I know that an image on a plane mirror is always upright and the same size, but is there any focal point? Could there actually be no focal point because of that?
[Physics] Is a focal point anywhere within a plane mirror
reflection
Related Solutions
As you mentioned, a plane mirror will produce a virtual image of a real object. But indeed, it is correct that a plane mirror will also produce a real image of a virtual object. This can occur when you have more than one optical element in the optical system. Then the object of one component becomes the image of the next optical component.
So let's give an example of a real and virtual images. And how, by combining a lens and a mirror we can get a virtual (intermediate) object. We'll use the standard convention (see https://apps.spokane.edu/InternetContent/AutoWebs/AsaB/Phys103/MirrorsThinLens.pdf).
On the top, we have an object to the left of the focal point in a converging lens. This forms a real image. In this example, we have $d_{o}=+2f$, if we solve the lens equation we get that $d_{i}=+2f$. Since the image distance is positive, it is real.
The upper middle image shows how we can form an imaginary image by moving the object closer to the lens than the focal length. In this particular example, $d_{o}=+f/2$. Solving the lens equation gives us $d_{i}=-f$. We have a virtual image at a distance of $f$ to the left of the lens. Note that the dashed gray lines aren't actual rays of light, but are backward extensions. If we were to put a screen where the virtual image was, we wouldn't see an image on the screen.
The lower middle image shows us how a mirror takes real object and generates virtual image.
Finally, on the bottom image, we see a two component system composed of 1) converging lens and 2) a mirror. The mirror is a distance of $3f/2$ to the right of the lens. The first component, the lens, acts the same way as on the top figure with $d_{o1}=+2f$ and $d_{i1}=+2f$. Now the image of the lens acts as the object for the mirror, but note that this image/object lies behind mirror. It is therefore a virtual object. The object distance is $d_{o2}=3f/2-2f=-f/2$. The image distance therefore is $d_{i2}=+f/2$ to the left of the mirror. It is a real image.
It is customary to define the focal point of a spherical mirror as that point on the principal axis where rays which are near to and parallel to the principal axis meet.
The reason for including the words near to can be shown as follows.
In the diagram below there is a concave mirror of radius $R$ and centre $C$.
The principal axis is $CP$ and the incident ray, parallel to the principal axis , is shown as $AB$.
Triangles $ACB$ and $CFB$ are similar so $$\dfrac {AB}{AC}= \dfrac {CB}{CF} \Rightarrow \dfrac {AB}{R}= \dfrac {R}{CF}$$
As the incident ray gets closer to the principal axis $AB \to 2R$ and so $CF \to \dfrac R2$.
So it is really all to do with how close the length of $AB$ to a diameter $2R$.
Using such a construction you could quantify the words near to by deciding how different $CF$ can be from $\dfrac R2$.
A parabola $y=\frac 14 x^2$ (blue) and a circle $(y-2)^2+x^2=4$ (red) are shown in the diagram below.
Curvature of both graphs is the same at point $P$ as is the centre of curvature $C_{\rm circle}$.
Just by eye you can see that the two graphs are "very" similar close to $P$.
All rays parallel to the principal axis arrive at $F$ at the same time because $BF= BD$ and so this is the focal pont of the parabola.
There is however a restriction in that this will only happen if the incoming parallel rays are parallel to the principal axis.
The position of the centre of curvature of the parabola $C_{{\rm parabola},B}$ depends on the position of $B$ and only coincides with that of the circle at position $P$.
The necessary construction to find the centre of curvature os a parabola is shown below and more details can be found here from where the gif image was taken.
Best Answer
A focal 'point' implies a convergence of light rays to some point in space whether it be a real or virtual point. And convergence of either transmissive or reflective optics requires curvature in the optics - so for plane mirrors, no there is no focal point that can occur by reflected light.