Finding the minimum deviation angle


From the graph of $D_ f(\alpha )$ we know that there is one minimum. We’ll find it by setting the derivative of $D_ f(\alpha )$ equal to 0. We have

  \[ D_ f(\alpha )=180^\circ +2\alpha -4\beta , \]    

where

  \[ \beta =\beta (\alpha )=arcsin{\frac{\sin {\alpha }}{n_{f,w}}.} \]    

To make things look simpler we’ll write $n$ for $n_{f,w}.$

Now

  \[ D_ f^\prime (\alpha )=2-4\beta ^\prime (\alpha ). \]    

The derivative of the function $f(x)=\arcsin (x)$ is

  \[ f^\prime (x)=\frac{1}{\sqrt {1-x^2}}. \]    

Using the chain rule gives

  \[ \beta ^\prime (\alpha )=\frac{\cos {\alpha }}{n\sqrt {1-\frac{\sin ^2{\alpha }}{n^2}}}=\frac{\cos {\alpha }}{\sqrt {n^2-\sin ^2{\alpha }}}. \]    

Thus

  \[ D_ f^\prime (\alpha )=2-\frac{4\cos {\alpha }}{\sqrt {n^2-\sin ^2{\alpha }}}. \]    

Setting $D_ f(\alpha )=0$ gives

  \[ 2-\frac{4\cos {\alpha }}{\sqrt {n^2-\sin ^2{\alpha }}}=0. \]    

Noting that $\sin ^2{\alpha }=1-\cos ^2{\alpha }$ and rearranging gives

  \[ \cos ^2{\alpha }=\frac{n^2-1}{3}. \]    

Thus, $D_ f^\prime (\alpha )=0$ for

  \[ \alpha _ m=\arccos {\sqrt {\frac{n^2-1}{3}}}. \]    

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