Calculate the angle of minimum deviation for a ray which is refracted through an equiangular prism of refractive index 1.4
A
600
B
290
C
990
D
900
Verified Correct Answer: B.
Physical Principles & Numerical Solution
From the relation refractive index(n) = \(\frac{sin(\frac{ A + Dm}{2})}{sin(\frac{A}{2})}\)
1.4 = \(\frac{sin(\frac{ 60º + Dm)}{2}}{sin(\frac{60}{2})}\)
1.4 = \(\frac{sin(\frac{ 60 + Dm}{2})}{sin30}\)
sin\(\frac{ 60 + Dm}{2}\) = sin 30 x 1.4
sin\(\frac{ 60 + Dm}{2}\) = 0.7
\(\frac{60 + Dm}{2}\) = sin\(^{-1}\)(0.7) = 44.427º
60 + Dm = 2 x 44.427 = 88.8
Dm = 88.8 - 60 = 28.8º ≈ 29º
Source: Myschool Past Questions Archive (Record #41445)