_{app}=d-\frac{d}{\mu }$
The object distance for the plane mirror will be,
$u=h-\Delta d$
or, $u=h-(d-\frac{d}{\mu })$
$\Rightarrow u=h-d+\frac{d}{\mu }$
For plane mirror,
$|u|=|v|$
$\therefore$ image will be formed at $(h-d+\frac{d}{\mu })$ behind the mirror.

Why this question? Tips: Imagine that the mirror is an observer in the rarer medium (air), thus it will see an object i.e. bottom of the beaker at a closer distance.