Finding the Velocity of Light through an Unknown substance

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Homework Statement



32-005.gif

In the above figure, a ray of light is incident upon the interface between two media having indices of refraction n1 = 1.2 and n2, which is unknown. The known angles are f2 = 60° and f3 = 29.9°. (Note that the angles in the diagram may not be accurate.) What is the speed of light in the second medium?


Homework Equations


Snell's Law n[itex]_{1}[/itex]sin θ[itex]_{1}[/itex] = n[itex]_{2}[/itex]sin θ[itex]_{2}[/itex]

And the velocity of light equation:

n = [itex]\frac{c}{v_{n}}[/itex]

The Attempt at a Solution


Since the Angles given aren't the "true" angles that are used in Snell's Law

I took each angle and subtract them from 90° to get θ[itex]_{2}[/itex] = 30° and θ[itex]_{3}[/itex] = 60.1°

From there, I calculated that θ[itex]_{1}[/itex] = 29.9°

I took the variables n[itex]_{1}[/itex] = 1.2 and θ[itex]_{1}[/itex] & θ[itex]_{2}[/itex] and plugged them into the Snell's Formula to get n[itex]_{2}[/itex] = 1.196 as the index of refraction of the unknown substance.

I then plugged that calculation into the velocity formula to get 2.5 x 10[itex]^{8}[/itex], but I get that as the wrong answer.

I don't know where I went wrong with this. Thank you for your help!
 

Answers and Replies

  • #2
TSny
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I took each angle and subtract them from 90° to get θ[itex]_{2}[/itex] = 30° and θ[itex]_{3}[/itex] = 60.1°

From there, I calculated that θ[itex]_{1}[/itex] = 29.9°

What does the "law of reflection" tell you about θ1 and θ3.
 
  • #3
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The Law's of reflection would state that θ[itex]_{1}[/itex] = θ[itex]_{3}[/itex]. I then got the angle as 60.1 degrees. I was looking at the picture incorrectly and used the wrong angles that were required for snell's Law. I got it now!

Thanks TSny!
 
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