Question

The refractive index of the material of a glass prism of refracting angle \(60^\circ\) and angle of minimum deviation \(30^\circ\) is

• A. \(\frac{1}{\sqrt{2}}\)
• B. \(\sqrt{2}\)
• C. \(\frac{1}{\sqrt{3}}\)
• D. \(\sqrt{3}\)

Hint:

Remember that the angle of incidence for minimum deviation is \(i = (A + \delta_m)/2\). In this case \(i = 45^\circ\), which is a common value for such problems.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When a prism is in its minimum deviation position, the refracted ray inside the prism is parallel to its base (for an isosceles prism).
The refractive index of the prism material is related to the prism angle and the angle of minimum deviation.

Step 2: Key Formula or Approach:
Prism Formula: \(\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\)
where \(A\) is the angle of the prism and \(\delta_m\) is the angle of minimum deviation.

Step 3: Detailed Explanation:
Given:
Prism angle \(A = 60^\circ\)
Minimum deviation \(\delta_m = 30^\circ\)
Applying the formula:
\[ \mu = \frac{\sin\left(\frac{60 + 30}{2}\right)}{\sin\left(\frac{60}{2}\right)} \]
\[ \mu = \frac{\sin(45^\circ)}{\sin(30^\circ)} \]
Substitute the values \(\sin 45^\circ = 1/\sqrt{2}\) and \(\sin 30^\circ = 1/2\):
\[ \mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

Step 4: Final Answer:
The refractive index of the prism material is \(\sqrt{2}\).