Question

A ray of light is incident normally on a refracting face of a prism of prism angle $A$ and suffers a deviation of angle $\delta$. Prove that the refractive index $n$ of the material of the prism is given by: \[ n = \frac{\sin(A + \delta)}{\sin A}. \]

Hint:

For triangular prisms, the angle of minimum deviation provides a simple way to calculate the refractive index using the geometry of the prism and Snell's law.

Answer 1

Derivation of Refractive Index of the Prism 

 When a ray of light is incident normally on a refracting face of a prism, the angle of incidence $i = 0$, and the ray enters the prism without deviation. Inside the prism, the ray undergoes refraction at the second face, resulting in a total deviation angle $\delta$. Using the geometry of the prism: \[ \delta = i_e - i_r, \] where: $i_e$ is the angle of emergence,  $i_r$ is the angle of refraction inside the prism. From Snell's law at the second face: \[ n = \frac{\sin i_e}{\sin i_r}. \] At the point of minimum deviation ($\delta = \delta_m$), the angles of incidence and emergence are equal: \[ i_e = A + \frac{\delta}{2}. \] Substituting: \[ n = \frac{\sin(A + \delta)}{\sin A}. \] Thus, the refractive index of the prism is: \[ \boxed{n = \frac{\sin(A + \delta)}{\sin A}}. \]