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:
- \(π-3A\)
- \(π-2A\)
- \(A\)
- \(\frac{A}{2}\)
The Correct Option is B
Approach Solution - 1
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 \(\pi - 2A\).
Approach Solution -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 \( \pi - 2A \).
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