A ray of white light traveling through fused quartz is incident at a quartz–air interface at angle${\mathit{\theta }}_{\mathbf{1}}$.Assume that the index of refraction of quartz is $\mathit{n}\mathbf{=}\mathbf{1}\mathbf{.456}$at the red end of the visible range and$\mathit{n}\mathbf{=}\mathbf{1}\mathbf{.470}$ at the blue end.If${\mathit{\theta }}_{\mathbf{1}}$ is(a) ${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{42}\mathbf{.00}\mathbf{°}$, (b),${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{43}\mathbf{.10}\mathbf{°}$and (c),${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{44}\mathbf{.00}\mathbf{°}$is the refracted light white, white dominated by the red end of the visible range, or white dominated by the blue end of the visible range, or is there no refracted light?

• The index of refraction of quartz at the red end of the visible range is,$n=1.456$.
• The index of refraction of quartz at the blue end of the visible range is,$n=1.470$ .

Use the concept of refraction of light. Calculate the critical angle for the quartz at both the red end of the visible range and at the blue end. So, you can compare the given incidence angle with the critical angle and can determine the nature of refracted light.

Formula:

The critical angle of a refracted light with the incident light,

${\theta }_c={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$….. (1)

Here, the ray of white light is incident at a quartz-air interface. Therefore,

The index of refraction of air, ${\text{n}}_2=1$

The index of refraction, ${\text{n}}_1=n$

Thus, using this value in equation (1), you can get the equation of critical angle as follows:

${\theta }_c={\sin }^{-1}\left(\frac{1}{n}\right)$ ….. (2)

Thus, the critical angle for the red end of the visible range is given using the given data in equation (2) as:

$\begin{array}{c}{\left({\theta }_c\right)}_{red}={\sin }^{-1}\left(\frac{1}{1.456}\right)\\ =43.38°\end{array}$

Similarly, the critical angle for the blue end of the visible range is given using the given data in equation (a) as:

$\begin{array}{c}{\left({\theta }_c\right)}_{blue}={\sin }^{-1}\left(\frac{1}{1.470}\right)\\ =42.86°\end{array}$

Now, considering the incidence angle${\theta }_1$as$42.00°$ , you get that the incident angle is less than both critical angle for red end and critical angle for blue end.

So, all types of rays can be refracted. Hence, the refracted light will be white light.

Now, considering the incidence angle${\theta }_1$as $43.10°$, you get that the incident angle is greater than the critical angle for the blue end of the visible range and less than the red end of the visible range.

Hence, the refracted light will be white-dominated by the red end of the visible end.

Now, considering incidence angle${\theta }_1$as $44.00°$, you get that the incident angle is less than both the critical angle for the red end and the critical angle for the blue end.

Hence, no visible light can be refracted, and you can say that there is no refracted light.