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The refracting angle of a prism is AA, and 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 is related to the angle of the prism (AA) and the angle of minimum deviation (δm\delta_m) by the formula: μ=sin⁡(A+δm2)sin⁡(A2)\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}
  2. Given Condition: The refractive index is given as μ=cot⁡(A/2)\mu = \cot(A/2). We can write cot⁡(A/2)\cot(A/2) as: μ=cos⁡(A/2)sin⁡(A/2)\mu = \frac{\cos(A/2)}{\sin(A/2)}
  3. Equating Expressions: sin⁡(A+δm2)sin⁡(A2)=cos⁡(A/2)sin⁡(A/2)\frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\cos(A/2)}{\sin(A/2)}
  4. Solving for Deviation: sin⁡(A+δm2)=cos⁡(A/2)\sin\left(\frac{A + \delta_m}{2}\right) = \cos(A/2) Using the trigonometric identity cos⁡θ=sin⁡(90∘−θ)\cos \theta = \sin(90^{\circ} - \theta): sin⁡(A+δm2)=sin⁡(90∘−A/2)\sin\left(\frac{A + \delta_m}{2}\right) = \sin(90^{\circ} - A/2) Comparing the angles: A+δm2=90∘−A2\frac{A + \delta_m}{2} = 90^{\circ} - \frac{A}{2} A+δm=180∘−AA + \delta_m = 180^{\circ} - A δm=180∘−2A\delta_m = 180^{\circ} - 2A
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