JEE Advanced Model Paper 6 with solutions for online practice

The critical angle is given by
sin ie =

µe

µg

=

1.5

√3

⇒ sin ie =

√3

2

⇒ ie = 60°
For total internal reflection, the angle of incident at B must be greater than 60ϕ . Angle rwhich is (90° - ie) must be less than 30° i.e. rmax = 30° . Applying Snell’s law at A, we have 1 × sin imax = √3sinrmax = √3 sin30° =

√3

2

Which gives imax = 60°