Wednesday, 6 March 2019

homework and exercises - Transmission of Gaussian Beam Through Graded-Index Slab


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 $d$?



Answer



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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