If we place two plane mirrors facing each other wouldn't the image produced by the first mirror act as a virtual object to the second mirror? If so, would the image produced by the second mirror then be a real or virtual image? Please explain how to identify real and virtual images as I think my understanding might be flawed.
Do Virtual Objects always Produce Real Images
geometric-opticsopticsreflection
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The "rule" that real images are always inverted is not correct. That rule might work when you have only a single optical element (like a lens), but not necessarily when you have two or more.
Take a look at this:
That's a real, upright (aka erect) image labeled $I_B$. You can tell it's real because the rays at the final image actually converge at that physical location, unlike virtual images whose location of "convergence" does not actually have physical rays passing through it (only the "backtracking" rays one typically draws).
As for your projector and mirror, you can draw a ray diagram carefully (if you know the internal workings of a projector) and apply the same test to see if the image is virtual or real. But do remember that projectors can be modified so that the image you see is inverted to accommodate mirrors and such. Anyway, to get something projected on to a screen, I do believe the image needs to be real, so I would say it is indeed a real image you're seeing.
Finally, here's a simple example of two elements that again goes against the rules in your book:
The situation above has a real, inverted image using a lens and mirror (a mirror!). Thus, the rules you've been given aren't general enough to deal with multiple-element situations.
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.
Best Answer
All the images a plain mirror can make are virtual. i. e. you can make a real image only with for example a lens, like in your ey or in a camera. real images you can project on a screen or your retina or a photo plate . Virtual pictures are pictures only for the ey or camera "real". the light rays seem to come from points where you localize the virtual picture, You should learn how to find virtual pictures in a mirror.