Deriving Equation Relating Angle of Incidence to the Angle of Refracti

AI Thread Summary
The discussion focuses on deriving an equation that relates the angle of incidence (Θ1) to the angle of refraction (Θ2) as light transitions from air to glass, using Snell's law. Participants confirm that the equation n1sinΘ1 = n2sinΘ2 is the correct starting point, leading to the formulation of Θ2/Θ1 in terms of arcsin. However, challenges arise in simplifying the equation without retaining Θ1 and Θ2 on the right side. Suggestions include using small angle approximations and graphical analysis to interpret the relationship between the angles, emphasizing that the slope of the graph of sin(Θ2) vs. sin(Θ1) represents the refractive index. Ultimately, the discussion highlights the complexities of deriving a simple relationship for larger angles.
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Homework Statement


Derive an equation relating the angle of incidence and the angle of refraction as light travels from air to glass. Do it in the form of Θ2/Θ1 = expression


Homework Equations


I think the equation to use here is Snell's law. Right?
n1sinΘ1 = n2sinΘ2

n1 = nair = 1.0003 approximately 1
n2 = nglass
Θ1 = angle of incidence
Θ2 = angle of refraction

The Attempt at a Solution


n1sinΘ1 = n2sinΘ2
n1/n2 = (sinΘ2)/(sinΘ1)
1/n2 = (sinΘ2)/(sinΘ1)

However, I'm having problems on how to remove the sin on the right side of my equation. I know I can't just apply arcsin to both sides since that wouldn't remove the sin in both the numerator and the denominator of my fraction. Please help!
 
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Solve for Θ2, and divide it by Θ1.

ehild
 
For small angles sin x is approximately x (x in radians). For large angles, there really is no way to get a relation of the form you specified.
 
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ehild said:
Solve for Θ2, and divide it by Θ1.

ehild

I got Θ2/Θ1 = [arcsin((sinΘ1)/n2)]/Θ1

However, is this really it? Is there no way I can arrive at an expression wherein Θ1 and Θ2 would not appear in the right hand side of the equation?

Orodruin said:
For small angles sin x is approximately x (x in radians). For large angles, there really is no way to get a relation of the form you specified.

I've tried doing it in a calculator. I know I'm working with angles to pi/2. However, as you said, sin x really gets deviated from x as x increases. But it's a good thing you pointed that out. Really simplifies things.

I'm doing graphical analysis (angle of refraction(y) vs angle of incidence(x)) and I'm trying to figure out what the slope represents. I might use this as an explanation and just say that my slope represents the reciprocal of the refractive index of the glass.
 
frozonecom said:
I'm doing graphical analysis (angle of refraction(y) vs angle of incidence(x)) and I'm trying to figure out what the slope represents. I might use this as an explanation and just say that my slope represents the reciprocal of the refractive index of the glass.
Why don't you plot ##\sin\theta_1## vs. ##\sin\theta_2## instead?
 
vela said:
Why don't you plot ##\sin\theta_1## vs. ##\sin\theta_2## instead?

Yes I did that too, and I've got a linear graph. However, a part of the experiment was to analyze graphs of θ2 vs θ1 too. It's been giving me a hard time. I know that the graph of that one isn't perfectly linear but our instructor advised us to use linear regression on our data. Now it was asking us to know what the slope represents. I've been trying to derive an expression to figure out what it really is.
 
##\sin(θ_2)=\frac{n_1}{n_2}\sin(θ_1)##

Take the derivative of both sides with respect to θ1. Solve for the slope dθ2/dθ1. If θ1 is small, the slope can be approximated as n1/n2 as the cosines are close to 1.
I do not see any sense to worry about the slope of the θ2 vs θ1 graph. You can apply linear regression to the graph sinθ2 vs sinθ1.

ehild
 
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