Question

The refractive indices of glycerin and diamond with respect to air are $1.4$ and $2.4$ respectively. Calculate the speed of light in glycerin and diamond. From these results, calculate the refractive index of diamond with respect to glycerin. $(c = 3 \times {10^8}m/s)$
(A) $2.143 \times {10^8}m/s;1.250 \times {10^8}m/s;1.714$
(B) $0.2143 \times {10^8}m/s;12.50 \times {10^5}m/s;0.1714$
(C) $2.143 \times {10^6}m/s;1.250 \times {10^3}m/s;1.74$
(D) $21.43 \times {10^8}m/s;0.125 \times {10^5}m/s;17.14$

Answer 1

Hint : To solve this equation we will start with understanding the problem. The solution to this problem is given by an absolute refractive index. The absolute refractive index is given by the ratio of the speed of light in a vacuum to the speed of light in any medium in which light is propagating. It is a parameter that defines how optically denser a medium can be.

Formula used:
$\Rightarrow \eta = \dfrac{c}{v}$ ,
where $c$ is the velocity of light in vacuum and $v$ is the velocity of light in the given medium.
$\Rightarrow {\eta _{abs.}} = \dfrac{{{\eta _{diamond}}}}{{{\eta _{glycerin}}}}$

Complete step by step answer
Here, the refractive index of glycerin and diamond is given as $1.4$ and $2.4$ i.e. the absolute refractive index. Hence the speed of light will be reduced if it goes from diamond to glycerin and will be increased if the situation is reversed.
Now using the formula of refractive index which is given as
$\Rightarrow \eta = \dfrac{c}{v}$ , where $c$ is the velocity of light in vacuum and $v$ is the velocity of light in a given medium.
Now for glycerin, the refractive index is given as $\eta = 1.4$ .
$\Rightarrow {\eta _{glycerin}} = \dfrac{c}{{{v_{glycerin}}}}$
$\Rightarrow 1.4 = \dfrac{{3 \times {{10}^8}m/s}}{{{v_{glycerin}}}}$
Now rearranging the terms we get
$\Rightarrow {v_{glycerin}} = \dfrac{{3 \times {{10}^8}m/s}}{{1.4}}$
$\therefore {v_{glycerin}} = 2.143 \times {10^8}m/s$ ------------ Equation $\left( 1 \right)$
Similarly for the diamond, the refractive index is given as $\eta = 2.4$ .
$\Rightarrow {\eta _{diamond}} = \dfrac{c}{{{v_{diamond}}}}$
$\Rightarrow 2.4 = \dfrac{{3 \times {{10}^8}m/s}}{{{v_{diamond}}}}$
Now rearranging the terms we
$\Rightarrow {v_{diamond}} = \dfrac{{3 \times {{10}^8}m/s}}{{2.4}}$
$\therefore {v_{diamond}} = 1.250 \times {10^8}m/s$ ------------- Equation $\left( 2 \right)$
Now the refractive index of diamond with respect to the refractive index of glycerin is given by absolute refractive index which is
$\Rightarrow {\eta _{abs.}} = \dfrac{{{\eta _{diamond}}}}{{{\eta _{glycerin}}}}$
$\Rightarrow {\eta _{abs.}} = \dfrac{{2.4}}{{1.4}} = 1.714$ ---------------- Equation $\left( 3 \right)$
Now the Equation $\left( 1 \right)$ and Equation $\left( 2 \right)$ shows the speed of light in glycerin and diamond and Equation $\left( 3 \right)$ shows the absolute refractive index of diamond with respect to glycerin.
Hence option (A) is the correct answer.

Note
The velocity of the light in a vacuum is always constant which can be given as $c = 2.99 \times {10^8}m/s$ . Here we have to remember that the absolute refractive index can also be written as a refractive index which is given by the ratio of the velocity of light in a vacuum to the velocity of light in the given medium in which the light is propagating.