A fish $F$, in the pond is at a depth of $0.8m$ from the water surface and is moving vertically upwards with velocity $2m{s}^{-1}$. At the same instant a bird $B$ is at a height of $6m$ from the water surface and is moving downwards with velocity $3m{s}^{-1}$. At this instant both are on the same vertical as shown in the figure. Which of the following statements are correct?

If an object is at a distance $x$ from a plane refracting surface and is viewed normally then it appears at a distance $\frac{x}{\mu }$ from the surface where $\mu$ is refractive index of that medium (in which it is situated) with respect to the medium in which observer is situated. 

Suppose height of bird from water surface is $x$ and depth of fish from the surface is $y$ then depth of fish, observed by the bird, will be equal to 

$r_1=x+\frac{y}{\mu }$                                                                       …(1)

where $\mu$ is refractive index of water with respect to air. 

$\therefore r_1=6.60m$

Hence, option (B) is correct. 

Height of bird, observed by fish, will be equal to 

$r_2=\frac{x}{\mu \text{'}}+y$                                                                        …(2)

where $\mu \text{'}$is refractive index of air with respect to water. 

Now,                                             $\mu \text{'}=\frac{1}{\mu }=\frac{3}{4}$

Hence,                                            $r_2=8.80m$

Hence, option (A) is wrong while (C) is correct.