Question

If the refractive index of the material of a prism is \(cot\bigg(\frac{A}{2}\bigg)\), where A is the angle of the prism, then the angle of minimum deviation will be:

• A. \(π-3A\)
• B. \(π-2A\)
• C. \(A\)
• D. \(\frac{A}{2}\)

Answer 1

The Correct Option is B

To determine the angle of minimum deviation (\( \delta_m \)) for a prism when the refractive index (\( n \)) is given as \( \cot\left(\frac{A}{2}\right) \), we proceed as follows: 

First, let's understand the relation between the refractive index, the angle of the prism (\( A \)), and the angle of minimum deviation (\( \delta_m \)). The formula for the refractive index \( n \) in terms of the angle of deviation is:

\(n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\)

Given that the refractive index \( n \) is \( \cot\left(\frac{A}{2}\right) \), we can set up the equation:

\(\cot\left(\frac{A}{2}\right) = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\)

We know that \(\cot\left(\frac{A}{2}\right) = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}\). Therefore, the equation becomes:

\(\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\)

Canceling \(\sin\left(\frac{A}{2}\right)\) from both sides, we get:

\(\cos\left(\frac{A}{2}\right) = \sin\left(\frac{A + \delta_m}{2}\right)\)

Using the identity \(\sin(\theta) = \cos\left(\frac{\pi}{2} - \theta\right)\), we have:

\(\sin\left(\frac{A + \delta_m}{2}\right) = \sin\left(\frac{\pi}{2} - \frac{A}{2}\right)\)

Equating angles since both are sine functions:

\(\frac{A + \delta_m}{2} = \frac{\pi}{2} - \frac{A}{2}\)

Solving for \( \delta_m \):

• Multiply through by 2 to eliminate the fraction: \(A + \delta_m = \pi - A\)
• Add \( A \) to both sides: \(\delta_m = \pi - 2A\)

Thus, the angle of minimum deviation (\( \delta_m \)) is \( \pi - 2A \), confirming that the correct answer is \(\pi - 2A\).

Answer 2

To find the angle of minimum deviation \( \delta_{\text{min}} \):

Step 1. Given Relation:  
  \( \cot \frac{A}{2} = \frac{\sin \frac{A + \delta_{\text{min}}}{2}}{\sin \frac{A}{2}} \)

Step 2. Rearrange and Simplify: Take the cosine of both sides:  
  \( \cos \frac{A}{2} = \sin \frac{A + \delta_{\text{min}}}{2} \)

Step 3. Solve for \( \delta_{\text{min}} \): Equate the arguments, giving:  
  \( \frac{A + \delta_{\text{min}}}{2} = \frac{\pi}{2} - \frac{A}{2} \)

  - Solving, we get:  
    \( \delta_{\text{min}} = \pi - 2A \)

Thus, the angle of minimum deviation is \( \pi - 2A \).