Question

Calculate the speed of light in a medium, whose critical angle is \[{60^0}\]. (Speed of light in air \[{v_l} = 3 \times {10^8}\]m/s)

Answer 1

Hint: First we need to find the refractive index using snell’s law formula. We know that the refractive index of a material \[\left( n \right)\]is the ratio of speed of light in air (vacuum) to the speed of light in a material. Then using the above formula find the speed of the light in medium.

Complete step by step answer:
Snell's Law states that the ratio of the sine of the angles of incidence and transmission is equal to the ratio of the refractive index of the materials at the interface.
\[\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{\sin {\theta _2}}}{{\sin {\theta _1}}}\]
Where \[{n_1}\] and \[{n_2}\] be the refractive index of the materials at the interface. \[{\theta _1}\] and \[{\theta _2}\] are the angles the light ray makes to the normal of the interface surface.
Let \[{n_1}\] and \[{n_2}\] be the refractive index of air and given medium respectively. \[{\theta _1}\] and \[{\theta _2}\] be the angle of incidence and the angle of refraction. Suppose the light perpendicularly touches the interface surface i.e., \[{\theta _1} = {90^o}\]. Then \[{\theta _1} > {\theta _2}\] and \[{\theta _2}\] becomes a critical angle of refraction.
Hence \[{\theta _1} = {90^o}\] and also given \[{\theta _2} = {60^o}\].

For critical angle we will use snell’s law
\[\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{\sin {\theta _2}}}{{\sin {\theta _1}}}\]
\[ \Rightarrow \]\[\dfrac{1}{{{n_2}}} = \dfrac{{\sin {{60}^o}}}{{\sin {{90}^0}}}\]--(1)
Since \[\sin {90^0} = 1\], then the equation (1) become
\[\dfrac{1}{{{n_2}}} = \sin {60^o}\]--(2)
Rewriting the equation (2), we get
\[{n_2} = \dfrac{1}{{\sin {{60}^o}}}\]
\[ \Rightarrow {n_2} = \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}} \\
\Rightarrow {n_2} = \dfrac{2}{{\sqrt 3 }} \\
\Rightarrow {n_2} = 1.1547\]----(3)
We know that refractive index of light in medium,
\[\left( n \right) = \dfrac{{Speed\;of\;light\;in\;air(vacuum)}}{{Speed\;of\;light\;in\;medium}}\]---(4)
Given the speed of light in air \[{v_l} = 3 \times {10^8}\] m/s
Then the equation (4) becomes
\[1.1547 = \dfrac{{3 \times {{10}^8}}}{v}\]
\[\Rightarrow v = \dfrac{{3 \times {{10}^8}}}{{1.1547}} \\
\therefore v = 2.5981 \times {10^8}\,m/s\].

Hence the speed of light in a medium is \[2.5981 \times {10^8}\] m/s.

Note: The smallest angle of incidence at which total internal reflection occurs is called the critical angle. Snell's law is not applicable when angle of incidence is zero as the angle of refraction will also be zero. Two main types of reflection are reflection-in-action and reflection-on-action.