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:

Remember the useful relation: \[ \mu=\frac{\text{Real Depth}}{\text{Apparent Depth}} \] and \[ v=\frac{c}{\mu}. \] Whenever apparent depth is given, first calculate the refractive index and then determine the speed of light in the medium.

Answer 1

Concept: When an object is viewed from air through a denser medium, the object appears to be raised above its actual position. This phenomenon occurs due to refraction of light at the interface between the two media. The refractive index of a medium is related to the real depth and apparent depth by \[ \mu=\frac{\text{Real Depth}}{\text{Apparent Depth}}. \] Also, the refractive index of a medium is defined as \[ \mu=\frac{c}{v}, \] where
• \(c\) is the speed of light in vacuum,
• \(v\) is the speed of light in the medium. Combining these relations allows us to determine the speed of light in the liquid.

Step 1: Write the given data. Real depth of the liquid tank: \[ d=12.5\,\text{m} \] Apparent depth of the needle: \[ d'=9.0\,\text{m} \]

Step 2: Calculate the refractive index of the liquid. Using \[ \mu=\frac{\text{Real Depth}}{\text{Apparent Depth}}, \] we get \[ \mu=\frac{12.5}{9.0}. \] \[ \mu=1.389. \] Thus, \[ \boxed{\mu\approx1.39}. \]

Step 3: Use the relation between refractive index and speed of light. The refractive index is \[ \mu=\frac{c}{v}. \] Therefore, \[ v=\frac{c}{\mu}. \] Substituting \[ c=3\times10^8\,\text{m s}^{-1} \] and \[ \mu=1.389, \] we obtain \[ v=\frac{3\times10^8}{1.389}. \] \[ v=2.16\times10^8\,\text{m s}^{-1}. \]

Step 4: Write the final answer. Hence, the speed of light in the liquid is \[ \boxed{2.16\times10^8\,\text{m s}^{-1}}. \]