Physics - Geometrical Optics Question with Solution | TestHub

Let the apparent position of the object $O$ as seen by the mirror be $O^{\prime }$.

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

The distance of $O^{\prime }$ 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 $I^{\prime }$.

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

Apparent depth of image $I^{\prime }$ from the water surface is $d_{ap}^{\prime }=\frac{20}{4/3}=15\text{ cm}$.

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

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

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

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

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

Answer: $-13.2cm$