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Physics Question on Refraction of Light

Find minimum deviation in a thin prism if refractive index μ=cot(A2)μ = cot\bigg(\frac{A}{2}\bigg). Here AA represents the angle of prism.

Answer

The Correct Answer is : δ=π−2Aδ = π-2A
It is given μ=cot⁡A2μ=\cot \frac{A}{2}
We know that:
μ=sin⁡(δm+A2)sin⁡A/2\mu=\frac{\sin(\frac{δm+A}{2})}{\sin A/2}
⇒cos⁡A/2sin⁡A/2=sin⁡(δm+A2)sin⁡A/2⇒\frac{\cos A/2}{\sin A/2}=\frac{\sin(\frac{δm+A}{2})}{\sin A/2}
⇒sin⁡(π2−A2)=sin⁡(δm+A2)⇒\sin(\frac{\pi}{2}-\frac{A}{2})=\sin(\frac{δm+A}{2})
⇒π2−A2=δm+A2⇒\frac{\pi}{2}-\frac{A}{2}=\frac{δm+A}{2}
⇒π−2A=δm⇒\pi-2A=δ m