No. | Angle of incidence |
Angle of deviation |

Figure shows the path of a ray of light through an equiangular prism. The incident ray makes an angle \( i_1 \) with the normal to face AB. The refracted ray bends towards the normal making an angle \( r_1 \) with it. It is then incident at \(r_2 \) on the face AC and and emerges at \( i_2 \).

As the ray refracts at the face AB it undergoes a deivation of \( i_1-r_1 \) towards the base BC and when it refracts the second time it deviates by \( i_2-r_2 \) again towards the base. The ray suffers a total deviation of \(d = (i_1-r_1) + (i_2-r_2) \).

It can be shown that the sum of \(r_1 + r_2 \) is equal to the angle of the prism A. So the deviation \(d = i_1 +i_2 - A \).

Graph shows the variation of angle of deviation with the angle of incidence. As the angle of incidence increases, the angle of deviation decreases, reaches a minimum value \( (d_m) \) and then increases. \( i_1 \) and \( i_2 \) are the interchangeable angles of incidence and emergence. When the deviation is minimum these two angles \( i_1 \) and \( i_2 \) are equal ( \(i \) say )

So \( d_m = i+i-A \) or \( i = \, (A+d_m)/2 \)

Also the two corresponding angles of refraction are equal. \( r_1 = r_2 = r \) (say). So \( A = r+r = 2r \) or \( r = A/2 \)

\( \mu = \dfrac {sin \, i_1}{sin \, r_1} = \dfrac {sin \, i}{sin \, r} = \dfrac {sin \, (A+d_m)/2}{sin \, A/2} \)