In Fig. find the maximum angle `i` for which light suffers total internal reflection at the vertical surface.

Answer

`48.6^(@)`

Explanation

`sin C = (1)/(mu) = (1)/(1.25) = (4)/(5)`
As `C + r = 90^(@)` or `r = 90^(@) - C`
`:. Sin r = sin (90^(@) - C) = cos C`
`= sqrt(1 - sin^(2)C) = sqrt(1 - ((4)/(5))^(2)) = (3)/(5)`
From Snell's Law, `(sin i)/(sin r) = mu = 1.25`
`sin i = 1.25 sin r = 1.25 xx (3)/(5) = 0.75`
`i = sin^(-1)(0.75) = 48.6^(@)`
This is the maximum value of `i`, for which Total Internal Reflection would occur at the vertical surface.