Why is their a penetration depth in superconductors if the Meissner effect perfectly expels the applied external field? I’m a little confused about the Meissner effect and the penetration depth. Does the penetration depth only happen above critical values for Hc, Jc, and Tc? Does it happen in type two superconductors above Hc1? In other words does it only happen in certain regimes of type 1 and type 2 superconductors or does it happen even in the superconducting state?  If the Meissner effect entails that the induced Magnetization from surface currents perfectly cancels out the external field then why does the external field partially penetrate the superconductor. I understand this is a lot of information so I apologize for my jumbled thoughts I have just yet to find a clear answer.
 A: The penetration depth is always non-zero, for both type 1 and type 2 superconductors. This is even the case for a perfect superconductor at 0 temperature. This happens below Hc, Tc, and Jc.
The perfect cancellation of the magnetic field only happens in the bulk of the superconductor. The magnetic field actually decays exponentially from the exterior to the interior with a characteristic length. That length is the penetration depth.
A: According to Maxwell's equations, it is not possible for the magnetic field to vanish suddenly at the surface of the superconductor. This corresponds to the differential equation,
$$\nabla \cdot \vec{B} = 0$$
As mentioned above, the magnetic field is expelled in the bulk of the superconductor. How does this happen? The supercurrent screens the magnetic field. For a time independant field,
$$\vec{\nabla}\times\vec{B}=\mu_0\vec{J}$$
There is a useful vector identity,
$$\vec{\nabla}\times(\vec{\nabla}\times\vec{B})= -\nabla^2\vec{B}$$
The left-hand side of the above equation describes the rotation of the supercurrent. For low energies, the magnitude of the supercurrent is proportional to the microscropic change in vector potential across its direction of flow minus the microscopic change in its phase (ref. page 176). Another vector identity states that the curl of a gradient is zero. Hence,
$$\vec{\nabla} \times \vec{J} \sim \vec{\nabla} \times (\delta \vec{A} - \delta \vec{\nabla}\phi) = \vec{\nabla} \times (\delta \vec{A}) = \vec{B}$$
Finally we arrive at the expression,
$$\nabla ^2 \vec{B} \sim \vec{B}$$
Since we know there is no magnetic field inside the bulk of the superconductor, this implies that the magnetic field decays expoentially as $ B = B_0 e^{-x/{\lambda_L}}$. The London penetration depth $\lambda_L$ defines how deep the magnetic field can penetrate.
Source: The Classical Condensates, Chapter 6.1
