Question

Calculate the critical angle for a glass air interface, if a ray of light incident in air on the surface is deviated through `15^@`, when its angle of incidence is `40^@`.

Answer 1

Here, `I = 40^(@)`
`r = 40^(@) - 15^(@) = 25^(@)`
`mu = (sin i)/(sin r) = (sin 40^(@))/(sin 25^(@)) = (0.6428)/(0.4226) = 1.52`
if `C` is critical angle for glass air interface, then from
`sin C = (1)/(mu) = (1)/(1.52) = 0.6579`
`C = sin^(-1)(0.6579) = 41.14^(@)`