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?