Collect terms and rearrange in terms of \(G\) and \(\frac{\partial G}{\partial n}\)
\begin{equation} -\frac{1}{4}\left[\frac{\rho_{1\rightarrow 2,\nu}}{n_{1}^{2}}+\frac{1}{n_{2}^{2}}\right]G-\frac{1}{4}\left[\frac{\rho_{1\rightarrow 2,\nu}+2}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\frac{2}{3-A_{1}\omega}\frac{\partial G}{\partial n}=\int_{0}^{2\pi}\int_{\frac{\pi}{2}}^{\pi}[1-\rho_{2\rightarrow 1,\nu}]\frac{I^{-}_{2,\nu}}{n_{2}^{2}}\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\psi.\\ \end{equation}Now let’s check the case of two media with the same refractive index.
\begin{equation} -\frac{1}{4}\left[\frac{\rho_{1\rightarrow 2,\nu}}{n^{2}}+\frac{1}{n^{2}}\right]G-\frac{1}{4}\left[\frac{\rho_{1\rightarrow 2,\nu}+2}{n^{2}}-\frac{1}{n^{2}}\right]\frac{2}{3-A_{1}\omega}\frac{\partial G}{\partial n}=\int_{0}^{2\pi}\int_{\frac{\pi}{2}}^{\pi}[1-\rho_{2\rightarrow 1,\nu}]\frac{I^{-}_{2,\nu}}{n^{2}}\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\psi.\\ \end{equation}For simplicity let \(n=1\). This should give the same result as the original boundary condition…
\begin{equation} -\frac{1}{4}\left[\rho_{1\rightarrow 2,\nu}+1\right]\left(G+\frac{2}{3-A_{1}\omega}\frac{\partial G}{\partial n}\right)=\int_{0}^{2\pi}\int_{\frac{\pi}{2}}^{\pi}[1-\rho_{2\rightarrow 1,\nu}]I^{-}_{2,\nu}\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\psi.\\ \end{equation}Lovely.