Question

Monochromatic light is incident on a plane interface AB
between two media of refractive indices $n_1$ and $n_2(n_2>n_1)$ at
an angle of incidence $\theta$ as shown in the figure.
The angle $\theta$ is infinitesimally greater than the critical angle
for the two media so that total internal reflection takes place.
Now if a transparent slab DEFG of uniform thickness and of
refractive index $n_3$is introduced on the interface (as shown in
the figure), show that for any value of $n_3$ all light will
ultimately be reflected back again into medium II.
Consider separately the cases :
(a) $n_3<n_1$ and
(b) $n_3>n_1$

Answer 1

Given $\theta$ is slightly greater than $si{n}^{-1}\left(\frac{n_1}{n_2}\right)$
(a) When $n_3<n_1$
i.e $n_3<n_1<n_2$
or $\frac{n_3}{n_2}<\frac{n_1}{n_2}$
or $si{n}^{-1}\left(\frac{n_3}{n_2}\right)<si{n}^{-1}\left(\frac{n_1}{n_2}\right)$
Hence, critical angle for III and II will be less than the
critical angle for II and I. So, if TIR is taking place between I
and II, then TIR will definitely take place between I and III.
(b) When $n_3>n_1$ Now two cases may arise :
Case 1 $n_1<n_3<n_2$
In this case there will be no TIR between II and III but TIR
will take place between III and I. This is because
Ray of light first enters from II to III ie, from denser to rarer.
$\therefore i>\theta$
Applying Snell's law at P
$n_{2}sin\theta =n_{3}sini$
or $sini=\left(\frac{n_2}{n_3}\right)sin\theta$
Since, sin $\theta$ is slightly greater than $\frac{n_1}{n_2}$
$\therefore sini$ is slightly greater than $\frac{n_2}{n_3}\times \frac{n_1}{n_2}$
or $\frac{n_1}{n_3}$
but $\frac{n_1}{n_3}$ is nothing but sin $({\theta }_c{)}_{I,III}$
$\therefore sin(i)$ is slightly greater than sin $({\theta }_c{)}_{I,III}$
or TIR will now take place on I and III and the ray will be
reflected back.
Case 2 $n_1<n_2<n_3$
This time while moving from II to III, ray of light will bend
towards normal. Again applying Snell's law at P
$n_{2}sin\theta =n_{3}sini$
$sini=\frac{n_2}{n_3}sin\theta$
Since, sin $\theta$ slightly greater than $\frac{n_1}{n_2}$
sin i will be slightly greater than $\frac{n_2}{n_3}\times \frac{n_1}{n_2}or\frac{n_1}{n_3}$
but $\frac{n_1}{n_3}issin({\theta }_c{)}_{I,III}$
i.e. $sini>sin({\theta }_c{)}_{I,III}$
or $i>({\theta }_c{)}_{I,III}$
Therefore, TIR will again take place between I and III and
the ray will be reflected back.
NOTE Case I and case 2 of $n_3>n_1$ can be explained by one equation
only. But two cases are deliberately formed for better understanding
of refraction, Snell's law and total internal reflection (TIR).