8614smith
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Homework Statement
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]
Homework 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]
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, I've multiplied them both by each other both ways round and neither way got the right answer, I am 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 don't know how to use it...