Question

The critical angle of glycerine-air is $43^\circ $. Find the refractive index of glycerine $(\sin 43^\circ = 0.68)$

Answer 1

Hint: The critical angle of a medium is the angle such that if the incident angle of a ray of light going from a denser to a rarer medium is greater than the critical angle, the ray of light will be reflected back in the medium. It depends on the refractive index of the denser medium and can be determined from Snell’s law.
Formula used: In this solution, we will use the following formula:
Snell’s law ${\mu _1}\sin {\theta _1} = {\mu _2}\sin {\theta _2}$ where ${\mu _1}$ is the refractive index of the first medium and ${\mu _2}$ , the second. ${\theta _1}$ and ${\theta _2}$ are the incident and the refractive index of the ray of light.

Complete step by step answer:
We’ve been given the critical angle of the glycerine-air medium as $43^\circ $. Let us start by finding the formula of critical angle from Snell’s law:
For the critical angle, the ray of light will be completely perpendicular to the normal. So, ${\theta _2} = 90^\circ $. As the second medium is air, so ${\mu _2} = 1$, we can write
$\sin {\theta _1} = \dfrac{1}{{{\mu _1}}}$
Or
$\theta = {\sin ^{ - 1}}\left( {\dfrac{1}{{{\mu _1}}}} \right)$
$ \Rightarrow {\mu _1} = \dfrac{1}{{\sin {\theta _1}}}$
Since the critical angle for the glycerine-air medium as $43^\circ $, we can find the refractive index of the glycerine as
${\mu _1} = \dfrac{1}{{\sin 43^\circ }} = \dfrac{1}{{0.68}}$
Which gives us
${\mu _1} = 1.47$

Hence the refractive index of glycerine is ${\mu _1} = 1.47$

Note: The critical angle is only defined for a transfer from a denser to a rare medium. In this case, glycerine is the denser medium and air is the rare medium. The critical angle will depend only on the refractive of the denser medium if the rarer medium is air as we can consider the refractive index of air to be 1.