In Fig. 33-65, a light ray enters a glass slab at point A at an incident angle ${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{45}\mathbf{.}\mathbf{0}\mathbf{°}$and then undergoes total internal reflection at point B. (The reflection at A is not shown.) What minimum value for the index of refraction of the glass can be inferred from this information?

The angle of incidence for glass slab,${\theta }_1=45.0°$

Use the concept of the total internal reflection and Snell’s law of refraction. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant. This constant is the refractive index of the medium.

Formulae:

Snell’s law is given by,

$\frac{\sin i}{\sin r}=\frac{n_2}{n_1}$

The equation for the total internal reflection is,

$n\sin {\theta }_3⩾1$

The trigonometric equation is as below:

${\sin }^2{\theta }_2+{\cos }^2{\theta }_2=1$

Let ${\theta }_1$be the angle of incidence and${\theta }^2$be the angle of refraction for the first surface.

Let${\theta }^3$be the angle of incidence for the second surface.

As the equation for the total internal reflection is,

$n\sin {\theta }_3⩾1$..............(1)

And according to Snell’s law,

$\frac{\sin i}{\sin r}=\frac{n_2}{n_1}$

Here, the refractive index of glass is, $n_2=n$

The refractive index of air,$n_1=1$

By substituting these values into Snell’s law, you get

$\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{n}{1}$

$\sin {\theta }_1=n\sin {\theta }_2$

$\frac{\sin {\theta }_1}{n}=\sin {\theta }_2$.....................(2)

The figure shows the triangle in which, you can use the geometry as,

$\begin{array}{rcl}{\theta }_2+{\theta }_3+90°& =& 180°\\ {\theta }_3& =& 90°-{\theta }_2\end{array}$

The equation (1) becomes as,

$n\sin \left(90°-{\theta }_2\right)⩾1$

By using the identity,

$\begin{array}{rcl}\sin \left(90°-{\theta }_2\right)& =& \sin 90°\cos {\theta }_2-\sin {\theta }_2\cos 90°\\ & =& \cos {\theta }_2\\ & & \end{array}$

By equation (1),

$n\cos {\theta }_2⩾1$

Squaring of both the sides as,

$n^2{\cos }^2{\theta }_2⩾1$...................(3)

According to the trigonometry identity,

$\begin{array}{rcl}{\sin }^2{\theta }_2+{\cos }^2{\theta }_2& =& 1\\ {\cos }^2{\theta }_2& =& 1-{\sin }^2{\theta }_2\end{array}$

The equation (3) becomes as,

$$\begin{gathered} n^2\left(1-{\sin }^2{\theta }_2\right)⩾1 \\ {n}^2-n^2{\sin }^2{\theta }_2⩾1 \end{gathered}$$

Putting equation (2) in the above equation,

$$\begin{gathered} n^2-n^2\left(\frac{{\sin }^2{\theta }_1}{n^2}\right)⩾1 \\ {n}^2-{\sin }^2{\theta }_1⩾1 \end{gathered}$$

For the largest value of $n$, this equation is

role="math" localid="1662975795297" $\begin{array}{rcl}{n}^2-{\sin }^2{\theta }_1& =& 1\\ n^2& =& 1+{\sin }^2{\theta }_1\end{array}$

$\begin{array}{rcl}n& =& \sqrt{1+{\sin }^2{\theta }_1}\\ & =& \sqrt{1+{\sin }^2\left(45.0°\right)}\\ & =& \sqrt{1+0.499}\\ & =& 1.22\end{array}$

Hence, the minimum value for the index of refraction of the glass is$1.22$.