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

Explanation

Here, `C = ?, i= 40^(@)`,
deviation, `delta = 15^(@)`
As ray deviates towards normal, therefore,
`r =I - delta = 40^(@) - 15^(@) = 25^(@)`
As `mu = (sin i)/(sin r) = (1)/(sin C)`
`:. sin C = (sin r)/(sin r) = (sin 25^(@))/(sin 40^(@)) = (0.4226)/(0.6428) = 0.6574`
`C = sin^(-1) (0.6574) = 41.1^(@)`.