Question

The refracting angle of a prism is \({}'A'\) and its refractive index is \(\cot \left(\frac{A}{2}\right)\). The minimum deviation produced by the prism is

• A. \((90)^{\circ }-\frac{3A}{2}\)
• B. \((180)^{\circ }-2A\)
• C. \((180)^{\circ }-A\)
• D. \((90)^{\circ }-2A\)

Hint:

Put the refractive index into the minimum deviation prism formula and solve for the deviation.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For a prism at minimum deviation, \(\mu=\frac{\sin\frac{A+\delta_{m}}{2}}{\sin\frac{A}{2}}\).

Step 2: Calculation
With \(\mu=\cot\frac{A}{2}\): \(\sin\frac{A+\delta_{m}}{2}=\cot\frac{A}{2}\sin\frac{A}{2}=\cos\frac{A}{2}=\sin\left(90^{\circ}-\frac{A}{2}\right)\). Hence \(\frac{A+\delta_{m}}{2}=90^{\circ}-\frac{A}{2}\).

Step 3: Compare the options
Solving, \(A+\delta_{m}=180^{\circ}-A\), so \(\delta_{m}=180^{\circ}-2A\), which is the second option.

Final Answer:
The minimum deviation is \(180^{\circ}-2A\). \[ \boxed{\text{(B) }180^{\circ}-2A} \]