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