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Snell's law, critical angle & refraction

by sveioen
Tags: angle, critical, refraction, snell
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sveioen
#1
May13-10, 03:04 PM
P: 14
1. The problem statement, all variables and given/known data
Given a three layer model

-------------------------------------------------
[tex]v_1=1.5[/tex]km/s
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[tex]v_2=1.3[/tex]km/s
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[tex]v_3=2.0[/tex]km/s

Assume a ray goes through layer 1 and hits the interface between layer 1 and layer 2. What is the critical angle?

2. Relevant equations

Snells law
[tex]\frac{\sin \theta_1}{\sin \theta_2}=\frac{v_1}{v_2}[/tex]

3. The attempt at a solution

To find the critical angle, you normally take [tex]\sin \theta_c = \frac{v_1}{v_2}=\frac{1.5}{1.3}[/tex]. But in this case that means I have to take [tex]\sin^{-1}[/tex] of a value that is over 1! How do I solve this?
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rl.bhat
#2
May14-10, 07:46 AM
HW Helper
P: 4,439
According to Snell's law
n1sin(θ1) = n2sin(θ2)

If θ1 is θc, then θ2 = 90 degrees.

So sin(θc) = n2/n1
sveioen
#3
May14-10, 10:21 AM
P: 14
Quote Quote by rl.bhat View Post
According to Snell's law
n1sin(θ1) = n2sin(θ2)

If θ1 is θc, then θ2 = 90 degrees.

So sin(θc) = n2/n1
When I look up Snell's law on Wikipedia it says

[tex]
\frac{\sin \theta_1}{\sin \theta_2}=\frac{v_1}{v_2}=\frac{n_2}{n_1}
[/tex]

Why does the subscript change in the [tex]n_n[/tex] ? Isnt [tex]v_1=n_1[/tex] and [tex]v_2=n_2[/tex]?

Thanks for answering

rl.bhat
#4
May15-10, 12:17 AM
HW Helper
P: 4,439
Snell's law, critical angle & refraction

According to the definition,
refractive index n = c/v. where c is the velocity of light in vacuum and v is the velocity in the refracting medium.
So v = c/n
Or v1 = c/n1 and v2 = c/n2
then v1/v2 = .....?


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