The $ABCD$ matrix of a glass graded-index slab with refractive index $n(y)=n_0(1-\frac{1}{2}\alpha^{2}y^{2})$ and length $d$ is $A=\cos(\alpha d)$, $B=\frac{1}{\alpha}\sin(\alpha d)$, $C=-\alpha \sin(\alpha d)$, $D=\cos(\alpha d)$ for paraxial rays along the z axis. Usually, $\alpha$ is chosen to be sufficiently small so that $\alpha^{2}y^{2} << 1$. A Gaussian beam of wavelength $\lambda_0$, waist radius $W_0$ in free space, and axis in the z direction enters the slab at its waist. How can I use the $ABCD$ law to get an expression for the beam width in the $y$ direction as a function of $d$?
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The ABCD law can be used for Gaussian beam propagation using the complex beam radius $q$. Defining $\frac{1}{q} = \frac{1}{R}-i\frac{2}{kW^2}$, $R = R(z)$ being the radius of curvature of the beam and $W = W(z)$ the halfwidth at point $z$ and $k = 2\pi/\lambda_0$, the complex beam radius transforms as $q \to \frac{Aq+B}{Cq+D}$. In your case (waist at the beginning of the medium, radius of curvature at the waist being infinite), so that $q = ikW_0^2/2$ at the front of the medium.
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$\begingroup$ Thank you for your reply. Wouldn't it be $q1=-kW_0^{2}/(2i)$ though? And this would just give us an expression for q1, from which we can get an expression for q2. However, we still need to substitute out for q2, and to do this, we need to get the radius of curvature for the beam leaving the slab. I'm still not sure how to do this. $\endgroup$ Commented Oct 1, 2012 at 23:57
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1$\begingroup$ $i$ and $-1/i$ are the same, so both expressions are correct, expect fot the square of $W_0$ which I missed. Now, you can get $q_2$ from $q_1$ using the ABCD transformation of complex beam radius as above and then using the definition $1/q_2 = 1/R_2 - 2i/(kW_2^2)$, you can determine both the radius of curvature and the beam width lleaving the slab. $\endgroup$ Commented Oct 2, 2012 at 11:41
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1$\begingroup$ Because all the parameters (except $q$) are real, you can split $1/q_2$ to real and imaginary part. The real part is then equal to $1/R_2$ while the imaginary part is $-2/(kW_2^2)$, which will then be expressed only in terms of $\alpha, d, W_0, \lambda_0$. $\endgroup$ Commented Oct 2, 2012 at 14:08
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1$\begingroup$ To get $1/R_2$ in terms of those parameters, you just take the real part of $1/q_2$. $\endgroup$ Commented Oct 2, 2012 at 15:36
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