The refractive index of the material of a glass prism is $\sqrt{3}$. The angle of minimum deviation is equal to the angle of the prism. What is the angle of the prism?

<p>$\textbf{Step 1: Symmetry at Minimum Deviation}$</p> <p>In a prism, when the deviation is minimum, the path of light is symmetric. This means that the angle of incidence ($i$) is equal to the angle of emergence ($i'$), and the light inside the prism makes equal angles with the prism faces. If the prism angle is $A$, then the refracted angle at each interface is:</p> <p>$r = \frac{A}{2}$</p> <p>$\textbf{Step 2: Relating Deviation to the Angles}$</p> <p>The formula for the deviation ($D$) in a prism is given by:</p> <p>$D = i + i' - A$</p> <p>At minimum deviation, $i = i'$, so:</p> <p>$D_{\text{min}} = 2i - A$</p> <p>The special condition given is that the minimum deviation is equal to the prism angle:</p> <p>$D_{\text{min}} = A$</p> <p>Thus:</p> <p>$A = 2i - A \quad \Longrightarrow \quad 2i = 2A \quad \Longrightarrow \quad i = A$</p> <p>$\textbf{Step 3: Applying Snell's Law}$</p> <p>At the first surface, Snell's law gives:</p> <p>$\sin i = \mu \sin r$</p> <p>Substitute the values $i = A$ and $r = \frac{A}{2}$:</p> <p>$\sin A = \mu \sin\frac{A}{2}$</p> <p>Given that the refractive index $\mu = \sqrt{3}$, we have:</p> <p>$\sin A = \sqrt{3} \sin\frac{A}{2}$</p> <p>$\textbf{Step 4: Solving the Equation}$</p> <p>Recall the double-angle formula for sine:</p> <p>$\sin A = 2 \sin\frac{A}{2} \cos\frac{A}{2}$</p> <p>Substitute this into the previous equation:</p> <p>$2 \sin\frac{A}{2} \cos\frac{A}{2} = \sqrt{3} \sin\frac{A}{2}$</p> <p>Assuming $\sin\frac{A}{2} \neq 0$, we can divide both sides by $\sin\frac{A}{2}$:</p> <p>$2 \cos\frac{A}{2} = \sqrt{3}$</p> <p>Solve for $\cos\frac{A}{2}$:</p> <p>$\cos\frac{A}{2} = \frac{\sqrt{3}}{2}$</p> <p>Since:</p> <p>$\cos 30^\circ = \frac{\sqrt{3}}{2}$</p> <p>It follows:</p> <p>$\frac{A}{2} = 30^\circ \quad \Longrightarrow \quad A = 60^\circ$</p> <p>$\textbf{Final Answer:}$</p> <p>The angle of the prism is $60^\circ$.</p>