Question

The diagram shows a glass prism of a critical angle 42°.

1. Redraw the diagram and complete the path of the light ray PQ till it emerges out of the prism.
2. Also, calculate the net angle of deviation of the ray PQ when it emerges from the prism.

Answer 1

(a) Path of the ray:

1. Ray PQ falls normally on face AC, so it enters at P without bending.
2. It meets face AB at R with an angle of incidence of 48°. This is greater than the critical angle (42°), so the ray is totally internally reflected.
3. The reflected ray meets face BC at S with an angle of incidence of 42°, which equals the critical angle.
4. So at S the angle of refraction is 90°, and the ray emerges grazing along face BC.

(b) Net angle of deviation:

\[ \text{Deviation at R (total internal reflection)} = 180^\circ - 2 \times 48^\circ = 84^\circ \]

\[ \text{Deviation at S (refraction)} = 90^\circ - 42^\circ = 48^\circ \]

\[ \text{Net deviation} = 84^\circ + 48^\circ = 132^\circ \]

The ray emerges from the prism with a net deviation of 132°.