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 strikes a prism's refracting face perpendicularly, the deviation \( \delta \) is linked to the prism's angle \( A \) and its refractive index \( n \) by the following relationship:

Normal Incidence Condition

With an incidence angle of \( 0^\circ \) on the first face, refraction exclusively occurs at the prism's second face.

Snell's Law Application at the Second Face

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

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

Here, \( i_2 \) denotes the angle of incidence at the second face. Given that the total deviation \( \delta \) is calculated as:

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

And for normal incidence, \( r_1 = 0 \). Therefore:

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

Substitution into Snell's Law

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

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

Refractive Index Formula

Consequently, the refractive index of the prism material is:

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