geometric-opticsopticsreflection
Can a plane mirror form a real image?
I came across this question in an examination.
We know that a plane mirror forms virtual images of objects,
But how can a plane mirror form a real image of an object?
I came to know that the answer is yes; but I don't see how.
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.
Any discussion of concave/convex mirrors needs to begin with a statement of the particular version of the mirror equation to be used, along with the convention for setting and interpreting the signs of focal lengths, and object/image positions.
For example, from http://scienceworld.wolfram.com/physics/MirrorFormula.html:
Unmentioned in this is the convention that virtual images and objects are found behind the mirror and have negative values of $d$.
In the particular example you present, the image formed by the primary mirror is a real image. If you put infinity for the object distance and a positive focal length, you find a positive image distance.
But when you insert a convex mirror, with a negative focal length, into the optical path, you must also consider the position of the real image (now an object) relative to the convex mirror. The object is behind the convex mirror; it is a virtual object, and its distance from the convex mirror is negative.
With appropriate positioning of the convex mirror, the formula will produce a positive value for the image distance. There will be real image formed in front of the convex mirror.
You've just designed a Cassegrain telescope...
Best Answer
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.