D. CSP-depth migration in one-layer medium

We suggest that a reflector is flat: z=H.

First step: We represent travel-time dependence for common source gathered in a true model in the following form:

$$t(\theta )={2H\over v \cos \theta}$$

 

$$x(\theta )=2H\tan \theta$$ (76)

where $\theta$ is incident angle.

Second step: Reverse Eiconal's continuation can be determined with help of ray's equations  

$$\xi (l)=x(\theta )- l\sin \theta ^{\prime}$$ (77)

 

$$\zeta (l)=l\cos \theta^{\prime}$$ (78)

where $\theta ^{\prime}$ is incident angle of a ray which is correspondent to reverse Eiconal's continuation:  

$$\sin \theta ^{\prime} = {v^{\prime} \over v} \sin \theta , {... ...t{1-{\left( {v^{\prime} \over v}\right) }^{2}\sin^{2} \theta },$$ (79)

l is a length of the ray from the point $(x(\theta ),0)$ to the point $(\xi ,\zeta )$. So

$$\tau^{(-)}=t(\theta ) - {l \over v^{\prime}}.$$

Third step: The location of the reflector's image can be found from the condition $\tau^{(-)}=\tau _{d}$, that is

$$t(\theta ) - {l \over v^{\prime}}= {1 \over v^{\prime}} \sqrt{\xi ^{2} + \zeta ^{2}}.$$

Having usage of equations (76),  (77),  (78), and  (79), we derive

$$l={v^\prime \over v}{H \over \cos^{3} \theta } \cdot \left[... ...t( {v^{\prime} \over v}\right) }^{2} \sin ^{2} \theta \right].$$

Now we substitute this formula into equations (77) and  (78) and finally get parametric descriptions of the reflector image:

$$\xi (\theta )=2H\tan \theta \left[ 1-{{({v^{\prime} \over v})}^{2} - \sin^{2} \theta \over 2\cos^{2} \theta} \right]$$

$$\zeta (\theta )= {v \over v^{\prime}}H \cdot {{({v^{\prime} ... ...{1-{\left( {v^{\prime} \over v}\right) }^{2} \sin^{2} \theta }.$$