Question

Consider the situation as shown in figure. A plane mirror is fixed at a height $h$ above the bottom of a beaker containing water (refractive index $\mu$ ) up to a height $d$. Find the position of the image of the bottom formed by the mirror.

• A. $h-d+\frac{d}{\mu }$
• B. $h+d+\frac{d}{\mu }$
• C. $h-d-\frac{d}{\mu }$
• D. $d+\frac{d}{\mu }$

Answer 1

The correct option is A $h-d+\frac{d}{\mu }$

The bottom of the beaker appears to be shifted up by a distance
$△t=(1-\frac{1}{\mu })d$

Thus, the apparent distance of the bottom from the mirror is $h-△t=h-(1-\frac{1}{\mu })d=h-d+\frac{d}{\mu }.$ The image is formed behind the mirror at a distance $h-d+\frac{d}{\mu }.$