Two optical media have a plane boundary between them. Suppose `theta_(1 cr)` is the critical angle of incidence of a beam and `theta_(1)` is the angle of incidence at which the refracted beam is perpendicular to the reflected one (the beam is assumed to come from an optically denser medium). Find the relective index of these media if `sin theta_(1 cr)//sin theta_(1) = eta = 1.28`.

Explanation

Let two optical mediums of `R.I. n_(1)` an d`n_(2)` respectively be such that `n_(1) gt n_(2)`. In the case when angle of incidence is `theta_(1cr)` (Fig.) from the law of refraction
`n_(1) sin theta_(1cr) = n_(2)` (1)
In the case, when the angle o fincidence is `theta_(1)`, from the law of refraction at the interface o fmediums `1` and `2`.
`n_(2) sin theta_(1) = n_(2) sin theta_(2)`
But in accordance with the problem `theta_(2) = (pi//2 -theta_(1))`
so, `n_(1) sin theta_(1) = n_(2) cos theta_(1) (2)`
Dividing Eqn (1) and (2)
`(sin theta_(1cr))/(sin theta_(1)) = (1)/(cos theta_(1))`
or, `eta = (1)/(cos theta_(1))`, so `cos theta_(1) = (1)/(eta)` and `sin theta_(1) = (sqrt(eta^(2) - 1))/(eta)` (3)
But `(n_(1))/(n_(2)) = (cos theta_(1))/(sin theta_(1))`
So, `n_(1)/(n_(2)) = (1)/(eta) (eta)/sqrt(eta^(2) - 1)` (Using 3)
Thus `(n_(1))/(n_(2)) = (1)/(sqrt(eta^(2) - 1)`