Problem: Reflectivity coefficient

Question 1

Prove this equation from the Neutron reflectivity page, \(r=\frac{k^2_{1z}-k^2_{2z}}{\left(k_{1z}+k_{2z}\right)^2}=\frac{q^2_c}{4\left(k_{1z}+k_{2z}\right)^2}\).

Solution

Starting with this equation from the Neutron reflectivity page, \(n_1 \cos \theta_1 = n_2 \cos \theta_2\), \begin{equation} n^2_1\left(1-\sin^2\theta_1\right)=n^2_2\left(1-\sin^2\theta_2\right) \\ \end{equation} Substitute this equation, \(n_m=\frac{\lambda_0}{\lambda_m}=\frac{k_m}{k_0}\): \begin{equation} k^2_{1\perp}-k^2_{2\perp}=k^2_0\left(n^2_1-n^2_2\right) \end{equation} Substitute this equation, \(n^2=1-\frac{4\pi \rho\overline{b}}{k_0^2}=1-\frac{\lambda_0^2}{\pi}\rho\overline{b}\): \begin{equation} k^2_{1\perp}-k^2_{2\perp}=4\pi\left(\rho_2\overline{b}_2-\rho_1\overline{b}_1\right) \end{equation} Substitute this equation, \(q_c=\sqrt{16\pi\left(\rho_2\overline{b}_2-\rho_1\overline{b}_1\right)}\): \begin{equation} k^2_{1\perp}-k^2_{2\perp}=\frac{q^2_c}{4} \end{equation} Note that: \begin{equation} r=\left(\frac{k_{1z}-k_{2z}}{k_{1z}+k_{2z}}\right)\left(\frac{k_{1z}+k_{2z}}{k_{1z}+k_{2z}}\right)=\frac{k^2_{1z}-k^2_{2z}}{\left(k_{1z}+k_{2z}\right)^2} \end{equation}