A concave mirror of focal length $f=-20\text{ cm}$ is placed at the bottom of a tank filled with water ($\mu =4/3$) up to a height $h=20\text{ cm}$. Find the position of the image of an object placed at the water surface.

Solution

1. The apparent depth of the object at the surface as seen by the mirror is $$\begin{array}{rl}{h}^{\prime }& =h/\mu \\ & =20/(4/3)\\ & =15\text{ cm}\end{array}$$.
2. The object distance for the mirror is $u=-15\text{ cm}$.
3. Using the mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$, we get $\frac{1}{v}+\frac{1}{-15}=\frac{1}{-20}$.
4. Solving for $v$ gives $$\begin{array}{rl}\frac{1}{v}& =\frac{1}{15}-\frac{1}{20}\\ & =\frac{4-3}{60}\\ & =\frac{1}{60}\end{array}$$, so $v=60\text{ cm}$ below the mirror.
5. The real depth is $60\times (4/3)=80\text{ cm}$ below the mirror, which is $80-20=60\text{ cm}$ below the surface, hence the answer is (D).