A container is filled with water ($\mu = 4/3$) up to a height of 24 cm. A concave mirror is placed 10 cm above the water level, and the image (I) of an object placed at the bottom is formed 20 cm below the water level, as shown in Figure. The focal length of the mirror is nearly

Let the apparent position of the object $O$ as seen by the mirror be $O'$.

Apparent depth of object $O$ from the water surface is $d_{ap} = \dfrac{d_{real}}{\mu} = \dfrac{24}{4/3} = 18 \text{ cm}$.

The distance of $O'$ from the concave mirror is $u = - (10 + 18) = -28 \text{ cm}$.

Let the apparent position of the image $I$ as formed by the mirror, seen from the water surface, be $I'$.

The image $I$ is formed $20 \text{ cm}$ below the water level.

Apparent depth of image $I'$ from the water surface is $d'_{ap} = \dfrac{20}{4/3} = 15 \text{ cm}$.

The distance of $I'$ from the concave mirror is $v = - (10 + 15) = -25 \text{ cm}$.

Using the mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{1}{-25} + \dfrac{1}{-28} = \dfrac{1}{f}$

$\dfrac{1}{f} = -\left(\dfrac{28 + 25}{25 \times 28}\right) = -\dfrac{53}{700}$

$f = -\dfrac{700}{53} \approx -13.2 \text{ cm}$

Answer: $-13.2\ cm$