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

Concept: Refractive index of a medium: \[ n = \frac{\text{Real depth}}{\text{Apparent depth}} \] Also: \[ n = \frac{c}{v} \] Where:

\( c = 3 \times 10^8 \, \text{m/s} \) (speed of light in vacuum)
\( v \) = speed of light in medium

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

Answer 2

Concept: The relation between real depth, apparent depth and refractive index can be obtained directly from Snell's law by considering how a nearly-vertical ray bends on leaving the liquid surface, rather than starting from the ready-made apparent-depth formula.

Step 1: Set up the refraction geometry.
Let a ray from the needle at the bottom travel almost vertically and strike the liquid surface at a small angle \( \theta_1 \) from the normal, refracting into air at angle \( \theta_2 \). Since the ray originates from a fixed horizontal distance \( x \) from the observer's eye, the real depth \( h_r \) and apparent depth \( h_a \) can be written as \[ \tan\theta_1 = \frac{x}{h_r}, \qquad \tan\theta_2 = \frac{x}{h_a} \]

Step 2: Apply Snell's law for near-normal rays.
By Snell's law, \( n \sin\theta_1 = \sin\theta_2 \), and for small angles \( \sin\theta \approx \tan\theta \), so \[ n \approx \frac{\tan\theta_2}{\tan\theta_1} = \frac{x/h_a}{x/h_r} = \frac{h_r}{h_a} \]

Step 3: Substitute the given depths.\[ n = \frac{h_r}{h_a} = \frac{12.5}{9.0} \approx 1.39 \]

Step 4: Obtain the speed of light in the liquid.
Since refractive index is defined as the ratio of the speed of light in vacuum to that in the medium, \( n = \dfrac{c}{v} \), so \[ v = \frac{c}{n} = \frac{3 \times 10^{8}}{1.39} \approx 2.16 \times 10^{8} \, \text{m/s} \]

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