Minimum angle of deviation - Thin prism

Question

I have two questions here.

The first is why can we get the minimum angle of deviation of a trianglar glass prism when only the angle of incidence equals the angle of emergence.

It seems it's a maths-related problem and I know this question was asked before several times but never found a mathematical proof for it.

I tried to figure out a formula which relates the angle of deviation directly to the angle of incidence, assuming we are working with the same prism (which means constant refractive index and apex angle)

We already know that $\alpha = \phi + \theta - A$

Where $\alpha$ is the angle of deviation, $\phi$ is the angle of incidence, $\theta$ is the angle of emergence and $A$ is the apex angle

And I found that $\sin\theta = \sin A\cdot\sqrt{n^2 - \sin^2\phi} - \sin\phi \cos A$

Where $n$ is the refractive index of the prism.

So the angle of deviation can be calculated from this relation:

$$\alpha = \arcsin\left(\sin A\cdot\sqrt{n^2 - \sin^2\phi} - \sin\phi \cos A\right) + \phi - A$$

This is where my maths stopped, I can't go further in my proof. What's next? How can I prove it?

My second question is, the thin prism (The prism whose apex is less than 10 degrees) is always set at the position of minimum deviation.

My question is why? Can anyone explain this?

Analytic solution for angle of minimum deviation here. I do not understand your second question. In what context is the thin prism set at minimum deviation. — Farcher, Dec 19 '18 at 16:24
That's what I found written in my textbook, that the thin prism is always set at the position of minimum deviation. To be honest, I don't understand what it means by that but that's exactly what's written! — Yousef Essam, Dec 19 '18 at 16:32
No, it was not an experiment but it's just a description of the thin prism "A prism whose apex is less than 10 degrees and always set at the position of minimum deviation" — Yousef Essam, Dec 19 '18 at 16:36
I am a high school student and it was in my high school physics book — Yousef Essam, Dec 19 '18 at 17:04
Differentiate the equation and put derivative equal to zero. You will get $i=e$ — Bully Maguire, Nov 08 '19 at 08:53

score 1 · Answer 1 · edited Jun 04 '20 at 16:03

For the first question the simple answer is reversibility principle in optics. This principle states that when you trace a special ray through an optical component from right to left, or from left to right the ray path must be same.

$$\phi_1=\phi_2$$ $$\phi_1^{'}=\phi_2^{'}$$ $$\beta=\gamma$$

For a prism if incidence and exit ray angles be different at minimum deviation, so there is two minimum deviation angles when you trace ray from right to left and when left to right. And because there is only when minimum for each prism, both incidence and exit angle must be same.

For the last question the condition for a thin prism satisfies when $\alpha$ is too small. For example about 0.8 degree for flint glass (n=1.6). For this small angle applying snell's law states that both entrance and existence angles are equal, and prism always acts in minimum deviation condition.

Minimum angle of deviation - Thin prism

1 Answers1