A fish is rising up vertically inside a pond with velocity 4 cms^–1 and notices a bird, which is diving vertically downward and its velocity appears to be 16 cms^–1 (to the fish). What is the actual velocity of the diving bird, in cms^–1, if refractive index of water is 4/3.

Let at some instant bird is at a height of x from the water surface and it is diving downwards with $\mathrm{v} {\mathrm{cms}}^{-1}$.
At this instant fish is at a depth y below water surface. Then the distance between fish and image of bird at this instance will be, 

$\mathrm{s}=\mathrm{\mu x}+\mathrm{y}\Rightarrow \mathrm{s}=\frac{4}{3}\mathrm{x}+\mathrm{y}$

Differentiating w.r.t time, we get,

$\left(-\frac{\mathrm{ds}}{\mathrm{dt}}\right)=\frac{4}{3}\left(-\frac{\mathrm{dx}}{\mathrm{dt}}\right)+\left(-\frac{\mathrm{dy}}{\mathrm{dt}}\right)$

$\Rightarrow 16=\frac{4}{3}\left(\mathrm{v}\right)+\mathrm{u}$ 

$\Rightarrow \mathrm{v}=9 {\mathrm{cms}}^{-1}$