Q. A concave mirror is placed at the bottom of an empty tank with face upwards and axis vertical. When sunlight falls normally on the mirror, it is focussed at distance of $32\ cm$ from the mirror. If the tank filled with water $\left( \mu = \frac{4}{3} \right)$ upto a height of $20\ cm$, then the sunlight will now get focussed at
A) 16 $cm$ above water level
B) $9\ cm$ above water level
C) $24\ cm$ below water level
D) 9 $cm$ below water level
Explanation
(b) Sun is at infinity, $i.e.,$ $u = \infty$ so from mirror formula we have $\frac{1}{f} = \frac{1}{- 32} + \frac{1}{( - \infty)} \Rightarrow f = - 32\ cm$
When water is filled in the tank upto a height of $20\ cm$, the image formed by the mirror will act as virtual object for water surface. Which will form it’s image at $I$ such that $\frac{Actual\ height}{Apparent\ height} = \frac{\mu_{\omega}}{\mu_{a}},\ i.e.,\frac{BO}{BI} = \frac{4/3}{1}$
$\Rightarrow BI = BO \times \frac{3}{4} = 12 \times \frac{3}{4} = 9\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.