Ray matrix - refraction at a curved surface

[8614smith]

Homework Statement

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

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

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

Homework Equations

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

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

The Attempt at a Solution

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

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...

 

[nickjer]

You will have to use the refraction equation, but you need to be careful when computing the refraction at the interface since the interface is at an angle. So it is dependent on y_1, assuming y is your interface axis.

Also since it is an interface, you know y_1=y_2 so the top of the matrix is (1 0). The bottom of the matrix will take some work.