B Deriving Snell's Law

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    Snell's law
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The discussion focuses on deriving Snell's Law by comparing two paths and using perpendiculars for analysis, questioning whether an isosceles triangle should be used instead. It confirms that the approximation made is acceptable, as the error is of higher order than first order. The conversation also addresses the angle approximation, clarifying that as point X approaches C, the angles FCX and BCN' differ only to first order. The relationship between angles and distances is emphasized, showing that using the zeroth order approximation for small angles yields accurate results. This highlights the validity of the method used in deriving Snell's Law.
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Why are perpendiculars used to compare the two paths, and where does this angle approximation come from?
Here we are deriving snell's law. The text compares the length of the two paths (which are assumed to be close enough such that the time difference is negligible), and uses the perpendicular to compare them. They should technically be making an isoceles triangle to compare them right? Is this an approximation? Also the text says that angle FCX is approximately equal to angle BCN', but this doesn't seem to be true in the limit as X approaches C. What am I missing?

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WheatNeat said:
TL;DR Summary: Why are perpendiculars used to compare the two paths, and where does this angle approximation come from?

Here we are deriving snell's law. The text compares the length of the two paths (which are assumed to be close enough such that the time difference is negligible), and uses the perpendicular to compare them. They should technically be making an isoceles triangle to compare them right? Is this an approximation?
Yes, that's right. You can show that the error associated with this approximation is of order higher than first order. So, it's ok.

WheatNeat said:
Also the text says that angle FCX is approximately equal to angle BCN', but this doesn't seem to be true in the limit as X approaches C. What am I missing?

Label the angles as shown.
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The angle ε is small of first order when distance XC is small compared to distance CB.

Show that ##\phi = \theta_2 + \varepsilon##. So, as X approaches C, ##\phi## differs from ##\theta_2## to first order. However, you are interested in the distance ##XF##, which equals ##XC \sin \phi##. Since ##XC## is already a small, first order quantity, ##XC\sin \phi## will be accurate to first order if you use the "zeroth order" approximation ##\phi \approx \theta_2##.
 
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