A monochromatic light is incident on the plane interface AB between two media of refractive indices `mu_(1)` and `(mu_(2)gtmu_(1))` at an angle of incidence `theta` as shown in Fig.
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 `mu_(3)` is introduced on the interface (as shown in the figure), show that for any value of `mu_(3)` all light will ultimately be reflected back into medium II.

Answer

A

Explanation

KEY CONCEPT: For total internal reflection, the condition are
The object should be in the denser medium.
The angle of incidence should be greater than the critical angle
Case (i): When `n_(3)ltn_(1)`
Obviously `n_(3)ltn_(2)` and the angle `theta` is greater than the critical angle required for the ray passing from medium II to medium III. Therefore total internal reflection will also take place when a ray strikes with the same angle at the interface of medium II and medium III.
Case (ii): `n_(3)gtn(1)` but `n_(3)ltn_(2)`
The ray will get refracted in medium III as the angle `theta` will now be less than the critical angle required for medium II
`(sintheta)/(sinr)=(n_(3))/(n_(2))`..(Applying Snell's law at P)
`:'` `sinr=(n_(2))/(n_(3))sintheta`
As `n_(2)gtn_(3)` So, `rgttheta`
When the refracted ray PR meets the boundary DE, it is travelling from a denser medium to a rarer medium. Therefore the ray will be totally internally reflected at DE if its angle of incidence r is more than the critical angle for med III and I.
`sini^(n)=(n_(1))/(n_(3))`
Since, `sinrgt(n_(1))/(n_(3))impliessinrgtsini^(``)impliesrgti^(``)`
Therefore ray PR will be totally internal reflected along RQ. On reaching Q, the ray will be refracted in med II. Thus, the ray will ultimately be reflected back in medium II.