The refractive index of a prism with apex angle \( A \) is \( \cot \frac{A}{2} \). The angle of minimum deviation is:
- \( \delta_m = 180^\circ - A \)
- \( \delta_m = 180^\circ - 3A \)
- \( \delta_m = 180^\circ - 4A \)
- \( \delta_m = 180^\circ - 2A \)
The Correct Option is D
Approach Solution - 1
To determine the angle of minimum deviation \( \delta_m \) for a prism, we begin with the fundamental equation relating the refractive index \( n \), the apex angle \( A \), and the angle of minimum deviation.
The relation for a prism is given by:
\(n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\)
We are given that the refractive index of the prism is \( n = \cot \frac{A}{2} \). Using trigonometric identities, we can rewrite \( \cot \frac{A}{2} \) as:
\(\cot \frac{A}{2} = \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}\)
Setting these expressions equal, we have:
\(\frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}\)
This simplifies to:
\(\sin\left(\frac{A + \delta_m}{2}\right) = \cos \frac{A}{2}\)
Using the trigonometric identity \( \sin(\theta) = \cos(90^\circ - \theta) \), we can equate:
\(\frac{A + \delta_m}{2} = 90^\circ - \frac{A}{2}\)
Solving the equation for \( \delta_m \):
\(\frac{A + \delta_m}{2} + \frac{A}{2} = 90^\circ\)
\(\frac{A + \delta_m + A}{2} = 90^\circ\)
\(A + \delta_m = 180^\circ - A\)
\(\delta_m = 180^\circ - 2A\)
Hence, the angle of minimum deviation is:
The correct answer is: \(\delta_m = 180^\circ - 2A\)
Approach Solution -2
Given:
\[ \mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \frac{A}{2}} \]
Using the relation:
\[ \cos \frac{A}{2} = \sin \left( \frac{A + \delta_m}{2} \right) \]
We get:
\[ \delta_m = \pi - 2A \]
Therefore:
\[ \delta_m = 180^\circ - 2A \]
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