Question

The refracting angle of prism is A and refractive index of material of prism is $\cot \frac{A}{2}$ The angle of minimum deviation is

• A. $180^\circ -2 A$
• B. $90^\circ - A$
• C. $180^\circ +2 A$
• D. $180^\circ -3 A$

Answer 1

The Correct Option is A

To determine the angle of minimum deviation for a prism with given conditions, we follow these steps:

Step 1: Understanding the Problem:

The question provides that the refracting angle of the prism is \(A\) and the refractive index of the material of the prism is \(\cot \frac{A}{2}\). We need to find the angle of minimum deviation, denoted as \(\delta\).

Step 2: Formula for Minimum Deviation:

The formula for the refractive index (\( \mu \)) in terms of the angle of minimum deviation (\(\delta\)) and the refracting angle (\(A\)) is given by:

\(\mu = \frac{\sin \left(\frac{A + \delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)

Step 3: Substitute the given refractive index:

From the question, we have \(\mu = \cot \frac{A}{2}\). Hence, we can equate:

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

Since \(\cot \theta = \frac{\cos \theta}{\sin \theta}\), we rewrite the refractive index in terms of trigonometric components:

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

By simplifying, we have:

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

Step 4: Solve for \( \delta \):

The identity \(\sin x = \cos (90^\circ - x)\) is useful here:

Therefore, \(\sin \left(\frac{A + \delta}{2}\right) = \cos \left(90^\circ - \frac{A + \delta}{2}\right)\)

Thus, equating inside the arcs:

\(\frac{A + \delta}{2} = 90^\circ - \frac{A}{2}\)

Solving for \(\delta\):

\(\frac{A + \delta}{2} = \frac{180^\circ - A}{2}\)

Cross multiplying gives:

\(A + \delta = 180^\circ - A\)

\(\delta = 180^\circ - 2A\)

Conclusion:

Therefore, the correct option is \(\delta = 180^\circ - 2A\).