Question

A coin placed at the bottom of a beaker appears to be raised by $4\mathrm{cm}$. If the refractive index of water is $\frac{4}{3}$, then find the depth of the water in the beaker.

Answer 1

Step 1: Given data

A coin placed at the bottom of beaker appears to be raised by $4\mathrm{cm}$.

Refractive index of water is $\frac{4}{3}$.

Let the real depth of the water in beaker is $\left(x\right)$.

So the apparent depth of the water is $\left(x-4\right)$

Step 2: Find the real depth.

$\mathrm{Refractive}\mathrm{index}=\frac{\mathrm{real}\mathrm{depth}}{\mathrm{apparent}\mathrm{depth}}$.

Substitute the given values in the above formula.

$\begin{array}{rcl}\frac{4}{3}& =& \frac{x}{x-4}\\ 3\left(x\right)& =& 4\left(x-4\right)\\ 3x& =& 4x-16\\ 4x-3x& =& 16\\ x& =& 16\mathrm{cm}\end{array}$

The depth of the water in the beaker is $16\mathrm{cm}$.