# Reflection and Refraction, much index of refraction

by davidelete
Tags: index, reflection, refraction
 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$$\vartheta$$i=nrsin$$\vartheta$$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.
 P: 141 Use Snell's law: $$n_1\sin\theta_1 = n_2\sin\theta_2\ .$$ Note: It's not additive like you suggested. At the critical angle, $$\theta_2$$ 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: $$\theta_{\mathrm{crit}} = \sin^{-1} \left( \frac{n_2}{n_1} \right)$$ 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.
P: 28
 Quote by astrorob Use Snell's law: $$n_1\sin\theta_1 = n_2\sin\theta_2\ .$$ Note: It's not additive like you suggested. At the critical angle, $$\theta_2$$ 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: $$\theta_{\mathrm{crit}} = \sin^{-1} \left( \frac{n_2}{n_1} \right)$$ 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
$$\theta_{\mathrm{crit}} = \sin^{-1} \left( \frac{n_2}{n_1} \right)$$, would it not be possible to put 1.33 (n2) over 1.5 (n1)? This would look like
$$\theta_{\mathrm{crit}} = \sin^{-1} \left( \frac{1.33}{1.5} \right)$$ and if plugged into a calculator, would return 62.45732485 degrees.

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.
 P: 141 Yes, that's correct as you've stated. It also gives you the answer to your second question as $$n_2$$ represents the refractive index of the medium that the light travelling in medium $$n_1$$ is incident on.
P: 28
 Quote by patriots1049 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.
P: 28
 Quote by astrorob Yes, that's correct as you've stated. It also gives you the answer to your second question as $$n_2$$ represents the refractive index of the medium that the light travelling in medium $$n_1$$ is incident on.
Thank you very much. You have been a big help today, astrorob.

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