distances $h$. Equation (\[eq:integralsechhyperbolfunction\]) can be integrated numerically. Alternatively, an analytical solution can be achieved by replacing the vertical ${\, \mathrm{sech} }^{2}$ distribution by an exponential profile, which is a good approximation [@1983ApJ...267...31E; @1987ApJ...312..626E; @Kim:2002gl; @Wang:2010wr; @2011ApJ...737...10E]. The reduction factor is then $$\label{eq:integralexponentialfunction} 
F_{\mathrm{exp}}(\lambda, z_0)  = \int_{-\infty}^{\infty}  \mathrm{e}^{- \frac{2 \upi}{\lambda} \mid h \mid} \; \frac{\mathrm{e}^{-\mid h \mid /z_{0}} }{2 z_{0}} \; dh,$$ and the solution $$\label{eq:exp_reductionfactor}
F_{\mathrm{exp}}(\lambda, z_{0}) = \left(  1+ \frac{2 \upi }{ \lambda } \; z_{0} \right) ^{-1} .$$ If we substitute $x = h /z_0$ in equations (\[eq:integralsechhyperbolfunction\]) and (\[eq:integralexponentialfunction\]), and use the symmetry in vertical direction, the integral can be written as $$\label{eq:F_sech2_rephrased}
F_{\mathrm{sech^2}}(\lambda, z_0)  =  \int_{0}^{\infty}  \mathrm{e}^{- \frac{2 \upi }{\lambda} z_0 x} \; {\, \mathrm{sech} }^{2}\left(x \right) \; dx,$$ and $$\label{eq:F_exp_rephrased}
F_{\mathrm{exp}}(\lambda, z_0)  =  \int_{0}^{\infty} \mathrm{e}^{- \frac{2 \upi }{\lambda} z_0 x}  \; \exp \left(- x \right) \; dx,$$ respectively. Both equations (equations \[eq:F\_sech2\_rephrased\] and \[eq:F\_exp\_rephrased\]) show a similar dependence on $\lambda$ and $z_0$, the comparison is illustrated in Fig. \[fig:reductionfactor\]. In principle, for a given $z_{0}$, a smaller wavelength $\lambda$ leads to smaller $F$. Much larger wavelengths $\lambda$ on the other hand are required to reach the upper range of $F \approx 1$. The reduction 