Question

A ray of light passing through an equilateral prism is having velocity $2.12 \times 10^8 \text{ m/s}$ in the prism material, then the minimum angle of deviation is _______ degrees.

• A. 45
• B. 30
• C. 28
• D. 58

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the refractive index of the prism material using the speed of light and then use the prism formula for minimum deviation.
Step 2: Key Formula or Approach:
1. $\mu = \frac{c}{v}$.
2. For a prism, $\mu = \frac{\sin[(A+\delta_m)/2]}{\sin(A/2)}$.
Step 3: Detailed Explanation:
First, calculate the refractive index $\mu$:
$c = 3 \times 10^8 \text{ m/s}$, $v = 2.12 \times 10^8 \text{ m/s}$.
\[ \mu = \frac{3 \times 10^8}{2.12 \times 10^8} \approx 1.414 = \sqrt{2} \]
For an equilateral prism, angle $A = 60^\circ$.
Using the prism formula:
\[ \sqrt{2} = \frac{\sin \left( \frac{60^\circ + \delta_m}{2} \right)}{\sin \left( \frac{60^\circ}{2} \right)} \]
\[ \sqrt{2} = \frac{\sin \left( \frac{60^\circ + \delta_m}{2} \right)}{\sin 30^\circ} = \frac{\sin \left( \frac{60^\circ + \delta_m}{2} \right)}{0.5} \]
\[ \sin \left( \frac{60^\circ + \delta_m}{2} \right) = \sqrt{2} \times 0.5 = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \]
We know $\sin 45^\circ = \frac{1}{\sqrt{2}}$, so:
\[ \frac{60^\circ + \delta_m}{2} = 45^\circ \implies 60^\circ + \delta_m = 90^\circ \]
\[ \delta_m = 30^\circ \]
Step 4: Final Answer:
The minimum angle of deviation is $30^\circ$.