Question

A fish rising vertically with speed $3$ m/s, to the surface of the water sees a bird diving vertically towards it with a speed of $9$ m/s. Given, $_a{\mu _w} = \dfrac{4}{3}$. The actual velocity of the dive of the bird is:
A.6m/s
B.4m/s
C.8.4m/s
D.4.5m/s

Answer 1

Here in the given question we have given two media then the speed of one object to the other the speed will be related and the fish will watch the image of the bird instead of the bird itself.
The above scenario given in the question is depicted by the following diagram,

Complete answer:
Given quantities,
Velocity of the fish, ${v_f} =$3m/s
Velocity of the bird, ${v_b} =$9m/s,
Refractive index of water with respect to the air, $_a{\mu _w} = \dfrac{4}{3}$.
So now we will calculate the image of the bird with respect to the fish,
${v_{ib}} = (9 - 3) = 6$m/s
Due to the change of the medium from air to water the actual velocity of dive of the bird will be different.
Let the actual velocity of dive of the bird$= v$,
Then due to the presence of the refractive index, we will have the following relation,
$_a{\mu _w}v = {v_{ib}}$
Now we will put the values which we have in the above relation,
$\Rightarrow \dfrac{4}{3}v = 6$
$\Rightarrow {v_{actual}} = \dfrac{{6 \times 3}}{4}$
$\Rightarrow {v_{actual}} = \dfrac{{18}}{4}$
$\Rightarrow {v_{actual}} = 4.5$m/s
Hence, the actual velocity of the dive of the bird is 4.5m/s.

So, option (D) is the correct answer.

Note:
Here we have seen the effect of the change of the medium. This change of the medium results in refraction of the light. So, some of the effects of the refractions are as follows,
An object appears to be raised when placed under water.
Stars appear to twinkle at night due to the refraction.
If a lemon is kept in a glass of water it appears to be bigger when viewed from the sides of the glass.
Depth of the water appears to be less in a pool.