A fish F, in the pond is at a depth of $0.8m$ from the water surface and is moving vertically upward 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 downward with velocity $3m{s}^{-1}$. At this instant, both are on the same vertical line as shown in the figure. Which of the following statement(s) are correct?

The correct option is C Height of B, observed by F (from itself), is equal to 8.80 m
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, as observed by the bird will be equal to $r_1=x+\frac{y}{\mu }$ (i)
where $\mu$ is refractive index of water with respect to air.
$r_1=6.60m$
Hence, option (b) is correct.
Height of bird, as observed by the fish will be equal to
$r_2=\frac{x}{{\mu }^{\prime }}+y$ (ii)
Where ${\mu }^{\prime }$ is refractive index of air with respect to water.
Now, ${\mu }^{\prime }=\frac{1}{\mu }=\frac{3}{4}$
Hence, $r_2=8.80m$
Hence, option (a) is wrong while (c) is correct .