91Ó°ÊÓ

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

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,

$n\sin {\mathrm{θ}}_3â\copyright ¾1$

The trigonometric equation is as below:

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

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

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

As the equation for the total internal reflection is,

$n\sin {\mathrm{θ}}_3â\copyright ¾1$..............(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

$\frac{\sin {\mathrm{θ}}_1}{\sin {\mathrm{θ}}_2}=\frac{n}{1}$

$\sin {\mathrm{θ}}_1=n\sin {\mathrm{θ}}_2$

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

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

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

The equation (1) becomes as,

$n\sin \left(90°-{\mathrm{θ}}_2\right)â\copyright ¾1$

By using the identity,

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

By equation (1),

$n\cos {\mathrm{θ}}_2â\copyright ¾1$

Squaring of both the sides as,

$n^2{\cos }^2{\mathrm{θ}}_2â\copyright ¾1$...................(3)

According to the trigonometry identity,

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

The equation (3) becomes as,

$$\begin{gathered} n^2\left(1-{\sin }^2{\mathrm{θ}}_2\right)â\copyright ¾1 \\ {n}^2-n^2{\sin }^2{\mathrm{θ}}_2â\copyright ¾1 \end{gathered}$$

Putting equation (2) in the above equation,

$$\begin{gathered} n^2-n^2\left(\frac{{\sin }^2{\mathrm{θ}}_1}{n^2}\right)â\copyright ¾1 \\ {n}^2-{\sin }^2{\mathrm{θ}}_1â\copyright ¾1 \end{gathered}$$

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

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

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

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