Light Refraction Problem – Snells Law

1. The problem statement, all variables and given/known data
After swimming, you realise you lost your room key whilst in the pool. You go back at night and use a flashlight to try and find it. The flashlight shines on it, which is at the bottom of the pool, when the flashlight is held 1.2m above the surface and is directed towards a point on the surface which is 1.5m from the edge of the pool. If the pool is 4m deep, how far is the key from the edge of the pool? Take the refractive index of the water to be 1.33.


2. Relevant equations
[itex]\frac{sin\theta_1}{sin\theta_2}=\frac{n_2}{n_1} [/itex]
Law of Sines

3. The attempt at a solution
Well, what I did first was trying to find out the angle of θ_1 , the angle denoted θ_i in the diagram.

To do that I used the law of sines and the fact that boths the angles are encompassed by right angled trianlges. First I found the length of the hypontenuse (the line of sight of the flashlight)

l=\sqrt{1.5^2+1.2^2}=1.921 [/itex] , then with that length use the law of sines

[itex]\frac{sin90}{1.921}=\frac{\theta_1}{1.5}=0.5206 \\
\therefore \theta_1=sin^{-1}(1.5 \times 0.5206) = 51.343° [/itex]

Now used snells law to find θ_2 …

\frac{sin51.343}{sin\theta_2}=\frac{1.33}{1}=1.33 \\
sin\theta_2=\frac{sin51.343}{1.33}=0.587 \\
\therefore \theta_2=sin^{-1}0.587=35.94°

Then used the law of sines again to find the value for x on the diagram

\frac{sin(180-90-35.94)}{4}=\frac{sin35.94}{x} \\
\frac{sin54.06}{4}=\frac{sin35.94}{x} \\
0.2024=\frac{sin35.94}{x} \\
\therefore x=\frac{sin35.94}{0.2024}=2.9m

And therefore the answer would be 2.9+1.5=4.4m

Im sure there must be loads of rounding errors in this, but I used 4 decimal places to try and minimize them. And there is probably easier ways to go about it, but would really appreciate it if someone could see if I have done it correctly please?

Thanks 🙂

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