Question

The light ray in the figure below strikes the hypotenuse of the prism at an angle of incidence greater than the critical angle. Find the angle phi of the ray as it leaves the base of the prism. The incident ray is perpendicular to the left side of the prism. (The index of refraction of the prism is 1.5. Take θ to be 44.9°.)

Answer 1

angle of incidence on inside = sin^-1(1/n)= 41.81

angle of incidence on base = 44.9-41.81 = 2.99

n*sin2.99 = 1*sin phi

phi = 4.48 = 5(approx)

Answer 2

Internal reflection follows rules of reflection

Given that incident ray is perpendicular

Angle of incidence on inclined side is 45.1 w.r.t to normal

So Angle of reflection is 45.1

now but vertical from inclined side is 44.9 degrees from normal

so reflected ray is 0.2 degrees from vertical which is the normal for the base (45.1 - 44.9) .

So 1.5 sin (0.2) = 1 sin (phi)

phi = 0.3 degrees