Mathematical derivation for the minimum deviation angle for a prism

[leright]

I am looking for a mathematical derivation for the idea that symmetry of the light beam path through a prism gives rise to a minimum deviation.

Thanks.

 

[leright]

and I just put this in the WRONG Forum. Sorry about that.

 

[Integral]

I think this forum is fine for such a question.

May I ask, deviation of what?

 

[Pythagorean]

http://www.nasatech.com/Briefs/Sept99/LAR14637.html"?

 

[jtbell]

May I ask, deviation of what?

I think he's referring to this:

http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/prism.html#c2

leright, are you asking why it is that the path shown in the diagram (the one in which section AB is parallel to the base of the prism) must be the one with minimum deviation (\delta)?

I think the only rigorous way to do it is to find an equation for \delta in terms of the incident angle (or some other convenient angle), then find the minimum via the usual calculus technique: take the derivative and set it equal to zero. See for example

http://scienceworld.wolfram.com/physics/Prism.html

Some books use a "symmetry argument" which goes something like this: Suppose for the sake of argument that the minimum deviation occurs when the entrance and exit angles are not equal. In a ray diagram, you can always reverse the direction of a light ray and get another valid light ray. In this case, reversing the ray switches the values of the entrance and exit angles. So there are two different values for the entrance angle that give minimum deviation. But if there's only one minimum, this can't be true. Therefore the initial supposition must be false, and the entrance and exit angles must be equal at minimum deviation.

Of couse, in order to make the assumption that I've put in boldface above, you have to know something in advance about how the deviation angle varies with entrance angle, for example by measuring it experimentally and making a graph of deviation angle versus entrance angle. Otherwise, how do you know the graph isn't actually W-shaped, with two minima?

 

[leright]

I think he's referring to this:

http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/prism.html#c2

leright, are you asking why it is that the path shown in the diagram (the one in which section AB is parallel to the base of the prism) must be the one with minimum deviation (\delta)?

I think the only rigorous way to do it is to find an equation for \delta in terms of the incident angle (or some other convenient angle), then find the minimum via the usual calculus technique: take the derivative and set it equal to zero. See for example

http://scienceworld.wolfram.com/physics/Prism.html

Some books use a "symmetry argument" which goes something like this: Suppose for the sake of argument that the minimum deviation occurs when the entrance and exit angles are not equal. In a ray diagram, you can always reverse the direction of a light ray and get another valid light ray. In this case, reversing the ray switches the values of the entrance and exit angles. So there are two different values for the entrance angle that give minimum deviation. But if there's only one minimum, this can't be true. Therefore the initial supposition must be false, and the entrance and exit angles must be equal at minimum deviation.

Of couse, in order to make the assumption that I've put in boldface above, you have to know something in advance about how the deviation angle varies with entrance angle, for example by measuring it experimentally and making a graph of deviation angle versus entrance angle. Otherwise, how do you know the graph isn't actually W-shaped, with two minima?

Yeah, I figured the best way to do it would be to find the derivative of the deviation angle function, which is a function of the angle of incidence and the apex angle and then set it equal to zero, and then solve for the angle of incidence.

Thanks a lot.