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

Answer 1

The Correct Option is D

To solve the problem of determining the angle of minimum deviation for a prism with a refracting angle \( A \) and a refractive index of the material of the prism given as \( \cot \frac{A}{2} \), we can use the formula for the angle of deviation of a prism.

The angle of minimum deviation \( D \) for a prism is given by the formula:

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

where \( n \) is the refractive index of the material of the prism.

Substituting the refractive index \( n = \cot \frac{A}{2} \) into the formula, we get:

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

Using the identity \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), the equation becomes:

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

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

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

Now, using the co-function identity, \( \sin(90^\circ - x) = \cos x \), rewrite the equation as:

\sin\left(\frac{A + D}{2}\right) = \sin\left(90^\circ - \frac{A}{2}\right)

This implies:

\frac{A + D}{2} = 90^\circ - \frac{A}{2}

Solving for \( D \), we multiply each term by 2:

A + D = 180^\circ - A

Rearranging gives:

D = 180^\circ - 2A

Thus, the angle of minimum deviation is 180^\circ - 2A.