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The refracting angle of a prism is AA, and the refractive index of the material of the prism is cot⁡(A/2)\cot(A/2). The angle of minimum deviation is:

A

180∘−3A180^\circ - 3A

B

180∘−2A180^\circ - 2A

C

90∘−A90^\circ - A

D

180∘+2A180^\circ + 2A

Step-by-Step Solution

  1. Prism Formula: The refractive index (μ\mu) of a prism with refracting angle AA and angle of minimum deviation δm\delta_m is given by: μ=sin⁡(A+δm2)sin⁡(A/2)\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin(A/2)}
  2. Substitute Given Data: We are given μ=cot⁡(A/2)\mu = \cot(A/2). cot⁡(A/2)=sin⁡(A+δm2)sin⁡(A/2)\cot(A/2) = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin(A/2)}
  3. Trigonometric Identity: Recall that cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta}. cos⁡(A/2)sin⁡(A/2)=sin⁡(A+δm2)sin⁡(A/2)\frac{\cos(A/2)}{\sin(A/2)} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin(A/2)}
  4. Simplify and Solve: Canceling sin⁡(A/2)\sin(A/2) from the denominator (assuming A≠0A \neq 0): cos⁡(A/2)=sin⁡(A+δm2)\cos(A/2) = \sin\left(\frac{A + \delta_m}{2}\right) Using cos⁡θ=sin⁡(90∘−θ)\cos \theta = \sin(90^\circ - \theta): sin⁡(90∘−A/2)=sin⁡(A+δm2)\sin(90^\circ - A/2) = \sin\left(\frac{A + \delta_m}{2}\right) 90∘−A2=A+δm290^\circ - \frac{A}{2} = \frac{A + \delta_m}{2} Multiplying by 2: 180∘−A=A+δm180^\circ - A = A + \delta_m δm=180∘−2A\delta_m = 180^\circ - 2A
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