Consider the situation shown in figure. Find the maximum angle theta for which the light suffers total internal reflection at the vertical surface.

Explanation

At AC:
`1xxsintheta=5/4sintheta^'`
`sintheta^'=4/5sintheta` (i)
`theta^'+theta^('')=90^@` i.e. `theta^('')=90^@-theta^'` (ii)
`mu_2sinC=mu_1sin90^@`
`sinC=mu_1/mu_2=(1)/(5//4)=4/5` (iii)
For T.I.R. on AB
`theta^('')geC`
`sintheta^('')gesinC`
`sin(90^@-theta^')gesinC`
`costheta^'gesinC`
`cos^2theta^'gesin^2C`
`1-sin^2theta^'gesin^2C`
`1-(4/5)^2sin^2thetage(4/5)^2`
`sin^2thetale[(5/4)^2-1=(3/4)^2]`
`sinthetale3/4`
`thetalesin^-1(3//4)`
`theta_(max)=sin^-1(3//4)`