A ray of light is incident on an equilateral triangular glass prism of refractive index 3/2, Calculate the angle through which the ray is minimally deviated in the prism
A
30.0\(^{\circ} \)
B
37.2\(^{\circ} \)
C
42.0\(^{\circ} \)
D
48.6\(^{\circ} \)
Verified Correct Answer: B.
Physical Principles & Numerical Solution
For the equilateral glass prism, A = 60º
n = \(\frac{sin \frac{Dm + A}{2}}{sin \frac{A}{2}}\)
\(\frac{3}{2}\) = \(\frac{sin \frac{Dm + 60}{2}}{sin \frac{60}{2}}\)
= \(\frac{sin \frac{Dm + 60}{2}}{sin 30}\)
\(\frac{3}{2}\) x \(\frac{1}{2}\) = sin\(\frac{Dm + 60}{2}\)
\(\frac{3}{4}\) = 0.75 = sin\(\frac{Dm + 60}{2}\)
sin\(^{-1}\)(0.75) = \(\frac{Dm + 60}{2}\)
48.590 = \(\frac{Dm + 60}{2}\)
48.590 x 2 = Dm + 60
97.180 = Dm + 60
Dm = 97.180 - 60 = 37.18 ≈ 37 .2º
Source: Myschool Past Questions Archive (Record #41453)