A fish F, in the pond is at a depth of 0.8m from the water surface and is moving vertically upward with velocity `2ms^(-1)`. At the same instant, a bird B is at a height of 6m from the water surface and is moving downward with velocity `3ms^(-1)`. At this instant. both are on the same vertical line as shown in figure . Which of the following statements are correct?

Options

A. Height of B, observed by F(from itself), is equal to 5.30m
B. Depth of F, Observed by B (from itselft), is equal to 6.60m
C. Height of B, observed by F (From itself), is equal to 8.80m
D. None of the above

Answer

b.,c.

Explanation

If an object is at a distance x from a plane refractinig surface and is viewed normally, then it apears at a distance `x//mu` from the surface, where `mu` is refractive index of that medium ( in which it is situated) with respect to the medium ini 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+(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)=(x)/(mu)+y` (ii)
where `mu^(')` is refractive index of air with respect to water.
Now, `mu^(')=(1)/(mu)=(3)/(4)`
Hence, `r_(2)=8.80m`
Hence, option (a) is wrong while (c) is correct.