Question

The length of a vertical pole at 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 at

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

Answer 1

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

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.
$\Rightarrow \frac{1}{4/3}=\frac{24}{\text{Apparent height}}$
$\Rightarrow \text{apparent height}=32\text{cm}$

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.