Does angle of minimum deviation depend on angle of prism? I know that angle of deviation does but I don't feel like the particular behavior is relevant for the angle of deviation as well? Please help me clarify it
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2 Answers
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For a prism,
$D = I + E - P$
where $D$ is the angle of deviation, $I$ is the incidence angle and E is the angle of emergence with $P$ being the physical angle of the prism.
And when you want the minimum angle of deviation, that is the angle of incidence is the same as the angle of emergence,
$D = 2I - P = 2E - P$
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Minimum deviation occurs when the ray passes symmetrically through the prism. You should use this to show for yourself that $$\sin \left[\frac 12 (A+D_{min})\right]=n \sin \left[\frac A2 \right]$$ in which $n$ is the refractive index of the prism material and $A$ is the angle between the faces via which the ray enters and leaves.
The answer to your question can easily be deduced from this equation.
answered Oct 16, 2020 at 7:36
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$\begingroup$ "Minimum deviation occurs when the ray passes symmetrically through the prism". Philip, what is the reasoning behind this assertion ??? If the answer is : but then we would have at least two minima, then how do we know a priori that there exists only one minimum ??? $\endgroup$– VoulkosCommented Oct 16, 2020 at 14:03
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$\begingroup$ If the ray's path is not symmetrical then the angle of incidence will be different from the angle of emergence. But rays are reversible, so we know that two different angles, $I$, of incidence will give the same angle of deviation, $D$. Imagine now a graph of $D$ against $I$. In the symmetrical case the two values of $I$ have merged into a single value. Clearly we have either a maximum or a minimum in the graph of $D$ against $I$. I'm sure that you can show for yourself that it is in fact a minimum. $\endgroup$ Commented Oct 16, 2020 at 16:58
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$\begingroup$ Philip, what I try to say in my previous comment is shown in this Figure : imgur.com/a/0lHgcvw. But now I realize that the deviation curve $\delta(i_1)$ must satisfy a condition with respect to the bisector of the 1rst quadrant. $\endgroup$– VoulkosCommented Oct 16, 2020 at 23:49
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$\begingroup$ And YES, you are absolutely right. This condition demands ...the ray to pass symmetrically through the prism when minimum deviation occurs. Anyway, in my answer here : Analytic solution for angle of minimum deviation? this symmetry is proved in an intermediate step, see equations (07)-(10). May be I must append an addendum there to show how this symmetry helps to make things simple. $\endgroup$– VoulkosCommented Oct 16, 2020 at 23:50
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1$\begingroup$ @Frobenius Many thanks. I'm an IT ignoramus, but I'll look into GeoGebra. $\endgroup$ Commented Oct 17, 2020 at 12:59