Question

When the rectangular metal tank in the Figure is filled to the top with an unknown liquid, the observer $O$, with eyes level with the top of the tank, can just see the corner. A ray that refracts toward$\mathit{O}$the top surface of the liquid is shown. If$\mathit{D}\mathbf{=}\mathbf{\text{85.0cm}}$so$\mathit{L}\mathbf{=}\mathbf{\text{1.10m}}$, what is the index of refraction of the liquid?

Figure:

Short answer

The refractive index of the liquid is$n_1=\text{1.26}.$

Step-by-step solution

Given

Consider the given parameters shown below:

$$\begin{gathered} L=\text{1.10m} \\ D=\text{85cm=0.85m} \\ {\text{θ}}_{\text{2}}\text{=90°} \end{gathered}$$

1. The Refractive index of vacuum is $n_2=\text{1}$.

Determine the concept

Use the formula of Snell’s law, which gives the relation between the angle of reflection and the angle of refraction. Before that, the angle of incidence using the equation of tangent ratio can be calculated

Determine the refractive index of liquid

Expression for an angle of incidence is as follows:

$$\begin{gathered} {\text{tanθ}}_{\text{1}}=\frac{L}{D} \\ \tan {\theta }_1=\frac{1.10}{0.85} \\ {\text{θ}}_{\text{1}}\text{=52.31°} \end{gathered}$$

Now use Snell’s law:

$$\begin{gathered} n_1{\text{sinθ}}_{\text{1}}=n_2{\text{sinθ}}_{\text{2}} \\ {n}_1\text{sin52.31=1sin90} \\ {n}_1=\text{1.26} \end{gathered}$$

So the refractive index of the liquid is$n_1=\text{1.26}$ .