Finding out the minimum deviation angle

Question

enter image description here

Let a light incident on a prisom at right angle. I want to determine the minimum deviation angle.

I know a relation $$ \frac{\cos i_1}{\cos r_1} = \frac{\sin i_1}{\sin r_1}$$

and minimum angle of deviation $$\delta_m = i_1 +i_2 -A$$ It can be seen that, $i_1$ is given but how to get $i_2$ and A. A is the prism angle. and $i_1 =90 $ and $i_2 $ is the angle of the outgoing ray.

Whenever something calls for a minimum or a maximum, you must remember calculus. — Torsten Hĕrculĕ Cärlemän, Dec 28 '13 at 09:39
If you want $A$ and $i_2$ for minimum $\delta_m$, you will need $\mu$ to reduce the number of variables by establishing a relation between the two of them. — stochastic13, Dec 28 '13 at 12:53
This contains a derivation of the minimum deviation angle: http://web.centre.edu/jason.neiser/Classes/Phy230Files/MinimumDeviation.pdf — Sandesh Kalantre, Dec 28 '13 at 14:32
possible duplicate of Analytic solution for angle of minimum deviation? — Stefan Bischof, Feb 03 '14 at 18:17

score 1 · Answer 1 · edited Jan 04 '14 at 12:03

Well first off you have to think about how much deviation occurs at the first face of the prism, and how much occurs at the second face. In general they will not be the same.

But if they are different, then one is larger than the other. But then simply reversing the ray, will flip the unequal deviations.

Ergo, one concludes, that there must be two different incidence angles on the first face, that yield the same total deviation. These two cases must be on either side of the symmetrical case where the ray path in the prism forms an isosceles triangle with the two sides meeting at A. This gives the same incidence ray angles on the two faces.

One also concludes, that this symmetrical case must be a stationary point, so if it isn't the minimum deviation case, then it must be a maximum deviation case.

It is in fact the minimum, and this symmetrical ray case, is commonly called the "minimum deviation prism".

So then what is the deviation in this case.

Luckily, I actually learned that in school, so you can check it for yourself.

If (A) is the prism angle, and (D) is the total deviation angle of the ray, and (N) is the refractive index of the prism at the test wavelength then we can show that:-

$$N = \dfrac{\sin\left(\dfrac{A+D}2\right)}{\sin(A/2)}$$

I'll let you calculate D min from that.

It's an important case since slight rocking of the prism about the minimum deviation condition gives very little change in the total deviation.

If you happen to have a certain model of Logitech laser mouse, you may find in it an off axis laser collimator, that has a thick lens built onto the faces of a minimum deviation prism, for just that exact reason. The lens increases the laser divergence angle at the first face, and bends it 15 degrees, then it collimates the beam at the second surface, and bends it another 15 degrees, giving a collimated laser beam at 30 degrees of tilt from the laser chip optical axis.

Finding out the minimum deviation angle

1 Answers1