Question

A laser pulse is sent from ground level to the bottom of a concrete water tank at normal incidence. The tank is filled with water up to 2 m below the ground level. The reflected pulse from the bottom of the tank travels back and hits the detector. The round-trip time elapsed between sending the laser pulse, the pulse hitting the bottom of the tank, reflecting back and sensed by the detector is 100 ns. The depth of the tank from ground level marked as \( x \) in metre is \(\underline{\hspace{2cm}}\) . 

• A. 9
• B. 10
• C. 11
• D. 12

Hint:

To solve such problems, remember that the round-trip time includes both the air and water travel times. Use the refractive index to calculate the effective speed of light in water.

Answer 1

The Correct Option is D

We are given the following data:
- The round-trip time of the pulse is \( t = 100 \, \text{ns} \).
- The refractive index of water is \( n_{\text{water}} = 1.3 \).
- The velocity of light in air is \( c_{\text{air}} = 3 \times 10^8 \, \text{m/s} \).
Let the depth of the tank from the ground level be \( x \) meters. The total time taken by the laser pulse to travel to the bottom of the tank and reflect back is the sum of the time taken in air and water. The time for the laser pulse to travel through a medium is given by: \[ t = \frac{\text{distance}}{\text{velocity}}. \] The total round-trip time is the sum of the times in air and water. The distance in air is 2 meters (from the laser to the top of the water) and the remaining distance in water is \( x - 2 \) meters. For the air: \[ t_{\text{air}} = \frac{2}{c_{\text{air}}}. \] For the water: \[ t_{\text{water}} = \frac{x - 2}{v_{\text{water}}} = \frac{x - 2}{\frac{c_{\text{air}}}{n_{\text{water}}}} = \frac{n_{\text{water}} (x - 2)}{c_{\text{air}}}. \] The round-trip time is the sum of these two times, doubled because the pulse travels to the bottom and back: \[ \text{Total time} = 2 \times \left(t_{\text{air}} + t_{\text{water}} \right). \] Substitute the known values: \[ 100 \times 10^{-9} = 2 \times \left( \frac{2}{3 \times 10^8} + \frac{1.3(x - 2)}{3 \times 10^8} \right). \] Simplifying this equation, we can solve for \( x \) and find that \( x = 12 \) meters. Final Answer: 12