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 \( A \) be the angle of the prism, \( i \) the angle of incidence on the face AB, and \( e \) the angle of emergence. The angle of deviation \( D \) is the angle between the incident and emergent rays. For this case, we are given that \( D = 15^\circ \) and the prism angle \( A = 30^\circ \). Using the formula for the deviation angle in a prism: \[ D = i + e - A \] Given that the emergent ray is normal to the face AC, we have \( e = 90^\circ \). Therefore, we can substitute into the formula: \[ 15^\circ = i + 90^\circ - 30^\circ \] Solving for \( i \): \[ i = 15^\circ \] Using Snell’s law to find the refractive index \( n \) of the material of the prism: \[ n = \frac{\sin(i)}{\sin(r)} \] Where \( r \) is the angle of refraction inside the prism. Since \( r = 90^\circ - A/2 \), we can substitute the known values to find the refractive index of the material of the prism.