Question

Cross-section view of a prism is the equilateral triangle ABC shown in the figure. The minimum deviation is observed using this prism when the angle of incidence is equal to the prism angle. The time taken by light to travel from P (midpoint of BC) to A is ________ $\times 10^{-10}$ s. (Given, speed of light in vacuum $= 3 \times 10^8$ m/s and $\cos 30^\circ = \frac{\sqrt{3}}{2}$)
\begin{center} \includegraphics[width=0.3\textwidth]{prism_diagram} \end{center}

Hint:

For any prism at minimum deviation, remember the symmetry: \(r = A/2\). This simplifies finding the refractive index significantly if the incidence angle is known.

Answer 1

Correct Answer: 5

Step 1: Understanding the Concept:
The time taken for light to travel a distance $d$ in a medium is $t = d/v$, where $v = c/\mu$ is the speed of light in the medium and $\mu$ is the refractive index.
Step 2: Key Formula or Approach:
1. For an equilateral triangle, prism angle \( A = 60^\circ \).
2. At minimum deviation: \( r_1 = r_2 = A/2 = 30^\circ \).
3. Snell's Law: \( \mu = \frac{\sin i}{\sin r} \).
Step 3: Detailed Explanation:
Given: \( i = A = 60^\circ \).
Calculate refractive index \(\mu\):
\[ \mu = \frac{\sin 60^\circ}{\sin 30^\circ} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3} \]
Calculate velocity of light in prism:
\[ v = \frac{c}{\mu} = \frac{3 \times 10^8}{\sqrt{3}} = \sqrt{3} \times 10^8 \text{ m/s} \]
Calculate distance PA:
In an equilateral triangle with side \( a = 10 \) cm, the height \( h \) (distance PA) is:
\[ h = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3} \text{ cm} = 5\sqrt{3} \times 10^{-2} \text{ m} \]
Calculate time taken:
\[ t = \frac{h}{v} = \frac{5\sqrt{3} \times 10^{-2}}{\sqrt{3} \times 10^8} = 5 \times 10^{-10} \text{ s} \]
Step 4: Final Answer:
The time taken is \( 5 \times 10^{-10} \) s. The numerical value is 5.