Question

There are different fish, monkeys, and water of the habitable planet of the star Proxima b. A fish swimming underwater feels that there is a monkey at $2.5\;m$ on the top of a tree. The same monkey feels that the fish is $1.6\;m$ below the water surface. Interestingly, the height of the tree and the depth at which the fish is swimming are exactly the same. Refractive index of that water must be?
A. $\dfrac{6}{5}$
B. $\dfrac{5}{4}$
C. $\dfrac{4}{3}$
D. $\dfrac{7}{5}$

Answer 1

Due to the difference in the view of both the animals, there is a creation of some sort of illusion wherein the distance between the monkey and the fish in water is said to vary in the point of views of the two animals even-though the distance between them is the same. Hence, the concept of real and apparent depths of water is applied and its relation between the refractive indexes of the combination of media is determined in order to find the refractive index of water.

Complete step by step answer:
The above problem revolves around the concept of real and apparent depths of an object in water which has a depth. Before going into this concept we must first know the principle behind refraction and the definition for the refractive index of the material.

The phenomenon of reflection and refraction are the two main properties of light. In this case the property of refraction plays a main role since light travels from one medium to another and this is when refraction of light takes place. Refraction is defined as the change in the path of light or more specifically the bending of the light rays due to the difference in the medium it is travelling in.

This is what happens in the above example as well wherein there are two mediums considered, that is, water the air medium. The amount of refraction of the light rays is dependent on the refractive indices of both the medium through which light is traversing. It measures the ability of the surface to refract light. Refractive index is the ratio between the speeds of light in a vacuum to that of a medium.

However, as per the laws of refraction total refractive index is given by the refractive index of one medium with respect to another. This applies in this problem as well and the refractive index will be given as the refractive index of air with respect to water. Hence the ratio of the refractive indices of these two mediums are considered here.

Now, let us look into the concept of real and apparent depths. When an object is placed in water having some depth the real depth is said to be the actual depth or the actual distance at which the object is residing in but when we observe from the rarer medium which is air, we see that the object appears to be slightly above its actual position. This is because of the phenomenon of refraction of light.

The apparent depth of the fish in water appears to be slightly raised from the viewpoint of the monkey which is viewing from the rarer medium. This is mainly because, as per the law of refraction we know that the light ray travelling from rarer to denser medium tends to bend towards the normal so a virtual image is formed above the actual position of the fish due to the bending of the rays. This happens from the viewpoint of the monkey sitting on top of the tree. The same happens in the viewpoint of the fish wherein the apparent height of the position of the monkey appears to be different from the real position of the monkey.

Let us now construct an equation relating the refractive indices of the medium and the real and apparent depths. The ratio of the refractive indexes of the two mediums is equivalent to the ratio of the real depth and the apparent depth of an object. The equation relating both is hence given by:
$\dfrac{{{n_{water}}}}{{{n_{air}}}} = \dfrac{{Real{\text{ depth}}}}{{{\text{apparent depth}}}}$ --------($1$)
The above equation is for the position of the fish according to the monkey. Now we apply the equation for the position of the monkey according to the fish’s viewpoint.
$\dfrac{{{n_{air}}}}{{{n_{water}}}} = \dfrac{{Real{\text{ height}}}}{{{\text{apparent height}}}}$ --------($2$)

Let us now extract the data given in the question. The apparent depth of the fish in the monkey’s point of view is given as $1.6\;m$ while the apparent height at which the monkey is residing in the fish’s point of view is given as $2.5\;m$. We are asked to find the ratio of the refractive indexes of the two mediums. The real depth of the fish in water or the real height at which the monkey is residing is said to be equal and let this unknown quantity be denoted by $x$. Hence, we can say that
Real height of the monkey on the tree = Real Depth of the fish in the water = $x$
Hence equation ($1$) becomes:
$\dfrac{{{n_{water}}}}{{{n_{air}}}} = \dfrac{{Real{\text{ depth}}}}{{{\text{apparent depth}}}} = \dfrac{x}{{1.6}}$ ---------($3$)
Similarly equation ($2$) becomes:
$\dfrac{{{n_{air}}}}{{{n_{water}}}} = \dfrac{{Real{\text{ height}}}}{{{\text{apparent height}}}} = \dfrac{x}{{2.5}}$ ---------($4$)
By multiplying equation ($1$) and equation ($2$) we get:
$\dfrac{{{n_{water}}}}{{{n_{air}}}} \times \dfrac{{{n_{air}}}}{{{n_{water}}}}$

The common terms get cancelled out and we get:
$\Rightarrow \dfrac{{{n_{water}}}}{{{n_{air}}}} \times \dfrac{{{n_{air}}}}{{{n_{water}}}} = 1$ --------($5$)
From equations ($3$) and ($4$) we can see that:
$\dfrac{{{n_{water}}}}{{{n_{air}}}} = \dfrac{x}{{1.6}}$ -------($6$)
$\Rightarrow \dfrac{{{n_{air}}}}{{{n_{water}}}} = \dfrac{x}{{2.5}}$
Hence these values are substituted into equation ($5$) to get:
$\dfrac{x}{{1.6}} \times \dfrac{x}{{2.5}} = 1$
Now, we solve the equation to find the value for $x$.
$\Rightarrow \dfrac{{{x^2}}}{{1.6 \times 2.5}} = 1$
$\Rightarrow {x^2} = 1.6 \times 2.5$
$\Rightarrow {x^2} = 4$
$\Rightarrow x = 2$
Hence the real height or the real depth is obtained as $2\;m$. We now substitute this value in equation ($6$) to get:
$\dfrac{{{n_{water}}}}{{{n_{air}}}} = \dfrac{x}{{1.6}}$
$\Rightarrow \dfrac{{{n_{water}}}}{{{n_{air}}}} = \dfrac{2}{{1.6}}$
$\therefore \dfrac{{{n_{water}}}}{{{n_{air}}}} = \dfrac{5}{4}$
Hence, the ratio of refractive indices of air and water is determined which is $\dfrac{5}{4}$.

Therefore, the correct answer is option B.

Note: There is a common misconception about the refractive indices of the mediums which the light ray is travelling through. The refractive index of one medium with respect to another is equivalent to the ratio of real and apparent depths and the refractive index of one medium to another is given by the ratio of the refractive indices of the two mediums.