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 () and use the relationships of these derivatives with $\tan{\alpha}$ and $\tan{\gamma}$.

$$\begin{eqnarray} \frac{\partial z}{\partial x} & = \tan{\alpha} = & \sqrt{\frac... ...ght) \frac{\frac{h_x}{h_0}} {\frac{h_x^2}{h_0^2} -1}. \ \nonumber\end{eqnarray}$$ (83) (84)

Substituting equations () and () into

$$\begin{eqnarray} \Delta_z I_{x_h}= \bar{z}-z & = & h_0\tan{\gamma}\tan{\alpha}\s... ...x & = & - h_0\tan{\gamma}\tan{\alpha}\cos{\alpha}. \ . \nonumber\end{eqnarray}$$ (85) (86)

and after some algebraic manipulations we prove the thesis.