Question

The triangular prism shown in Figure I above has index of refraction 1.5 and angles of 37°, 53°, and 90°.

The shortest side of the prism is set on a horizontal table. A beam of light, initially horizontal, is incident on the prism from the left.

a. On Figure I above, sketch the path of the beam as it passes through and emerges from the prism.

b. Determine the angle with respect to the horizontal (angle of deviation) of the beam as it emerges from the prism.

c. The prism is replaced by a new prism of the same shape, which is set in the same position. The beam experiences total internal reflection at the right surface of this prism. What is the minimum possible index of refraction of this prism?

The new prism having the index of refraction found in part (c) is then completely submerged in water (index of refraction = 1.33) as shown in Figure II below. A horizontal beam of light is again incident from the left.

d. On Figure II, sketch the path of the beam as it passes through and emerges from the prism.

e. Determine the angle with respect to the horizontal (angle of deviation) of the beam as it emerges from the prism.

Answer 1

(a)

(b) n_i sin θ_i = n_r sin θ_r, … 1.5 sin (37) = 1 sin θ_r θ_r = 65° … therefore angle α shown above is 28°.

(c) n_i sin θ_i = n_r sin θ_r, … n_i sin (37) = 1 n_i = 1.66, any n higher than this causes total internal.

(d) This ray would not totally reflect because by putting it in water we have reduced the difference between the n’s meaning less bending and the limit of total internal has not be reached.

(e) n_i sin θ_i = n_r sin θ_r, … (1.66) sin (37) = (1.33) sin θ_r … θ_r = 48.7° – 37° = 11.7°