Question: In the ray diagram of the Figure, where the angles are not drawn to scale, the ray is incident at the critical angle on the interface between materials 2 and 3. Angle $\mathbf{\varphi }\mathbf{=}\mathbf{60}\mathbf{.}{\mathbf{0}}^{\mathbf{o}}$, and two of the indexes of refraction are ${\mathbf{n}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{70}$and ${\mathbf{n}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{60}$.(a) Find index of refraction ${\mathbf{n}}_{\mathbf{3}}$and(b) Find angle $\mathbf{\theta }$. (c) If$\mathbf{\theta }$ is decreased, does light refract into material 3 ?

Figure:

• The ray is incident at the critical on the interface between materials 2 and 3.
• Angle, $\mathrm{\varphi }=60.0^{\mathrm{o}}$
• Refractive index of the first medium,${\mathrm{n}}_1=1.70$
• Refractive index of the second medium, ${\mathrm{n}}_2=1.60$

By using the concept of critical angle, you can find the refractive index of the third medium. By using Brewster’s law, you can find the required angle.

Brewster's law states that an unpolarized light beam will attain a polarized maximum when it strikes at a certain angle of incidence, called the Brewster angle, at the interface of another medium.

Formulae:

The critical angle is defined by,

$\varphi ={\sin }^{-1}\left(\frac{n_3}{n_2}\right)$

The Brewster’s law is,

${\mathrm{n}}_1\sin \left({\mathrm{\theta }}_{\mathrm{B}}\right)={\mathrm{n}}_2\sin \left({\mathrm{\theta }}_{\mathrm{r}}\right)$

As the ray is incident at the critical on the interface between materialsand,so you can write the equation for the critical angle as,

$\mathrm{\varphi }={\sin }^{-1}\left(\frac{{\mathrm{n}}_3}{{\mathrm{n}}_2}\right)$

Substituting the given values in the above equation.

$$\begin{gathered} 60.0^{\circ }={\sin }^{-1}\left(\frac{{\mathrm{n}}_3}{1.60}\right) \\ \sin 60.0^{\circ }=\left(\frac{{\mathrm{n}}_3}{1.60}\right) \\ 0.866\times 1.60={\mathrm{n}}_3 \\ {\mathrm{n}}_3=1.385 \end{gathered}$$

Using Snell’s law, you can write

${\mathrm{n}}_1{\mathrm{sin\theta }}_1={\mathrm{n}}_{2}sin{\mathrm{\theta }}_2$

Applying this equation to the interface between materialand material, and you have

$\begin{array}{l}{n}_{1}sin\left(\theta \right)=n_2\sin \left({90}^{\circ }-\varphi \right)\\ 1.70\times \sin \left(\theta \right)=1.60\times \sin \left({90}^{\circ }-{60}^{\circ }\right)\\ \sin \theta =\frac{1.60\times \sin {30}^{\circ }}{1.70}\\ \sin \theta =0.47\\ \theta ={\sin }^{-1}0.47\\ \theta =28.{07}^{\circ }\end{array}$

Decrease in $\theta$will increase in $\varphi$and thus, it causes the ray to strike the interface between materials 2 and 3, at an angle more than the critical angle. So, no transmission of light into material 3 can occur.

Hence, if $\theta$is decreased, light does not refract into the material.