A fish rising vertically up towards the surface of water with speed $3{\mathrm{ms}}^{-1}$ and a bird diving vertically downwards with speed $9{\mathrm{ms}}^{-1}$. The velocity of bird with respect to fish is :

Let the actual height of bird from surface of water be y and actual distance of fish from surface be y'.
Let the apparent height of bird as seen by fish be x.
Refractive index of water,

$\mu =\frac{x}{y}$

$x=\mu y$

Distance between fish and bird is given as

$h=x+y^{\mathrm{\prime }}$

$h=\mu y+y^{\mathrm{\prime }}$

$\frac{\mathrm{d}h}{\mathrm{d}t}=\mu \frac{\mathrm{d}y}{\mathrm{d}t}+\frac{\mathrm{d}{y}^{\mathrm{\prime }}}{\mathrm{d}t}$

$\frac{\mathrm{d}h}{\mathrm{d}t}=\frac{4}{3}\times 9+3=15{\mathrm{ms}}^{-1}$

Hence, relative velocity of bird with respect to fish is $15{\mathrm{ms}}^{-1}\text{. }$