Question

A ray of light is incident on a refracting face AB of a prism ABC at an angle of \( 45^\circ \). The ray emerges from face AC and the angle of deviation is \( 15^\circ \). The angle of prism is \( 30^\circ \). Show that the emergent ray is normal to the face AC from which it emerges out. Find the refraction index of the material of the prism.

Hint:

For a ray of light passing through a prism, the angle of deviation depends on the angle of incidence and the prism angle. Snell's law is used to find the refractive index of the material.

Answer 1

Let \( A \) denote the prism angle, \( i \) the angle of incidence on face AB, and \( e \) the angle of emergence. The deviation angle \( D \) is the angle between the incident and emergent rays. We are given \( D = 15^\circ \) and \( A = 30^\circ \). The formula for deviation in a prism is \( D = i + e - A \). Since the emergent ray is normal to face AC, \( e = 90^\circ \). Substituting these values: \( 15^\circ = i + 90^\circ - 30^\circ \). Solving for \( i \) yields \( i = 15^\circ \). Snell's law, \( n = \frac{\sin(i)}{\sin(r)} \), is used to determine the refractive index \( n \) of the prism material, where \( r \) is the angle of refraction within the prism. Given \( r = 90^\circ - A/2 \), substituting known values allows calculation of the refractive index.