Light Refraction Problem - Snells Law

  • Thread starter FaraDazed
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Homework Statement


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.

Diagram: http://oi58.tinypic.com/2s1qfqp.jpg


Homework Equations


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


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)

[itex]
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 ...

[itex]
\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°
[/itex]

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

[itex]
\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
[/itex]

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 :)
 
Last edited:

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
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Excellent work. Clear picture, good post. I get the same answer. No easier way.

If you really insist on improving something: for the calculation of the result, you need ##\sin\theta_1## and all you have is ##\tan\theta_1 = 1.5/1.2##, so ##\sin (\tan^{-1}(1.5/1.2) ) ## is what I would calculate. Not really different or easier. Law of sines is OK, but you want to pick out sines, cosines, tangents more or less directly without much ado.
 
  • #3
347
2
Excellent work. Clear picture, good post. I get the same answer. No easier way.

If you really insist on improving something: for the calculation of the result, you need ##\sin\theta_1## and all you have is ##\tan\theta_1 = 1.5/1.2##, so ##\sin (\tan^{-1}(1.5/1.2) ) ## is what I would calculate. Not really different or easier. Law of sines is OK, but you want to pick out sines, cosines, tangents more or less directly without much ado.
OK great to know it was correct. Thanks :)
 

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