Feynman's Lectures Refraction Angle question

[karkas]

Hey there, recently been studying Optics in Feynman's book vol.1 and came across his geometrical approach of Fermat's least time theorem and its application on refraction. So he illustrates refraction geometrically like this (attached jpeg for those without the book)

and so he goes on to say that :
"Therefore we see that when we have the right point, XC sin(EXC) = n XC sin(XCF) or, cancelling the common hypotenuse length XC and noting that : EXC = ECN = θ_i (angle) and XCF = BCN' = θ_r (angle) , we have sinθi = n sinθr.

However I am trying to figure out why the equivalence in bold stands, and I can't see it. In fact I'm pretty sure it shouldn't be so normally, but Feynman states something about an approximation later on (that the two different paths' travel time should be thought of as equal), so this may be why the angles are so. Anyone care to explain? It may be something really geometrically simple, but it bothers me !

 

[technician]

I think that you have identified a small fault in the analysis !
The angle XCN' is 90 degrees so angle XCF + FCN' = 90
So if BCN' = XCF then BCN' + FCN' should also = 90. On the diagram this is not the case Angle BFC is the right angle.
When you read the full analysis he makes a point that only small changes from the path AXB are being considered and the curve is fairly flat in that region so any deviations are of the SECOND ORDER.
This is not pointed out but I imagine ASSUMING BCN' equals XCF is good enought !

 

[karkas]

I think I get what you're saying, thanks. I'd be obliged if you'd elaborate on the second order deviations but nevertheless I can roughly see what's going on !

 

[technician]

An almost identical situation arises when deriving the equation for the separation of maxima in the 2 slit interference pattern. The diagram consists of 2 very close point sources (the slits) and a distant screen.
When you draw the straight lines and construct triangles to calculate angles you find a triangle that cannot be a right angle triangle but the DIFFERENCE is so small it can be ignored. This sort of approximation occurs frequently in physics. If the difference occurs in, let's say, the 4th sig figure then it is not significant.!
A good example is the approximation Sinθ = θ (in radians) which is good for angles up to about 20 degrees (difference only begins to show in the 3rd, 4th significant figure.
If I remember my calculus correctly '2nd order' means you can ignore (dy/dx)^2 for small values.
Hope this helps

 

[karkas]

Oh like in a Taylor expansion. I see, thanks!