Monochramatic light is incident on the pLane interface AB between two media of refractive indices `mu_1` and `mu_2(mu_2gtmu_1)` at an angle of incidence theta as shown in figure. The angle theta is infinitesimally greater thannte critical angle for the two media so thast total internal reflection takes place. Now, if a transparent slab DEFG of uniform thickness and of refractive inde `mu_2` is introduced on theinterface (as shown in figure ) , show that for any value of `mu_2` all light will ultimately be reflected back into medium II.

Explanation

Given, `theta` is slightly greater than `sin^-1(mu_1/mu_2)`
When `mu_3ltmu_1`
i.e. `mu_3ltmu_1ltmu_2`
or `mu_3/mu_2ltmu_1/mu_2` or `sin^-1(mu_3/mu_2)ltsin^-1(mu_1/mu_2)`
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`.
When `mu_3gtmu_1` two cases may arise: Case1 `mu_1ltmu_3ltmu_2`
In this case, there will be no TIR between `II` and `III`
but TIR will take place between `III` and `I`.
This is becuase
Ray of light first enters from `II` to `III`. i.e. from
denser to rarer.
`:. igttheta`
Applying Snell's law at P,
`mu_2sin theta=mu_3sini` or `sini=((mu_(2))/(mu_(3))sin theta`
Since, `sin theta` is slightly greater than `mu_1/mu_2`.
`:. sini` is slightly greater than `mu_2/mu_3xxmu_1/mu_2` or `mu_1/mu_3`
but `mu_1/mu_3` is nothing but `sin (theta_C)_(I,III)`
`:. sin(i)` is slightly greater than sin(theta_C)_(I,III)`
or TIR will now take place on I and III and the ray of light will be reflected back.

Case2 `mu_1ltmu_2ltmu_3`
This time while moving from `II` and `III`, ray of light will
bend towards normal. Again applying Snell's law at P,
`mu_2 sin theta =mu_3 sini`
`sini=mu_2/mu_3 sin theta`
Since, `sin theta` is slightly greater than `mu_1/mu_2`.
Therefore, `sin i` will be slightly than `(mu_2/mu_3xxmu_1/mu_2)` or `(mu_1/mu_3)`
But, `mu_1/mu_3` is `sin (theta_C)_(I,III)`
i.e. `sinigtsin(theta_C)_(I,III)`
`sinigtsin(theta_C)_(I,III)`
`igt(theta_C)_(I,III)`
Therefore, TIR will again take place between I and III and the ray light will be reflected back.