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NEET PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSMedium

Question

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

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from RAY OPTICS AND OPTICAL INSTRUMENTS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSrefractingrefractivematerialminimumdeviation

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