(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The diagram shows a curved convex surface with radius of curvature R separating two media with differing refractive indices of [tex]n_{1}[/tex] and [tex]n_{2}[/tex]

Show that the matrix representing refraction at this surface is [tex]\left[\stackrel{1}{\frac{1}{R}\left(\frac{n_1}{n_2}-1\right)}\stackrel{0}{\frac{n_1}{n_2}}\right][/tex]

In your derivation you may assume that [tex]\sqrt{1-{x^2}}{\approx}1-\frac{x^2}{2}[/tex] and that the slope of a curved surface [tex]\theta_s[/tex] may be approximated by [tex]\theta_{s}{\approx}\frac{dx}{dy}[/tex]

2. Relevant equations

refraction matrix: [tex]{r_t2}={R_2}{r_i2}[/tex] where [tex]{R_2}=\left[\stackrel{1}{0}\stackrel{-D_2}{1}\right][/tex] and [tex]{D_2}=\frac{{n_t2}-{n_i2}}{R_2}[/tex]

Transfer matrix: [tex]\left[\stackrel{1}{\frac{d_2_1}{n_t_1}}\stackrel{0}{1}\right][/tex]

3. The attempt at a solution

using the refraction equation i have got to [tex]\left[\stackrel{1}{0}\stackrel{-\frac{{n_2}-{n_1}}{R}}{1}\right][/tex] and using the transfer matrix i got [tex]\left[\stackrel{1}{\frac{R}{2n_2}}\stackrel{0}{1}\right][/tex]

I'm not too sure what to do from here, ive multiplied them both by eachother both ways round and neither way got the right answer, im unsure even whether this is the right way to do it.

i assume that the aprroximation it gives has something to do with the equation of a circle but i dont know how to use it...

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# Homework Help: Ray matrix - refraction at a curved surface

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