Question

Calculate the speed of light in a medium, whose critical angle is $45^\circ$.

Answer 1

Hint: Whenever we come across a critical angle, we can instantly say that the angle of refraction will be 90 degrees and the given critical angle will be our angle of incidence. So we have the angle of incidence and the angle of refraction; we can use Snell’s law to find the refractive index of the medium and consequently, we can find the speed of light in the medium.

Complete step-by-step solution:
Critical angle, \[C = 45^\circ \]and angle of refraction, \[R = {90^ \circ }\]
Now, by applying Snell’s law, \[{\mu _{medium}}\sin C = {\mu _{air}}\sin R\] where \[\mu \]implies the refractive index
Substituting the values, we get
\[ \Rightarrow {\mu _{medium}}\sin C = 1 \times \sin 90^\circ \]
$ \Rightarrow {\mu _{medium}}\sin C = 1$
Therefore, \[{\mu _{medium}} = \dfrac{1}{{\sin C}} = \dfrac{1}{{\sin 45^\circ }} = \sqrt 2 = 1.414\]
Now, speed of light in the medium = (speed of light in vacuum)/( refractive index of the medium)
\[v = \dfrac{c}{{{\mu _{medium}}}}\] where \[v\] is the speed of light in the medium and \[c\] is the speed of light in vacuum
\[ \Rightarrow v = \dfrac{{3 \times {{10}^8}}}{{1.414}} = 2.12 \times {10^8}m/s\]
Therefore, the speed of light in the medium is \[2.12 \times {10^8}m/s\].

Additional Information: For a given pair of media, the critical angle is inversely proportional to the refractive index. Total internal reflection is a phenomenon that is analogous with a critical angle. We know that if the angle of incidence is equal to the critical angle, the angle of refraction is 90 degrees and the refracted ray travels along the boundary separating the two media. But what if the angle of incidence is greater than the critical angle? In such a case, the incident ray is reflected into the first medium.

Note:- There’s a formula that directly relates the critical angle for a given pair of mediums to their refractive indices, which is as follows: \[\sin C = \dfrac{1}{{\mu _b^a}}\] but it might be confusing to recall it when solving a question which is why it’s safer and easier to start with a simple Snell’s equation and then take it from there.