Calculate the angle of minimum deviation for a ray which is refracted through an equiangular...

Calculate the angle of minimum deviation for a ray which is refracted through an equiangular prism of refractive index 1.4

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

There is an explanation video available below.

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