Question

The refractive index of water relative to air is $\dfrac{4}{3}$. Find the relative refractive index in the case of light being refracted from water to air.

Answer 1

Hint: Refractive index of a material is a dimensional number that describes how fast light travels through the material. It is defined as $n = \dfrac{c}{v}$, where n is the refractive index, c is the speed of light in vacuum and v is the phase velocity of light.

Complete step by step answer:
We are given that the refractive index of water relative to air is $\dfrac{4}{3}$.
We have to find the refractive index in the case of light being refracted from water to air.
The refractive index of water is 1.333 i.e. $\dfrac{4}{3}$ meaning that light travels 1.333 times faster in vacuum than in water. Increasing refractive index corresponds to decreasing speed of light in the material. Light takes a short path (reflects) as it goes from rarer to denser due to (more density, more obstruction from particles of medium). So the refractive index for a beam of light going from water to air will be the inverse of the refractive index of light going from air to water.
$
{n_f} = \dfrac{1}{n} \\
\Rightarrow n = \dfrac{4}{3} \\
\Rightarrow {n_f} = \dfrac{1}{{\left( {\dfrac{4}{3}} \right)}} \\
\Rightarrow {n_f} = \dfrac{3}{4} \\
\Rightarrow {n_f} = 0.75 \\
$
Therefore, the refractive index in the case of light being refracted from water to air is 0.75.

Note:When we measure speed of light in two different media, the relative refractive index of medium 2 with respect to medium 1 is deferred as the ratio of speed of light in medium 1 to speed of light in medium 2.
${n_{21}} = \dfrac{{spee{d_1}}}{{spee{d_2}}}$
Refractive index must be measured at standard temperatures. A higher temperature makes the liquid less viscous and less dense causing the light to move faster in the medium and resulting in a larger refractive index.