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 find the angle of minimum deviation (\( \delta_m \)) for a prism when the refractive index (\( n \)) equals \( \cot\left(\frac{A}{2}\right) \), follow these steps:

The relationship between refractive index (\( n \)), prism angle (\( A \)), and angle of minimum deviation (\( \delta_m \)) is given by:

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

Substitute the given refractive index, \( n = \cot\left(\frac{A}{2}\right) \):

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

Using the identity \( \cot\left(\frac{A}{2}\right) = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} \), the equation becomes:

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

Cancel \( \sin\left(\frac{A}{2}\right) \) from both sides:

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

Apply the identity \( \sin(\theta) = \cos\left(\frac{\pi}{2} - \theta\right) \):

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

Equate the arguments of the sine functions:

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

Solve for \( \delta_m \):

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

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