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

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.
$\Rightarrow {{n}_{blue}}=\sin \left( {{\theta }_{c}} \right)>\dfrac{1}{\sqrt{2}}$
$\begin{aligned} & {{\theta }_{c}}>{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right) \\\ & {{\theta }_{c}}>45{}^\circ \\\ \end{aligned}$
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.