Question

A coin is placed at the bottom of a beaker containing water $\left(\mathrm{refractive}\mathrm{index}=\frac{4}{3}\right)$ to a depth of $12\mathrm{cm}$. By what height the coin appears to be raised when seen from vertically above?

• A. $3\mathrm{cm}$
• B. $6\mathrm{cm}$
• C. $9\mathrm{cm}$
• D. $16\mathrm{cm}$

Answer 1

The correct option is A

$3\mathrm{cm}$

Step 1. Given parameters:

Refractive index of water, $RI=\frac{4}{3}$

A coin is placed at a depth of $12\mathrm{cm}$.

Step 2: Calculate the apparent depth as follows:

Use the following expression for calculating apparent depth

$\begin{array}{rcl}RI& =& \frac{\mathrm{Real}\mathrm{depth}}{\mathrm{Apparent}\mathrm{depth}}\end{array}$

Substitute the given values in the above formula.

$\begin{array}{rcl}RI& =& \frac{\mathrm{Real}\mathrm{depth}}{\mathrm{Apparent}\mathrm{depth}}\\ & \Rightarrow & \frac{4}{3}=\frac{12}{\mathrm{Apparent}\mathrm{depth}}\\ & \Rightarrow & \mathrm{Apparent}\mathrm{depth}=\frac{12\times 3}{4}\\ & \Rightarrow & \mathrm{Apparent}\mathrm{depth}=9\mathrm{cm}\end{array}$

Therefore, the height by which it appears to be raised is given by,

$\begin{array}{rcl}\mathrm{Real}\mathrm{depth}-\mathrm{Apparent}\mathrm{depth}& =& 12-9\\ & =& 3\mathrm{cm}\end{array}$

Thus, the height by which it appears to be raised is $3\mathrm{cm}$.

Hence, option (A) is correct.