Question

A$\mathbf{20}\mathbf{-}\mathit{m}\mathit{m}\mathbf{-}\mathit{t}\mathit{h}\mathit{i}\mathit{c}\mathit{k}\mathbf{}\mathit{l}\mathit{a}\mathit{y}\mathit{e}\mathit{r}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{w}\mathit{a}\mathit{t}\mathit{e}\mathit{r}$$\mathbf{(}\mathit{n}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{33}\mathbf{)}$ floats on a$\mathbf{40}\mathbf{-}\mathit{m}\mathit{m}$localid="1662979231067" $\mathbf{thick}\mathbf{}\mathbf{layer}\mathbf{}\mathbf{of}\mathbf{}\mathbf{carbon}\mathbf{}\mathbf{tetrachloride}\mathbf{}$localid="1662979325107" $(n=1.46)$in a tank. A coin lies at the bottom of the tank. At what depth below the top water surface do you perceive the coin? (Hint: Use the result and assumptions of Problem 112 and work with a ray diagram.)

Short answer

The apparent depth of the coin from top of the water surface$d_a=42.0mm$.

Step-by-step solution

Listing the given quantities

For water$n=1.33$

For carbon tetrachloride $n=1.46$

Thickness of water layer$20.0mm$

Thickness of carbon tetrachloride layer$40.0mm$

Understanding the concepts of apparent depth

We know that due to the change of medium of propagation of light, the speed of light will decrease resulting in the bending of rays. Observers in another medium will perceive the different depths of the object.

So, using the relative refractive index, we can calculate the apparent depth.

Formula:

Apparent depth

$d_a=d/n$

n is the relative refractive index of the medium in which the object is kept.

Calculation of the apparent depth of the coin from the top of the water surface

The relative refractive index of the carbon tetrachloride with respect to water is

$\begin{array}{c}n\text{'}=\frac{1.46}{1.33}\\ =1.1\end{array}$

So, the apparent depth of the coin from the bottom of the water layer is

$\begin{array}{c}{d}_a^{\text{'}}=\frac{40.0}{1.1}\\ =36.4mm\end{array}$

So, the distance of the coin perceived from the top of the water surface is

$\begin{array}{c}d\text{'}\text{'}=d\text{'}+\mathrm{thickness}\mathrm{of}\mathrm{water}\mathrm{layer}\\ =36.4+20\\ =56.4mm\end{array}$

Now, the apparent depth of the coin from the top of the water surface is

$\begin{array}{c}{d}_a=\frac{d\text{'}\text{'}}{1.33}\\ =\frac{56.4}{1.33}\\ =42.0mm\end{array}$

The apparent depth of the coin from the top of the water surface$d_a=42.0mm$

The above explanation can be illustrated using the following ray diagram.