Question

Obtain the formula for refractive index of material of prism in terms of angle of minimum deviation and angle of prism.

Hint:

At minimum deviation, light travels symmetrically through the prism. Use geometry and Snell’s law to derive the relation between refractive index, angle of prism, and angle of minimum deviation.

Answer 1

Consider a prism with refracting angle \( A \), and let \( \delta_m \) be the angle of minimum deviation. When light passes symmetrically through the prism at minimum deviation, the angle of incidence \( i \) and the angle of emergence \( e \) are equal, i.e.,
\[ i = e \] From the geometry of the prism, the relation between angle of incidence \( i \), angle of refraction \( r \), and prism angle \( A \) is:
\[ A = r_1 + r_2 \] But under minimum deviation condition, the path of the ray inside the prism is symmetrical, so:
\[ r_1 = r_2 = r \Rightarrow A = 2r \Rightarrow r = \frac{A}{2} \] The total deviation \( \delta \) is given by:
\[ \delta = i + e - A \Rightarrow \delta_m = 2i - A \Rightarrow i = \frac{A + \delta_m}{2} \] Now apply Snell's law at the first surface of the prism:
\[ n = \frac{\sin i}{\sin r} \] Substitute the values of \( i \) and \( r \):
\[ n = \frac{\sin\left( \frac{A + \delta_m}{2} \right)}{\sin\left( \frac{A}{2} \right)} \] Final Formula:
\[ \boxed{n = \frac{\sin\left( \frac{A + \delta_m}{2} \right)}{\sin\left( \frac{A}{2} \right)}} \] This is the required expression for the refractive index \( n \) of the material of the prism in terms of the prism angle \( A \) and angle of minimum deviation \( \delta_m \).