For a glass prism $(\mu=\sqrt{3})$ the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism.
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We know that, $\mu=\frac{\sin\Big[\frac{(\text{A}+\text{D}_\text{m})}{2}\Big]}{\sin\Big(\frac{\text{A}}{2}\Big)}$
We are given that, the angle of minimum deviation is equal to the angle of the ptism i.e., $D_m = A$
Substituring $\mu=\sqrt{3}\text{ and D}_\text{m}=\text{A in }\mu=\frac{\sin\Big[\frac{(\text{A}+\text{D}_\text{m})}{2}\Big]}{\sin\Big(\frac{\text{A}}{2}\Big)}$, we get
$\sqrt{3}=\frac{\sin\text{A}}{\sin\frac{\text{A}}{2}}$
$\Rightarrow\ \sqrt{3}=\frac{2\sin\frac{\text{A}}{2}\cos\frac{\text{A}}{2}}{\sin\frac{\text{A}}{2}}\ \ \Big[\because\ \sin \text{A}=2\sin\frac{\text{A}}{2}\cos\frac{\text{A}}{2}\Big]$
$\Rightarrow\ \sqrt{3}=2\cos\frac{\text{A}}{2}$
$\Rightarrow\ \frac{\sqrt{3}}{2}=\cos\frac{\text{A}}{2}$
$\Rightarrow\ \frac{\text{A}}{2}=30^\circ$
$\Rightarrow\ \text{A}=60^\circ$
We are given that, the angle of minimum deviation is equal to the angle of the ptism i.e., $D_m = A$
Substituring $\mu=\sqrt{3}\text{ and D}_\text{m}=\text{A in }\mu=\frac{\sin\Big[\frac{(\text{A}+\text{D}_\text{m})}{2}\Big]}{\sin\Big(\frac{\text{A}}{2}\Big)}$, we get
$\sqrt{3}=\frac{\sin\text{A}}{\sin\frac{\text{A}}{2}}$
$\Rightarrow\ \sqrt{3}=\frac{2\sin\frac{\text{A}}{2}\cos\frac{\text{A}}{2}}{\sin\frac{\text{A}}{2}}\ \ \Big[\because\ \sin \text{A}=2\sin\frac{\text{A}}{2}\cos\frac{\text{A}}{2}\Big]$
$\Rightarrow\ \sqrt{3}=2\cos\frac{\text{A}}{2}$
$\Rightarrow\ \frac{\sqrt{3}}{2}=\cos\frac{\text{A}}{2}$
$\Rightarrow\ \frac{\text{A}}{2}=30^\circ$
$\Rightarrow\ \text{A}=60^\circ$
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