Question

A slab of refractive index `mu` is placed in air and light is incident at maximum angle `theta_0` from vertical. Find minimum value of `mu` for which total internal reflection takes place at the vertical surface.

Answer 1

For vertical surface, `alpha gt C`
`rArr sinalpha gt sinC`
`sinalpha gt 1/(mu)` (i)
For horizontal surface,
`sintheta_(0)= mu cos alpha`
`rArr sin alpha=sqrt((mu^(2)-sin^(2)theta_0)/mu^(2))` (ii)
From Eqs. (i) and (ii), we get
`rArr sqrt((mu^(2)-sin^(2)theta_(0))/mu) gt 1/mu rArr mu^(2)-sin^(2)theta_(0) gt 1`
`rArr mu gt sqrt(1+sin^(2)theta_(0))`
So, minimum value of `mu=sqrt(1+sin^(2)theta_(0))`