Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm doing magnification and lens in class currently, and I really don't get why virtual and real images are called what they are.

A virtual image occurs the object is less than the focal length of the lens from the lens, and a real image occurs when an object is further than focal length.

By why virtual and real? What's the difference? You can't touch an image no matter what it's called, because it's just light.

share|improve this question

3 Answers 3

up vote 19 down vote accepted

You can project a real image onto a screen or wall, and everybody in the room can look at it. A virtual image can only be seen by looking into the optics and can not be projected.

As a concrete example, you can project a view of the other side of the room using a convex lens, and can not do so with a concave lens.


I'll steal some image from Wikipedia to help here:

First consider the line optics of real images (from http://en.wikipedia.org/wiki/Real_image):

Notice that the lines that converge to form the image point are all drawn solid. This means that there are actual rays, composed of photon originating at the source objects. If you put a screen in the focal plane, light reflected from the object will converge on the screen and you'll get a luminous image (as in a cinema or a overhead projector).

Next examine the situation for virtual images (from http://en.wikipedia.org/wiki/Virtual_image):

Notice here that the image is formed by a one or more dashed lines (possibly with some solid lines). The dashed lines are draw off the back of solid lines and represent the apparent path of light rays from the image to the optical surface, but no light from the object ever moves along those paths. This light energy from the object is dispersed, not collected and can not be projected onto a screen. There is still a "image" there, because those dispersed rays all appear to be coming from the image. Thus, a suitable detector (like your eye) can "see" the image, but it can not be projected onto a screen.

share|improve this answer

For a real image, rays from a single source point converge to a single point on the other side of the lens. This means that a point on the image remains well-defined after the optical transformation (refraction by lens).

For a virtual image, rays from a single source point diverge after they pass through the lens. This means that the a point on the image is no longer well-defined after the optical transformation. Realistically, what you'll see on a screen place at the other end of a diverging (concave) lens is a large haze of weak light (since the rays have diverged so greatly).

Note that convex (converging) single lenses always produce real images, while concave (diverging) single lenses always produce virtual images. For mirrors it's generally the other way round (there are exceptions though). Similarly for double lenses, things are reversed (compared to single lenses).

share|improve this answer

Operationally, one can distinguish real and virtual images by attaching a small but very powerful light bulb to the source.

If you insert your hand near the real image, the hand will get burned because the light rays from the source actually converge at the point of the real image. If you manage to insert your hand to the point of the virtual image, your hand won't burn because there are no light rays over there (and sometimes, there is a wall).

An ordinary flat mirror creates a virtual image because there are no light rays on the internal side of the mirror.

The optics in the human eye creates a real image where the retina is located and sensitive cells actually read the information about the original source. The same mechanism works in digital cameras.

Real images may be produced by converging lenses (eye, digital camera) and concave mirrors.

Virtual images may be produced by diverging lenses and convex mirrors, but also by concave mirrors and converging lenses if you place the source within the focal length.

share|improve this answer

protected by Qmechanic Mar 13 '13 at 16:15

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.
{"cc_re__cc_re__cc_by_sa_3_0": ["sites in the Stack Exchange network\">Area 51</a></li>\r\n                        <li><a href=\"//careers.stackoverflow.com?utm_source=stackoverflow.com&amp;utm_medium=site-ui&amp;utm_campaign=footerlink\">Stack Overflow Careers</a></li>\r\n                    \r\n                </ol>\r\n            </td>\r\n    </tr>\r\n</table>\r\n                </div>\r\n            </div>\r\n\r\n            <div id=\"copyright\">\r\n                site design / logo &#169; 2014 stack exchange inc; user contributions licensed under <a href=\"http://creativecommons.org/licenses/by-sa/3.0/\" rel=\"license\">cc by-sa 3.0</a> \r\n                with <a href=\"http://blog.stackoverflow.com/2009/06/attribution-required/\" rel=\"license\">attribution required</a>\r\n            </div>\r\n            <div id=\"svnrev\">\r\n                rev 2014.9.30.1898\r\n            </div>\r\n            \r\n        </div>\r\n    </div>\r\n    <noscript>\r\n        <div id=\"noscript-warning\">Physics Stack Exchange works best with JavaScript enabled<img src=\"http://pixel.quantserve.com/pixel/p-c1rF4kxgLUzNc.gif\" alt=\"\" class=\"dno\"></div>\r\n "], "copyright__copyright__copyright": ["d convex mirrors, but also by concave mirrors and converging lenses if you place the source within the focal length.\n\nshare|improve this answer\n\nprotected by Qmechanic Mar 13 '13 at 16:15\n\nThank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.\n\nWould you like to answer one of these unanswered questions instead?\n\nNot the answer you're looking for? Browse other questions tagged or ask your own question."]}