Question

The length of a vertical pole above the surface of a lake of water $(\mu =\frac{4}{3})$ is $24\text{cm}$. Then for a fish just below the water surface, the tip of the pole appears to be

• A. $18\text{cm}$ above the surface of water.
• B. $24\text{cm}$ above the surface of water.
• C. $30\text{cm}$ above the surface of water
• D. $32\text{cm}$ above the surface of water

Answer 1

The correct option is D $32\text{cm}$ above the surface of water

We know that,

${}_1{\mu }_2=\frac{{\mu }_2}{{\mu }_1}=\frac{\text{Real height}}{\text{Apparent height}}$

where ${\mu }_2$ is the refractive index of the medium in which the object is kept and ${\mu }_1$ is the refractive index of the medium in which the observer is present.
Here, ${\mu }_1$ is the refractive index of water and ${\mu }_2$ is the refractive index of air.

$\Rightarrow \frac{1}{4/3}=\frac{24}{\text{Apparent height}}$

$\Rightarrow$ Apparent height $=32\text{cm}$

$\therefore$ option $\left(d\right)$ is correct.

Why this question? The intent is to make you think about the bending of the light ray during refraction and its impact on image formation.