Register to reply

Reflection and Refraction, much index of refraction

by davidelete
Tags: index, reflection, refraction
Share this thread:
davidelete
#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.
Phys.Org News Partner Science news on Phys.org
Wildfires and other burns play bigger role in climate change, professor finds
SR Labs research to expose BadUSB next week in Vegas
New study advances 'DNA revolution,' tells butterflies' evolutionary history
astrorob
#2
Jan28-09, 12:46 PM
astrorob's Avatar
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
#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
#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
#5
Jan28-09, 01:08 PM
astrorob's Avatar
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
#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
#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.


Register to reply

Related Discussions
Index of Refraction Introductory Physics Homework 7
Index of refraction when total internal reflection ceases Introductory Physics Homework 4
Index of refraction Introductory Physics Homework 3
Index of refraction Introductory Physics Homework 4
Reflection,Refraction,Total Internal Reflection,Diffraction Introductory Physics Homework 2