Question

What will be the angle of minimum deviation by a thin prism of 10° and refractive index 2?

Hint:

The angle of minimum deviation for a prism is constrained by the sine function, which cannot exceed 1. Ensure the values for angle and refractive index are practical.

Answer 1

Step 1: Formula for Minimum Deviation.
The angle of minimum deviation \( D_{\text{min}} \) in a prism is related to the prism angle \( A \) and the refractive index \( n \) by the following formula: \[ \sin \left( \frac{D_{\text{min}}}{2} \right) = \frac{n-1}{\sin \left( \frac{A}{2} \right)} \]
Step 2: Substituting Given Values.
Given the prism angle \( A = 10^\circ \) and refractive index \( n = 2 \), we can substitute these values into the formula: \[ \sin \left( \frac{D_{\text{min}}}{2} \right) = \frac{2-1}{\sin \left( \frac{10^\circ}{2} \right)} = \frac{1}{\sin 5^\circ} \]
Step 3: Solving for \( D_{\text{min}} \).
Using \( \sin 5^\circ \approx 0.0872 \), we get: \[ \sin \left( \frac{D_{\text{min}}}{2} \right) = \frac{1}{0.0872} \approx 11.46 \] Since the sine function cannot exceed 1, this indicates that the given parameters result in an impractical situation, where the angle of minimum deviation cannot be achieved under normal conditions.

Step 4: Conclusion.
The given combination of prism angle and refractive index does not yield a physically valid solution for the angle of minimum deviation.