Infinitesimal, Snell's law, and ray tracing

AI Thread Summary
The discussion revolves around applying Snell's law to analyze the relationship between the angle of incidence (alpha1) and the velocity of light in fluid 1 (v1) when transitioning into fluid 2 with a constant velocity (v2). The user seeks an analytical method to calculate the derivative d(alpha1)/d(v1), rather than relying on numerical methods. Participants suggest using a relationship between alpha1 and alpha2 to eliminate alpha2 from Snell's law for differentiation. The importance of drawing a diagram to visualize the problem and confirm the calculations is emphasized. Ultimately, the user successfully resolves the issue by utilizing the diagram.
ytht100
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With FIXED SOURCE AND RECEIVER, I have a light incident from fluid 1 with velocity v1 into fluid 2 with velocity v2. Obviously, according to Snell's law, v1/v2=sin(alpha1)/sin(alpha2), where alpha1 and alpha2 are the angles with regard to the vertical line.

My question is: how to calculate d(alpha1)/d(v1)?

Obviously, I can obtain the result from numerical calculation, but I am seeking analytical method. I guess maybe I need some approximation to calculate that?

Please note v2 is always constant, please also note when v1 changes, alpha1 and alpha2 change. Please note direct differentiation of Snell's law, I can obtain d(alpha1)/d(alpha2). I am clueless and have already tried several methods but obtain no result, please help!
 
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Have you drawn a diagram? If you do so, you ought to be able to see a relationship between ##\alpha_1## and ##\alpha_2##.
 
Thank you very much for your kind answer. Yes, I have drawn it. I am NOT seeking a relationship between alpha1 and alpha2. I am seeking relationship between d(alpha1) and d(v1), in which d represents differentiation.
 
And you can get that by using a relationship between ##\alpha_1## and ##\alpha_2## to eliminate ##\alpha_2## from your Snell's Law expression.
 
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Ibix said:
And you can get that by using a relationship between ##\alpha_1## and ##\alpha_2## to eliminate ##\alpha_2## from your Snell's Law expression.

Sorry you answer is NOT clear. I have tried everything that I can think of for several days, can you write down specific equations? Thank you very much!
 
I can help you find the answer, but I can't give it to you. The aim of this site is to help you learn, not to do things for you.

You say the source and receiver are fixed. Where are they? Write down the coordinates. Then, work out where the ray traveling from source to receiver crosses from one medium into the other (this is where a diagram will help you). That will give you a second expression relating ##\alpha_1## and ##\alpha_2##.
 
ytht100 said:
Thank you very much for your kind answer. Yes, I have drawn it.

show us your diag. so we can confirm you are on the right path
 
Thank you all for your kind answer. I have figured it out by using the diagram. Kind regards!
 
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