The refractive index of prism is $\mu=\sqrt{3}$ and the ratio of the angle of minimum deviation to the angle of prism is one. The value of angle of prism is _________$^\circ$.

<p>To find the value of the angle of the prism given the refractive index of the prism $\mu = \sqrt{3}$ and the ratio of the angle of minimum deviation ($\delta_m$) to the angle of the prism ($A$) is 1 (i.e., $\frac{\delta_m}{A} = 1 \Rightarrow \delta_m = A$), we can use the prism formula relating these variables.</p> <p>The formula that relates the angle of deviation $\delta$, refractive index $\mu$, angle of prism $A$, and angle of minimum deviation $\delta_m$ is given by:</p> <p>$$\mu = \frac{\sin \left(\frac{\delta_m + A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$$.</p> <p>Given that $\delta_m = A$, our formula becomes:</p> <p>$$\sqrt{3} = \frac{\sin \left(\frac{A + A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$$</p> <p>Simplifying this, we have:</p> <p>$\sqrt{3} = \frac{\sin(A)}{\sin \left(\frac{A}{2}\right)}$</p> <p>Now, recall the trigonometric identity:</p> <p>$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$</p> <p>By setting $\theta = \frac{A}{2}$, we get:</p> <p>$\sin(A) = 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)$</p> <p>Substituting back into our equation:</p> <p>$$\sqrt{3} = \frac{2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$$</p> <p>$\sqrt{3} = 2\cos\left(\frac{A}{2}\right)$</p> <p>Dividing by 2 and solving for $\cos\left(\frac{A}{2}\right)$:</p> <p>$\cos\left(\frac{A}{2}\right) = \frac{\sqrt{3}}{2}$</p> <p>This corresponds to an angle $\frac{A}{2}$ of $30^\circ$ since the cosine of 30 degrees is $\frac{\sqrt{3}}{2}$. Thus:</p> <p>$\frac{A}{2} = 30^\circ$</p> <p>So, the angle of the prism $A$ is:</p> <p>$A = 2 \times 30^\circ = 60^\circ$.</p> <p>Therefore, the value of the angle of the prism is $60^\circ$. </p>