A. Homogenous layer with a planar reflector

Let equation

$$z=x\tan \phi + h_{0}$$

describe the location of a reflector.

First step: calculation of two-way zero-offset travel time in a true model:

$$t_{0} = {2 \sin \phi \over v} x + {2h_{0} \cos \phi \over v}.$$

Second step: reverse Eiconal's continuation of t₀(x) with wrong velocity $(v_{c}={v^{\prime} \over 2}, {\:}v^{\prime} \neq v)$: 

$$\tau^{(-)}(x,z)= {2h_{0} \cos \phi \over v} + {2 \sin \phi \... ...t{1-{\left( {v^{\prime} \over v}\right) }^{2} \sin ^{2} \phi }.$$ (73)

Third step: imaging from the condition  

$$\tau^{(-)} (x,z)=0.$$ (74)

Inserting equation (73) into equation (74) and solving with respect to z, we derive location of the image in depth-section:

$$z= {({v^{\prime} \over v}) \cos \phi \over \sqrt{1-{({v^{\prime}\over v})}^{2} \sin^{2} \phi }} (h_{0}+ x \tan \phi ).$$

The reflection image in time section we receive after substitution $z=tv^{\prime}$ is:

$$t= {1 \over v} {\cos \phi \over \sqrt{1-{({v^{\prime}\over v})}^{2}\sin^{2} \phi }} (h_{0} + x\tan \phi ).$$

Time-to-depth migration in this situation is depth scaling with true velocity $(t={z\over v})$:

$$z= {\cos \phi \over \sqrt{1-{({v^{\prime}\over v})}^{2} \sin^{2} \phi }}(h_{0} + x\tan \phi ).$$

We see that depth scaling does not supply correct imaging, although the error is not big: it is proportional to $(v_{c}-v){\sin }^{2} \phi$.