Derivation of the thin lens equation?

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SUMMARY

The discussion centers on the derivation of the thin lens equation as presented in chapter 27 of the Feynman Lectures on Physics. The equation is expressed as h²/2s + nh²/s' = (n-1)h²/2r, where s is the object distance, s' is the image distance, n is the index of refraction of the second medium, r is the radius of curvature, and h is the altitude of a triangle. The term (n-1) arises from the need to account for the excess time light travels in the medium compared to air, clarifying the distinction between total travel time and excess travel time. The discussion emphasizes the importance of understanding the refractive index's role in the equation.

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'm currently reading chapter 27 of the Feynman lectures, where he derives the thin lens equation for a single refracting surface. He arrives at the equation

h^2/2s + nh^2/s' = (n-1)h^2/2r

Where s is the object distance, s' is the image distance, n is the index of refraction of the second medium (the first is one), r is the radius of curvature, and h is the altitude of a triangle

Where does the (n-1) come from? Wouldn't it simply be n?

This is kind of vague, so please look at the second page of
http://student.fizika.org/~jsisko/Knjige/Opca%20Fizika/Feynman%20Lectures%20on%20Physics/Vol%201%20Ch%2027%20-%20Geometric%20Optics.pdf

I would be so grateful. Thanks.
 
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All expressions in the equation are excess times as compared with the idea that light travels from O to Q in the air and then from Q to O' in the glass.

(n-1)\frac{h^2}{2R} is about how much longer light travels between V and Q. n\frac{h^2}{2R} is a total time of travel between V and Q, not excess time, for the excess time you have to subtract time it would take for the light to travel between V and Q in the air - which is just \frac{h^2}{2R} (as we assumed the speed in air is just 1).
 

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