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Monochromatic light is incident on plane...

Monochromatic light is incident on plane interference AB between two media of refractive indices `mu_1` and `mu_2 (mu_2gtmu_1)` at 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 places. 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 again into medium II. Consider separately the cases.
(a) `mu_3ltmu_1` , (b) `mu_3gtmu_1`

Text Solution

Verified by Experts

We will use the symbol `le` to mean 'infinitesimally greater than'.
When the slab is not inserted,
`thetaletheta_(c)= sin^(-1)(mu_(1)//mu_(2)) ` or `sinthetage mu_(1)//mu_(2)`
When the slab is inserted, we have two cases
`mu_(3)lemu_(1)` and `mu_(3)gt mu_(1)`.
Case I. `mu_(3)lt mu_(1)` . We have `sintheta gemu_(1)//mu_(2)gemu_(3)//mu_(2)`
Thus, the light is incident on AB at an angle greater than the critical angle `sin^(-1)(mu_(3)//mu_(2))`. It suffers total internal reflection and goes back to medium II.
Case II. `mu_(3)gtmu_(1)`
`sin theta ge mu_(1)//mu_(2) lt mu_(3)//mu_(2)`
Thus, the angle of incidence `theta` may be smaller than the critical angle `sin^(-1)(mu_(3)//mu_(2))` and hence it may enter medium III. The angle of refraction `theta` is given by (figure).
`(sintheta)/(sintheta^')= (mu_(3))/(mu_(2))` (i)
`rArr sintheta^(')= (mu_(2))/(mu_(3))sinthetale(mu_(2))/(mu_(3))*(mu_(1))/(mu_(2))`
Thus, `sintheta^(')ge(mu_(1))/(mu_(3)) rArr theta^(')ge sin^(-1)((mu_(1))/(mu_(3)))` (ii)
As the slab has parallel faces, the angle of refraction at the face FG is equal to the angle of incidence at the face DE. Equation (ii) shows that this angle is infinitesimally greater than the critical angle here. Hence, the light suffers total internal reflection and falls at the surface FG at an angle of incidence `theta^(')`. At this face, it will refract into medium II and teh angle or refraction will be `theta` as shown by Eq. (i). Thus, the total ligh energy is ultimately reflected back into medium II.
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