# Angle of minimum deviation

A prism of angle $60^\circ$ deviates a ray of light through $40^\circ$ for two angles of incidence which differ by $11^\circ$. What is the refractive index of the glass of the prism?

If we are given two angles for which the prism gives the same amount of deviation, can we arrive at angle of minimum deviation by taking average of the two?

I thought it might be possible after looking at the graph and I was getting answers pretty close to the solutions in my textbook. Can someone explain?

I used the formula:

$$I_1 + I_2= a+ \delta$$

(Angle of incidence, emergence, angle of prism and angle of deviation respectively)

Then I got 44.5 degrees and 55.5 degrees respectively. I took the average and assumed that was angle of incidence for which I would get minimum deviation. It was close to the textbook value. Am I allowed to do that?

• Please give the textbook's problem and your approach. Commented Oct 12, 2015 at 9:17
• I don't think you're allowed to do that. The average value of $i_1$ and $i_2$ is same as the average value of $A$ and $\delta$. If the average value is equal to incidence angle for minimum deviation which is a constant, then that means that $\delta$ is a constant which is wrong. Commented Oct 12, 2015 at 10:27

To get the incidence angle for minimum deviation($\theta$), apply Snell's law for the two refractions and eliminate the unknown angles of refraction(use $r_1 + r_2 = A$) to get relation between $i_1$, $i_2$ and the refractive index and solve to get the value of refractive index. When $\delta$ is minimum, $r_1 = \frac{A}{2} = \frac{\pi}{6}$ which means $\sin \theta = n/2$.