Question

A tank is filled with a liquid to a height of \( 12.5 \, \text{m} \). The apparent depth of a needle lying at the bottom of the tank is measured to be \( 9.0 \, \text{m} \). Calculate the speed of light in the liquid.

Hint:

For apparent depth problems:

\( n = \frac{\text{real}}{\text{apparent}} \)
Then use \( v = \frac{c}{n} \)
If apparent depth is smaller, medium is optically denser.

Answer 1

Step 1: Calculate the refractive index of the liquid. \[ n = \frac{\text{Real depth}}{\text{Apparent depth}} = \frac{12.5}{9.0} \approx 1.39 \]
Step 2: Determine the speed of light in the liquid. \[ v = \frac{c}{n} = \frac{3 \times 10^8}{1.39} \approx 2.16 \times 10^8 \, \text{m/s} \]
Final Answer: \[ v \approx 2.2 \times 10^8 \, \text{m/s} \]

Answer 2

Step 1: Combine the two governing relations into a single expression.
The refractive index relates real and apparent depth as \( n = \dfrac{h_{\text{real}}}{h_{\text{apparent}}} \), and the speed of light in a medium is \( v = \dfrac{c}{n} \). Substituting the first relation into the second gives a direct formula \[ v = \frac{c \cdot h_{\text{apparent}}}{h_{\text{real}}} \] so the speed can be obtained in one step, without separately computing \( n \).

Step 2: Substitute the given values.\[ v = \frac{(3 \times 10^{8}) \times 9.0}{12.5} = \frac{27 \times 10^{8}}{12.5} = 2.16 \times 10^{8} \, \text{m/s} \]

Step 3: Sanity-check the implied refractive index.
This value of \( v \) corresponds to \( n = c/v \approx 1.39 \), which lies between that of water (\( 1.33 \)) and ordinary glass (\( 1.5 \)), a physically reasonable value for a transparent liquid, confirming the result.

Final Answer:\[ v \approx 2.2 \times 10^{8} \, \text{m/s} \]