Question

An astronaut is exploring an unknown planet when she accidentally drops an oxygen canister into a 1.50-m-deep pool filled with an unknown liquid. Although she dropped the canister 21 cm from the edge, it appears to be 31 cm away when she peers in from the edge. What is the liquid’s index of refraction? Assume that the planet’s atmosphere is similar to earth’s

Short answer

Refractive index of the liquid is 1.46

Step-by-step solution

Step 1. Given information is :Depth of the pool = 1.50 mCanister is 21 cm from the edgeCanister's image from the astronaut is 31 cm from the edgeRefractive index of earth's air, n = 1.00

We need to calculate the refractive index of the liquid.

Step 2. Using the concepts of divergence and snell's law.

Refraction of the rays away from the normal and diverging outwards, along with the image point of the canister is shown below :

As per the diagram,

$$\begin{gathered} \tan {\theta }_1=\frac{21cm}{150cm} \\ {\theta }_1={\tan }^{-1}\left(\frac{21}{150}\right) \\ {\theta }_1=7.97° \end{gathered}$$

${\theta }_2$is represented as :

$$\begin{gathered} \tan {\theta }_2=\frac{31cm}{150cm} \\ {\theta }_2={\tan }^{-1}\left(\frac{31}{150}\right) \\ {\theta }_2=11.7° \end{gathered}$$

Using Snell's Law,

$$\begin{gathered} n_{liquid}\sin {\theta }_1=n_{air}\sin {\theta }_2 \\ {n}_{liquid}=n_{air}\frac{\sin {\theta }_2}{\sin {\theta }_1} \\ {n}_{liquid}=(1.00)\frac{\sin 11.7°}{\sin 7.97°} \\ {n}_{liquid}=1.46 \end{gathered}$$