Question

A jar contains two immiscible liquids A and B with a coin at its bottom. The depths of liquid A and liquid B are 6 cm and 8 cm respectively. If the apparent depth of the coin is 9 cm and the refractive index of liquid A is 1.5, what is the refractive index of liquid B.

Answer 1

Hint: When light travels from a denser medium to rare medium or rare medium to denser medium the light deviates and therefore the image of the object which is in the other medium looks somewhat different from its original position.

Formula used:
The formula of the refractive index in terms of the depth of object to the depth of image that appears is given by,
$\mu = \dfrac{R}{I}$
Where the refractive index is $\mu $ and the depth of the real object is R and the image of the object depth is given as $I$.

Complete step by step answer:
It is given in the problem that there is a jar containing two immiscible liquids A and B with a coin at its bottom. The depths of liquid A and liquid B are 6 cm and 8 cm respectively and if the apparent depth of the coin is 9 cm and the refractive index of liquid A is 1.5 then we need to find the refractive index of liquid B.
The formula of the refractive index in terms of the depth of the object to the depth of image that appears is given by,
$\mu = \dfrac{R}{I}$
Where the refractive index is$\mu $and the depth of the real object is R and the image of the object depth is given as $I$.
The apparent depth of the liquid $A$ is,
$ \Rightarrow \mu = \dfrac{R}{I}$
$ \Rightarrow I = \dfrac{R}{\mu }$
$ \Rightarrow I = \dfrac{6}{{1 \cdot 5}}$ ……..………eq. (1)
The apparent depth of liquid $B$ is,
$ \Rightarrow \mu = \dfrac{R}{I}$
Let x be the refractive index of liquid B.
$ \Rightarrow I = \dfrac{8}{x}$ …………….………eq. (2)
As the total apparent image of the coin is 9 cm. Therefore adding the equation(1) and equation (2).
$ \Rightarrow {I_t} = \dfrac{6}{{1 \cdot 5}} + \dfrac{8}{x}$
$ \Rightarrow \dfrac{6}{{1 \cdot 5}} + \dfrac{8}{x} = 9$
$ \Rightarrow \dfrac{{6x + 12}}{{1 \cdot 5x}} = 9$
$ \Rightarrow 6x + 12 = 13 \cdot 5x$
$ \Rightarrow 12 = 7 \cdot 5x$
$ \Rightarrow x = \dfrac{{12}}{{7 \cdot 5}}$
$ \Rightarrow x = 1 \cdot 6$

The refractive index of the liquid B is equal to $x = 1 \cdot 6$.

Note:
The refractive index can be expressed in terms of speed of light in different mediums and also the refractive index can be expressed in the terms of the depth of the object and depth of the apparent image. It is advisable to remember these formulas in different terms so these kinds of problems can be solved easily.