A fish F, in the pond is at a depth of 0.8 m from the water surface and is moving vertically upward with velocity 2 ms^-1. At the same instant, a bird B is at a height of 6 m from the water surface and is moving downward with velocity 3 ms^-1. At this instant, both are on the same vertical line as shown in 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 $\mathrm{x}/\mathrm{\mu }$ from the surface, where $\mathrm{\mu }$ 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

           ${\mathrm{r}}_1=\mathrm{x}+\frac{\mathrm{y}}{\mathrm{\mu }}$               …(i)

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

           ${\mathrm{r}}_1=6.60 \mathrm{m}$

Hence, option (2) is correct.

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

           ${\mathrm{r}}_2=\frac{\mathrm{x}}{\mathrm{\mu }\text{'}}+\mathrm{y}$              …(ii)

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

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

Hence, ${\mathrm{r}}_2=8.80 \mathrm{m}$

Hence, option (1) is wrong while (3) is correct.