# Refraction through a Prism

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