The refracting angle of prism is \(A\) and refractive index of material of prism is \(\cot \frac{A}{2}\). The angle of minimum deviation is \(A\)

Given, angle of prism \(=A\)
Refractive index of prism, \(\mu=\cot \frac{A}{2}\)
As, we know, \(\mu=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
where, \(\delta_{m}=\) angle of minimum deviation.
\(\Rightarrow \cot \left(\frac{A}{2}\right)=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
\(\Rightarrow \frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
\(\Rightarrow \sin \left(90^{\circ}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta_{m}}{2}\right)\)
\(\Rightarrow \frac{180^{\circ}-A}{2}=\frac{A+\delta_{m}}{2}\)
\(\Rightarrow \delta_{m}=180^{\circ}-2 A\)