How fine a crack can you see

The reflection coefficients of (12.14) are for a crack with nothing in it, so that the two surfaces of the crack could move freely with respect to one another. In that case the calculations unequivocally give strong contrast from cracks regardless of their thickness (or thinness ). In the microscope there is water 

Suppose that the effect of the fluid is to transmit normal components of displacement and traction across the crack, but not tangential components. With the crack in the half plane x = 0, z 0, and with the notation 

A superposition of this scattered field and the incident field would satisfy the boundary conditions. An approximation to the scattered field can be generated by a distribution of body forces Qz acting in the z-direction in the half-plane x = 0, z 0, i.e. where the crack would be. Because of symmetry this distribution of body forces gives u = 0 at x = 0, and also rz(z) = Qz(z)/2 just to the left of the plane of the crack. Hence for x 0 the scattered field is found by selecting 

Now the field due to a delta function source Qz = S(z — zo)5(x) is known as Lamb s problem, and at some distance from x = 0 the solution is known (Achenbach 1973). If the surface displacement of the surface wave generated by the concentrated load is wz x, C) and the depth of the crack is d, then for the distributed body forces, by superposition, 

The transmitted field to the right of the crack can be calculated similarly, with a change of sign in eqn (12.38) and in the x-dependence of wc(x, () and wsc(x). In this way approximate values of 1r and Rr can be found for the two-dimensional crack contrast theory, and possibly for the three-dimensional theory as well. The calculated field is reasonably good near the free surface but not near the crack tip, so the approximations are better the deeper the crack is compared with the wavelength. 

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