Question

Find the angle of minimum deviation for an equilateral prism made of a material of refractive index 1.732. What is the angle of incidence for this deviation?

Answer 1

Given,
Refractive index (μ) of the material from which prism is made = 1.732
We know refractive index is given by:
\[\mu = \frac{\sin\left[ \frac{\delta_\min + A}{2} \right]}{\sin\left[ \frac{A}{2} \right]}\]

Where δ_min is the angle of minimum deviation and A is the angle of prism = 60˚
\[\Rightarrow   1 . 732 \times \sin  (30^\circ ) = \sin  \left( \frac{\delta_\min + 60^\circ }{2} \right)\]
\[\Rightarrow \frac{1 . 732}{2} = \sin \left( \frac{\delta_\min + 60^\circ }{2} \right)\]
\[\Rightarrow   \left( \frac{\delta_\min + 60^\circ }{2} \right) = 60^\circ\]
δ_min = 60°
δ_min = 2i − A
2i = 120°
i = 60°
Hence, the required angle of deviation is 60°.