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 δ. Prove that the refractive index n of the material of the prism is given by:n = sin(A+δ)/sin A

Answer 1

When light is incident normally on a refracting face of the prism, the deviation \( \delta \) is related to the prism angle \( A \) and the refractive index \( n \) as follows:

At Normal Incidence

The angle of incidence on the first face is \( 0^\circ \), and hence the refraction occurs only at the second face of the prism.

Snell's Law at the Second Face

The angle of refraction \( r_2 \) at the second face satisfies Snell's Law:

\[ n = \frac{\sin(i_2)}{\sin(r_2)} \]

where \( i_2 \) is the angle of incidence at the second face. Since the total deviation \( \delta \) is given by:

\[ \delta = i_2 + r_1 - A \]

and for normal incidence, \( r_1 = 0 \), we have:

\[ \delta = i_2 - A \quad \text{or} \quad i_2 = \delta + A \]

Substituting into Snell's Law

Substituting \( i_2 = \delta + A \) into Snell's Law:

\[ n = \frac{\sin(\delta + A)}{\sin A} \]

Refractive Index

Hence, the refractive index of the prism material is:

\[ n = \frac{\sin(\delta + A)}{\sin A} \]