Question

An underwater swimmer is at a depth of \[12\,m\] below the surface of water. A bird is at a height of $18\,m$ from the surface of water, delivery above his eyes. for the swimmer the birds appear to be at a distance from the surface of water equal to (refractive index of water is $4/3$)
(A) $24\,m$
(B) $12\,m$
(C) $18\,m$
(D) $\,9m$

Answer 1

Hint Here we know that the bird's underwater depth and height value also know the refractive index value so that we calculate the width of the water above the depth is referred to as the refractive index. By using the definition of the velocity in a medium.

Complete step by step answer
Given by,
Height of bird from surface of water $ = 18m$
Depth of swimmer below the surface of water $ = 12m$(Real depth)
Refractive index of water $ = 4/3$
We have to find apparent depth of swimmer
Apparent depth in a medium is the depth of an object in a denser medium as seen from the rarer medium.
Its value is smaller than the real depth.
The bird lies in Air medium and the observer is the swimmer which lies in water.
So that,

Refractive index of water $\mu = \dfrac{4}{3}$
Apparent depth can be calculated by using this formula

\[\mu = \dfrac{{real\,height}}{{apparent\,height}}\]
Substituting the given value in above equation,
We get,
$\dfrac{4}{3} = \dfrac{{real\,depth}}{{18}}$
Rearranging the given equation,
$real\,height = \dfrac{4}{3} \times 18$
On simplifying the above equation,
$real\,height = 24\,m$
Hence,
$24\,m$ above surface of water

Thus, option A is correct answer.

Note According to the above definition, there is a different refractive index as it moves at an angle towards a medium. Similarly, the light ray moves obliquely in the air entering the water, the light ray bends towards the normal, because water as a medium has an optically denser medium than air. This shift in speed results in a change in direction.