Are two lights waves that are out of phase mirror images? Is light shifted 180 degrees out of phase, when it is reflected off a mirror?
When light is shifted 180 degrees out of phase, is it the mirror image of the light wave?
 A: For an ideal mirror it is shifted by 180 degrees. For imaging this means a shift of $\lambda/2$ along the propagation direction. This is not related to the mirroring effect, which is caused by the change of sign of $k_\perp$.
A: When an electromagnetic wave is reflected from a perfectly conducting mirror,* the electric field ${\bf E}$ is phase shifted by $180^{\circ}$.  However, there is also the magnetic field ${\bf B}$, and the magnetic field is not phase shifted.  There is a tendency to identify the wave with its electric field, and this is entirely natural, since in most situations, the electric field is what determines the wave's interactions with matter—including the phenomena of emission, absorption, and scattering.  Yet the magnetic field is an intrinsic part of the wave and carries just as much energy as the electric field.
To make this definite, let us consider a plane wave approaching from $z=-\infty$, propagating toward a mirror. If the ($z<0$) fields of the incoming wave are (in SI units)
$${\bf E}_{i}({\bf r},t)=E_{0}\hat{{\bf x}}\cos(kz-\omega t+\phi)\\
{\bf B}_{i}({\bf r},t)=\frac{E_{0}}{c}\hat{{\bf y}}\cos(kz-\omega t+\phi)$$
(note that these are in phase, as is the case for propagating waves), and they are reflected by a conducting mirror plane located at $z=0$, the fields of the reflected wave are
$${\bf E}_{r}({\bf r},t)=E_{0}\hat{{\bf x}}\cos(-kz-\omega t+\phi+\pi)=-E_{0}\cos(-kz-\omega t+\phi)\\
{\bf B}_{r}({\bf r},t)=\frac{E_{0}}{c}\hat{{\bf y}}\cos(-kz-\omega t+\phi).$$
The phase shift of the electric field (by $\pi$ radians or $180^{\circ}$) is equivalent to an overall minus sign, as shown in the second formula for ${\bf E}$.
The phase shift (or minus sign) ensures that at the location of the mirror, $z=0$, the $x$-component of the electric field vanishes, as it must at the surface of a conductor,
$${\bf E}(z=0,t)={\bf E}_{i}(z=0,t)+{\bf E}_{r}(z=0,t)=0.$$
However, the magnetic field component does not vanish at the surface; instead
$${\bf B}(z=0,t)={\bf B}_{i}(z=0,t)+{\bf B}_{r}(z=0,t)=\frac{2E_{0}}{c}\hat{{\bf y}}\cos(-\omega t+\phi).$$
This makes sense, since the boundary condition for the magnetic field is not that ${\bf B}=0$ at the surface; however, since $\dot{{\bf E}}$ and $\nabla\times{\bf B}$ are related by the Ampere-Maxwell Law, there is a boundary condition that $\partial{\bf B}/\partial t=0$ at $z=0$.
The fact that ${\bf E}$ has a phase shift/sign change at the mirror while ${\bf B}$ does not is also crucial to understanding the energy flow in the system. If we calculate the Poynting vector (which gives energy transfer per unit time per unit area) purely for the incoming fields, the result is
$${\bf S}_{i}=\frac{1}{\mu_{0}}{\bf E}_{i}\times{\bf B}_{i}=\frac{E_{0}^{2}}{\mu_{0}c}\hat{{\bf z}}\cos^{2}(kz-\omega t+\phi).$$
This gives, not surprisingly, a consistently nonnegative energy flow, always oriented in the direction of propagation, the $z$-direction. The direction of ${\bf S}$ arises from the direction of the cross product $\hat{{\bf x}}\times\hat{{\bf y}}=\hat{{\bf z}}$, since ${\bf E}_{i}$ and ${\bf B}_{i}$ are aligned along $\hat{{\bf x}}$ and $\hat{{\bf y}}$, respectively. However, the Poynting vector for just the reflected wave has a sign change, because the sign of ${\bf E}_{r}$ is flipped, but the sign of ${\bf B}_{r}$ is not. The result is
$${\bf S}_{r}=\frac{1}{\mu_{0}}{\bf E}_{r}\times{\bf B}_{r}=-\frac{E_{0}^{2}}{\mu_{0}c}\hat{{\bf z}}\cos^{2}(-kz-\omega t+\phi),$$
which shows energy flow in the $-z$-direction.
The sum ${\bf S}_{i}+{\bf S}_{r}=0$. The total energy flow in the system is naturally zero; the incoming wave is carrying energy toward $z=0$ in the $+z$-direction, while the reflected wave is carrying it back toward $z=-\infty$ in the $-z$-direction. However, this does not tell quite the whole story. While the sum ${\bf S}_{i}+{\bf S}_{r}$ gives the total aggregate energy correctly, it does not capture the behavior of the energy on scales of a wavelength $\lambda=2\pi/k$ or smaller. The true Poynting vector must be calculated using the full electric and magnetic fields, with both the incoming and reflected wave included by superposition.
The full electric field is actually
$${\bf E}({\bf r},t)={\bf E}_{i}({\bf r},t)+{\bf E}_{r}({\bf r},t)=E_{0}\hat{{\bf x}}\left[\cos(kz-\omega t+\phi)-\cos(-kz-\omega t+\phi)\right]\\
=2E_{0}\hat{{\bf x}}\sin(kz)\sin(\omega t-\phi).$$
This electric field has the structure of standing wave. The two traveling waves, propagating in opposite directions, interfere to produce a standing wave. At every point, the field oscillates in the $\pm x$-direction with angular frequency $\omega$. However, the amplitude of the oscillations depends on the $z$-position. Where $kz$ is an integral multiple of $\pi$ the amplitude is zero; there is complete destructive interference at these nodes. Where $kz$ is a half-integral multiple of $\pi$, we instead have perfect constructive interference (anti-nodes).
The total magnetic field ${\bf B}={\bf B}_{i}+{\bf B}_{r}$ is similar, but because the magnetic fields of the incident and reflected wave have a different phase relation, the standing wave looks a little different,
$${\bf B}({\bf r},t)=\frac{2E_{0}}{c}\hat{{\bf y}}\cos(kz)\cos(\omega t-\phi).$$
The locations of the nodes and antinodes have been interchanged; ${\bf E}$ interferes constructively where ${\bf B}$ interferes destructively, and vice versa. If we calculate
the Poynting vector using the complete fields, including interference, the result it
$${\bf S}=\frac{1}{\mu_{0}}{\bf E}\times{\bf B}=\frac{4E_{0}^{2}}{\mu_{0}c}\hat{{\bf z}}
\left[\sin(kz)\cos(kz)\right]\left[\sin(\omega t-\phi)\cos(\omega t-\phi)\right]\\
=\frac{E_{0}^{2}}{\mu_{0}c}\hat{{\bf z}}\sin(2kz)\sin(2\omega t-2\phi).$$
So it turns out that the energy flow is not exactly zero, because of the mismatch between the electric and magnetic standing waves. However, because ${\bf S}$ varies sinusoidally, ${\bf S}$ will average to zero if it is averaged over either time or space. So there is energy moving around, but it is only oscillating back and forth on scales of less than half a wavelength.
*There are other arrangements of matter than can produce mirror, and with more exotic mirrors that are not just made of sheets of conductor, the behavior of both ${\bf E}$ and ${\bf B}$ can be considerably more complicated. However, I will limit my attention here to the purely conductive mirror, which is what the question appear to refer to.
