Question

Find minimum deviation in a thin prism if refractive index \(μ = cot\bigg(\frac{A}{2}\bigg)\). Here \(A\) represents the angle of prism.

Answer 1

The correct answer is: \(δ = π-2A\). It is given that \(μ=\cot \frac{A}{2}\). We know that \(μ=\frac{\sin(\frac{δm+A}{2})}{\sin A/2}\). This implies \(\frac{\cos A/2}{\sin A/2}=\frac{\sin(\frac{δm+A}{2})}{\sin A/2}\), which further implies \(\sin(\frac{\pi}{2}-\frac{A}{2})=\sin(\frac{δm+A}{2})\). Therefore, \(\frac{\pi}{2}-\frac{A}{2}=\frac{δm+A}{2}\), leading to \(\pi-2A=δ m\).