Reflection and Refraction, much index of refraction


by davidelete
Tags: index, reflection, refraction
davidelete
davidelete is offline
#1
Jan28-09, 12:06 PM
P: 28
1. The problem statement, all variables and given/known data
1. The index of refraction for water is 1.33 and that of glass is 1.50.

a. What is the critical angle for a glass-water interface?


b. In which medium is the light ray incident for total internal reflection?


2. Relevant equations
nisin[tex]\vartheta[/tex]i=nrsin[tex]\vartheta[/tex]r


3. The attempt at a solution
a. I think the answer to a. is 62.46 degrees, but I am not sure.
b. I think the answer is glass, simply because it is going to be moving from a less dense area to a more dense area.
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astrorob
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#2
Jan28-09, 12:46 PM
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P: 141
Use Snell's law:

[tex]n_1\sin\theta_1 = n_2\sin\theta_2\ .[/tex]

Note: It's not additive like you suggested.

At the critical angle, [tex]\theta_2[/tex] is 90 degrees (the light refracts along the boundary). That is to say, it's sin is 1. We can therefore rearrange for the critical angle:

[tex]\theta_{\mathrm{crit}} = \sin^{-1} \left( \frac{n_2}{n_1} \right)[/tex]

Now we have two refractive indices, the glass and the water. If you plug them in incorrectly, you're going to end up with trying to find the inverse sin of a value > 1, which isn't possible as this has no solution.
davidelete
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#3
Jan28-09, 12:59 PM
P: 28
Quote Quote by astrorob View Post
Use Snell's law:

[tex]n_1\sin\theta_1 = n_2\sin\theta_2\ .[/tex]

Note: It's not additive like you suggested.

At the critical angle, [tex]\theta_2[/tex] is 90 degrees (the light refracts along the boundary). That is to say, it's sin is 1. We can therefore rearrange for the critical angle:

[tex]\theta_{\mathrm{crit}} = \sin^{-1} \left( \frac{n_2}{n_1} \right)[/tex]

Now we have two refractive indices, the glass and the water. If you plug them in incorrectly, you're going to end up with trying to find the inverse sin of a value > 1, which isn't possible as this has no solution.
Quite sorry for the mistake in Snell's Law. I knew what it meant, I just messed up when writing it in the forum.

Anyway, I appreciate the input, but if you are using
[tex]\theta_{\mathrm{crit}} = \sin^{-1} \left( \frac{n_2}{n_1} \right)[/tex], would it not be possible to put 1.33 (n2) over 1.5 (n1)? This would look like
[tex]\theta_{\mathrm{crit}} = \sin^{-1} \left( \frac{1.33}{1.5} \right)[/tex] and if plugged into a calculator, would return 62.45732485 degrees.

patriots1049
patriots1049 is offline
#4
Jan28-09, 01:05 PM
P: 10

Reflection and Refraction, much index of refraction


For this problem I calculated 62.4 degrees, the same thing you got.

As for B, I put water.
astrorob
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#5
Jan28-09, 01:08 PM
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P: 141
Yes, that's correct as you've stated.

It also gives you the answer to your second question as [tex]n_2[/tex] represents the refractive index of the medium that the light travelling in medium [tex]n_1[/tex] is incident on.
davidelete
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#6
Jan28-09, 01:08 PM
P: 28
Quote Quote by patriots1049 View Post
For this problem I calculated 62.4 degrees, the same thing you got.

As for B, I put water.
Ah, thank you. I was actually thinking that it would be glass because my book details the fact that "Total internal reflection occurs when light passes from a more optically dense medium to a less optically dense medium at an angle so great that there is no refracted ray." This would mean that glass-water would be a more optically dense to a less optically dense scenario.
davidelete
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#7
Jan28-09, 01:10 PM
P: 28
Quote Quote by astrorob View Post
Yes, that's correct as you've stated.

It also gives you the answer to your second question as [tex]n_2[/tex] represents the refractive index of the medium that the light travelling in medium [tex]n_1[/tex] is incident on.
Thank you very much. You have been a big help today, astrorob.


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