For a prism, `A=60^(@), n=sqrt(7//3)` . Find the minimum possible angle of incidence, so that the light ray is refracted from the second surface. Also, find `delta_(max)`.

Explanation

In minimum incidence case the angles will be as shwon in figure
Applying snell's law:
`x xx sin i_(min) = sqrt((7)/(3)) sin (A-C)`
`= sqrt((7)/(3)) (sin A cos C - cos A sin C)`
`= sqrt((7)/(3)) (sin60 sqrt(1 - (3)/(7)) - cos 60 sqrt((3)/(7))) = (1)/(2)`
`:. i_(min) = 30^(@), :. delta_(max) = i_(max) + 90^(@) - A = 30^(@) + 90^(@) - 60^(@) = 60^(@)`