Question

Sol. [Key Concept: For an equilateral prism $A{= 60}^{\circ}$ with $\delta_{\min} = \frac{A}{2}$: $\mu = \frac{\sin\frac{A + \delta_{\min}}{2}}{\sin\frac{A}{2}} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \sqrt{2}$.]

$${\mu_{p} = \mu_{s}\frac{sin\left( \frac{A + \delta_{\text{min}\text{~}}}{2} \right)}{sin\left( \frac{A}{2} \right)} = \frac{sin\frac{3A}{4}}{sin\frac{A}{2}} }{\mu_{p} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \sqrt{2} = \sqrt{2} }$$