Question

The figure shows the cross-section of a hollow cylindrical tank, 2.2 m in diameter, which is half filled with water (refractive index of 1.33). The space above the water is filled with a gas of unknown refractive index. A small laser moves along the bottom surface and aims a light beam towards the center (see figure). When the laser moves a distance of \( S = 1.09\,\text{m} \) or beyond from the lowest point in the water, no light enters the gas. Identify the correct statement(s). (Speed of light = \( 3 \times 10^8\,\text{m/s} \)) 

• A. The refractive index of the gas is 1.05
• B. The time taken for the light beam to travel from the laser to the rim of the tank when \( S<1.09\,\text{m} \) is \( 8.9\,\text{ns} \)
• C. The time taken for the light beam to travel from the laser to the rim of the tank when \( S>1.09\,\text{m} \) is \( 9.7\,\text{ns} \)
• D. The critical angle for the water–gas interface is \( 56.77^\circ \)

Hint:

For total internal reflection, light must move from denser to rarer medium, and the critical angle is given by \( \sin\theta_c = \frac{n_2}{n_1} \).

Answer 1

The Correct Option is B, C, D

Step 1: Understand the situation.
The light from the laser travels from water (\( n_1 = 1.33 \)) to gas (\( n_2 = ? \)). When \( S = 1.09\,\text{m} \), the light strikes the interface at the critical angle. The tank radius is \( R = 1.1\,\text{m} \). At critical angle \( \theta_c \), \[ \sin\theta_c = \frac{S}{R} = \frac{1.09}{1.1} = 0.991. \] Step 2: Apply Snell’s law for critical angle.
\[ \sin\theta_c = \frac{n_2}{n_1}. \] \[ n_2 = n_1 \sin\theta_c = 1.33 \times 0.991 = 1.05. \] Step 3: Compute the critical angle.
\[ \theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right) = \sin^{-1}\left(\frac{1.05}{1.33}\right) = 56.77^\circ. \] Step 4: Final Answer.
Hence, the refractive index of the gas is \( 1.05 \), and the critical angle is \( 56.77^\circ. \)