Question

The critical angle of glass-air is \[45{}^\circ \] for the light of yellow colour. State whether it will be less than, equal to, or more than \[45{}^\circ \] for blue light?
\[A)\] more than \[45{}^\circ \]
\[B)\] less than \[45{}^\circ \]
\[C)\] same as \[45{}^\circ \]
\[D)\] can’t say

Answer 1

Hint: We must know that the critical angle is the angle of incidence that provides an angle of refraction equal to \[90{}^\circ \]. The relation between refractive indices and angle of incidence and angle of refraction is given by Snell’s law as \[\dfrac{\sin \left( i \right)}{\sin \left( r \right)}=\dfrac{{{n}_{2}}}{{{n}_{1}}}\]. The refractive index of a particular medium may differ for light waves with different frequencies. The relation between this is that refractive index is directly proportional to frequency.

Complete step by step answer:
Firstly we will calculate the refractive index of this particular glass for yellow light. The expression is given by snell’s law as,
      \[\dfrac{\sin \left( i \right)}{\sin \left( r \right)}=\dfrac{{{n}_{2}}}{{{n}_{1}}}\]
Here, the angle of refraction is equal to \[90{}^\circ \] , because incident angle is critical angle which is equal to \[45{}^\circ \] for yellow light. Now, this is given for glass-air medium. So our refractive index if the first medium will be one.
      \[\Rightarrow \dfrac{\sin \left( 45{}^\circ \right)}{\sin \left( 90{}^\circ \right)}=\dfrac{{{n}_{2}}}{1}\]
Therefore, refractive index for yellow light will be,
 \[{{n}_{yellow}}=\sin \left( 45{}^\circ \right)=\dfrac{1}{\sqrt{2}}\]
Now, if we change this with yellow light with blue, the refractive index of this glass will increase a little. We can understand this from the relation that refractive index is directly proportional to frequency. We know blue light has a higher frequency than that of yellow light. So the refractive index increases.
That means, \[{{n}_{blue}}>\dfrac{1}{\sqrt{2}}\] .
So, from this we can understand the critical angle when we use blue light.
 \[\Rightarrow {{n}_{blue}}=\sin \left( {{\theta }_{c}} \right)>\dfrac{1}{\sqrt{2}}\]
       \[\begin{align}
  & {{\theta }_{c}}>{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right) \\
 & {{\theta }_{c}}>45{}^\circ \\
\end{align}\]
Therefore, the critical angle when we use blue light here will be more than \[45{}^\circ \] .

So, the correct answer is “Option A”.

Note:
We can also solve this question on the basis of wavelength. As we know, wavelength is given by \[\lambda =\dfrac{c}{\upsilon }\] . So, as the wavelength increases, the refractive index will be decreasing. We can clearly understand the dependence of the refractive index on frequency by looking at dispersion of white light by a prism. There, the light waves are dispersed because different frequencies will have different refractive indices. Blue wave will be at the bottom because it will be having the largest refractive index.