Question

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 statements are corret?

• A. Height of $B$, observed by $F$ (from itself), is equal to $5.30m$
• B. Depth of $F$, observed by $B$ (from itself), is equal to $6.60m$
• C. Velocity of $B$, observed by $F$ (from itself), is equal to $8.80m$
• D. Velocity of $F$, observed by $F$ (from itself), is equal to $4.50m$

Answer 1

Answer: (B), (C)

Velocity of $B$, observed by $F$ (from itself), is equal to $8.80m$
Depth of $F$, observed by $B$ (from itself), is equal to $6.60m$

If an object is at a distance $x$ from a plane refracting surface and is viewed normally, then it appears 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 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
$r_1=x+\frac{y}{\mu }...\left(i\right)$
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
$r_2=\frac{x}{\mu }+y....\left(ii\right)$
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.