One side of an equilateral prism is painted by a transparent material of refractive index \( n_2 \). The refractive index of prism is 1.6. The minimum value of \( n_2 \) required for total internal reflection from the painted face is _______.
Step 1: Understanding the Concept:
For a light ray to undergo Total Internal Reflection (TIR) at an interface, it must travel from a denser medium to a rarer medium.
The angle of incidence must be greater than or equal to the critical angle (\(i \ge \theta_c\)).
Based on the geometric setup of an equilateral prism, a horizontally entering ray hits the second face at a specific angle of incidence, allowing us to find the required refractive index. Step 2: Key Formula or Approach:
The critical angle is given by \(\sin \theta_c = \frac{n_2}{n_1}\), where \(n_1 = 1.6\).
For TIR to occur, the condition is \(\sin i \ge \sin \theta_c \implies \sin i \ge \frac{n_2}{n_1}\).
Rearranging gives the maximum possible value for \(n_2\) to allow TIR: \(n_2 \le n_1 \sin i\). Step 3: Detailed Explanation:
The prism is equilateral, meaning all its angles are \(60^\circ\).
The light ray is incident normally on the first face, meaning it passes through without any deviation.
It then hits the second face. The normal to the second face makes a \(30^\circ\) angle with the horizontal.
By geometry, the angle of incidence \(i\) at the second face is \(60^\circ\).
Apply the TIR condition:
\[ \sin(60^\circ) \ge \frac{n_2}{1.6} \]
Substitute the value of \(\sin(60^\circ)\):
\[ \frac{\sqrt{3}}{2} \ge \frac{n_2}{1.6} \]
Solve for \(n_2\):
\[ n_2 \le 1.6 \times \frac{\sqrt{3}}{2} \]
\[ n_2 \le 0.8\sqrt{3} \]
Convert the decimal to a fraction to match the options:
\[ 0.8 = \frac{8}{10} = \frac{4}{5} \]
So, \(n_2 \le \frac{4\sqrt{3}}{5}\).
The limiting (or maximum boundary) value for \(n_2\) is \(\frac{4\sqrt{3}}{5}\). Step 4: Final Answer:
The required limiting value of \(n_2\) is \(\frac{4\sqrt{3}}{5}\).
