Evaluation of the image shift as a function of $\alpha$ ad $\gamma$

The final step is to take the derivative of the impulse response of equation (18) and use the relationships of these derivatives with $\tan{\alpha}$ and $\tan{\gamma}$.

$$\begin{eqnarray} \frac{\partial z}{\partial x} & = \tan{\alpha} = & \sqrt{\frac... ...ght) \frac{\frac{x_h}{h_0}} {\frac{x_h^2}{h_0^2} -1}. \\ \nonumber\end{eqnarray}$$ (19) (20)

Substituting equations (19) and (20) into

$$\begin{eqnarray} \Delta_z I_{x_h}= \bar{z}-z & = & h_0\tan{\gamma}\tan{\alpha}\s... ...x & = & - h_0\tan{\gamma}\tan{\alpha}\cos{\alpha}. \\ . \nonumber\end{eqnarray}$$ (21) (22)

and after some algebraic manipulations we prove the thesis.