Question

The light beam in the figure below strikes surface 2 at the critical angle, θc = 42°. Determine the angle of incidence, θi.
°

Answer 1

Concepts and reason

The concept required to solve the given problem is Snell’s law of refraction.

Initially, find the refractive index of glass when light strikes surface 2 using equation of Snell’s law. After that, use the triangle law to find the angle of refraction when light strikes surface 1. Use the angle of refraction and refractive index value to find the angel of incidence.

Fundamentals

The Snell’s law of refraction relates the angle of incidence and refraction with the refractive index value. Mathematically, it is given as follows:

${n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}$

Here, ${n_1}$ is the refractive index of first medium, ${n_2}$ is the refractive index of second medium, ${\theta _1}$ is the angle that ray makes with the normal in medium 1 and ${\theta _2}$ is the angle that ray makes with the normal in medium 2.

The following figure shows the angle made by light beam with surface 1 and 2.

Here, ${n_1}$ is the refractive index of first medium (air), ${n_2}$ is the refractive index of second medium(glass), ${\theta _r}$ is the angle of refraction when light strikes surface 1 and ${\theta _c}$ is the critical angle.

Snell’s law equation when light beam strikes surface 2 is,

$\begin{array}{c}\\{n_2}\sin {\theta _c} = {n_1}\sin 90^\circ \\\\{n_2} = \frac{{{n_1}}}{{\sin {\theta _c}}}\\\end{array}$

Substitute 1 for ${n_1}$ and $42^\circ$ for ${\theta _c}$ in equation ${n_2} = \frac{{{n_1}}}{{\sin {\theta _c}}}$ .

$\begin{array}{c}\\{n_2} = \frac{1}{{\sin 42^\circ }}\\\\ = 1.49\\\end{array}$

In $\Delta ABC$ ,

$\begin{array}{c}\\{\theta _1} + 60^\circ + {\theta _2} = 180^\circ \\\\{\theta _1} + {\theta _2} = 120^\circ \\\end{array}$

From figure,

$\begin{array}{c}\\{\theta _c} + {\theta _1} = 90^\circ \\\\{\theta _1} = 90^\circ - {\theta _c}\\\end{array}$

Substitute $42^\circ$ for ${\theta _c}$ in equation ${\theta _1} = 90^\circ - {\theta _c}$ .

$\begin{array}{c}\\{\theta _1} = 90^\circ - 42^\circ \\\\ = 48^\circ \\\end{array}$

Substitute $48^\circ$ for ${\theta _1}$ in equation ${\theta _1} + {\theta _2} = 120^\circ$ .

$\begin{array}{c}\\48^\circ + {\theta _2} = 120^\circ \\\\{\theta _2} = 72^\circ \\\end{array}$

From figure,

$\begin{array}{c}\\{\theta _r} + {\theta _2} = 90^\circ \\\\{\theta _r} = 90^\circ - {\theta _2}\\\end{array}$

Substitute $72^\circ$ for ${\theta _2}$ in equation ${\theta _r} = 90^\circ - {\theta _2}$ and calculate the angle of refraction as follows:.

$\begin{array}{c}\\{\theta _r} = 90^\circ - 72^\circ \\\\ = 18^\circ \\\end{array}$

Snell’s law equation when light beam strikes surface 1 is,

$\begin{array}{c}\\{n_1}\sin {\theta _i} = {n_2}\sin {\theta _r}\\\\{\theta _i} = {\sin ^{ - 1}}\left( {\frac{{{n_2}}}{{{n_1}}}\sin {\theta _r}} \right)\\\end{array}$

Substitute 1 for ${n_1}$ , 1.49 for ${n_2}$ and $18^\circ$ for ${\theta _r}$ in equation ${\theta _i} = {\sin ^{ - 1}}\left( {\frac{{{n_2}}}{{{n_1}}}\sin {\theta _r}} \right)$ .

$\begin{array}{c}\\{\theta _i} = {\sin ^{ - 1}}\left( {\frac{{1.49}}{1}\sin 18^\circ } \right)\\\\ = 27.4^\circ \\\end{array}$

Ans:

The angle of incidence is $27.4^\circ$ .