Question

Suppose you are inside the water in a swimming pool near an edge. A friends is standing on the edge. Do you find your friend taller or shorter than his usual height?

Answer 1

When viewed from the water, the friend will seem taller than his usual height.

Let actual height be h and the apparent height be h'.
Here, the refraction is taking place from rarer to denser medium and a virtual image is formed.
Using
$\frac{{\mu }_1}{-u}+\frac{{\mu }_2}{v}=\frac{{\mu }_2-{\mu }_1}{R}$
Where refractive index of water is μ₂ and refractive index of air is μ₁.
u and v are object and image distances, respectively.
R is the radius of curvature, here we will take it as ∞.
$$\begin{gathered} \frac{{\mu }_1}{-u}+\frac{{\mu }_2}{v}=\frac{{\mu }_2-{\mu }_1}{\infty } \\ \frac{{\mu }_1}{u}=\frac{{\mu }_2}{v} \\ v=\frac{{\mu }_2}{{\mu }_1}\times u \end{gathered}$$
As μ₁ = 1
v = u × μ₂
We know magnification is given by:
$m=\frac{v}{u}$
Putting the value of v in the above equation:
$$\begin{gathered} m=\frac{u\times {\mu }_2}{u} \\ m={\mu }_2 \end{gathered}$$
As the magnification is greater than 1, so the apparent height seems to be greater than actual height.