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, directly above his eyes. For the swimmer the bird appears 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) 9 m

Answer 1

Hint : The refractive index of water can also be given as the apparent height divided by the real height. The real height is less than the apparent height. So by substituting the values real height and the refractive index, we can get the value of the apparent height.

Formula used: In this solution we will be using the following formula;
 $\Rightarrow n = \dfrac{{h'}}{h} $ where $ n $ is the refractive index of a transparent medium, $ h' $ is the apparent height as seen from observer in the medium, and $ h $ is the real height as seen from an observer in air or vacuum.

Complete step by step answer
According to the question, a swimmer is at a depth of 12 m below the surface of water, and a bird is said to be at a height of 18 m from the surface of the water. Hence, the real height of the bird is 18 m. Now the swimmer, from inside the water observes the bird directly above him, we are to find the height of which he perceives the bird to be.
In such cases the refractive index of water can be given as,
 $\Rightarrow n = \dfrac{{h'}}{h} $ where $ n $ is the refractive index of a transparent medium, $ h' $ is the apparent height as seen from observer in the medium, and $ h $ is the real height as seen from an observer in air.
Hence, inserting all given values from the question in the above equation, we have
 $\Rightarrow \dfrac{4}{3} = \dfrac{{h'}}{{18}} $ ,
By cross multiplication, and dividing both sides by 3, we have
 $\Rightarrow h' = \dfrac{{18 \times 4}}{3} = 24m $
Hence, the correct option is A.

Note
For clarity, any object in a less dense medium will always appear farther to an observer in a less dense medium because as light rays reflected from the object cross the boundary between the media, it bends towards the normal, this allows the rays of light to appear to be coming from a distance farther than the real distance.