Are Real Images Always Inverted and Virtual Always Erect in Ray Optics?

[Krushnaraj Pandya]

Homework Statement

My question is, are real images always inverted and virtual always erect?

Homework Equations

1/v+1/u=1/f for mirrors
and 1/v-1/u=1/f for lenses

The Attempt at a Solution

Consider a concave lens, with object at -x. the condition for a virtual image is v<0; i.e on the same side of the lens as the object. The condition for image being erect is m>0 or v/u > 0. This suggests that a virtual image is always erect; can I generalize this for every case in ray optics?

 

[YoungPhysicist]

Homework Statement

My question is, are real images always inverted and virtual always erect?

Homework Equations

1/v+1/u=1/f for mirrors
and 1/v-1/u=1/f for lenses

The Attempt at a Solution

Consider a concave lens, with object at -x. the condition for a virtual image is v<0; i.e on the same side of the lens as the object. The condition for image being erect is m>0 or v/u > 0. This suggests that a virtual image is always erect; can I generalize this for every case in ray optics?

For a single convex or concave lens,yes.Real images are always inverted and virtual images are always erect,but when there are multiple lenses that doesn’t hold true.

 

[YoungPhysicist]

Here is a chart I made on our physics class:
Every possible way of a convex lens forming images(F is the assumed focus point,2F is two times the distance than the focus )

Concave:

 

[jtbell]

Every possible way of a convex lens forming images

What about virtual objects?

 

[YoungPhysicist]

What about virtual objects?

Virtual objects are also shown in the image,the bottom right one of convex lens and every image in concave lens are virtual. I drew then as dotted lines. (They may not be clear)

 

[jtbell]

I said virtual object, not virtual image. In all of your diagrams, the object is real, to the left of the lens, with u < 0 in the Cartesian sign convention that @Krushnaraj Pandya is using. A virtual object is to the right of the lens (more precisely, on the side opposite the side the light is coming from), with u > 0 in the Cartesian sign convention. You can see a diagram of one example of a virtual object here:

https://www.physicsforums.com/threads/what-is-a-lens.763141/

(edit: I corrected my original "v < 0" and "v > 0" above to "u < 0" and "u > 0". I got my u's and v's mixed up. )

My question is, are real images always inverted and virtual always erect?

By "always", do you include virtual objects, as described above?

 

[Krushnaraj Pandya]

I said virtual object, not virtual image. In all of your diagrams, the object is real, to the left of the lens, with u < 0 in the Cartesian sign convention that @Krushnaraj Pandya is using. A virtual object is to the right of the lens (more precisely, on the side opposite the side the light is coming from), with u > 0 in the Cartesian sign convention. You can see a diagram of one example of a virtual object here:

https://www.physicsforums.com/threads/what-is-a-lens.763141/By "always", do you include virtual objects, as described above?

The purpose of my question was to get informed of any cases at all (if they exist)- virtual objects or real- to be forming a real erect OR virtual inverted image. Is that possible in a single lens/mirror system?

 

[Krushnaraj Pandya]

For a single convex or concave lens,yes.Real images are always inverted and virtual images are always erect,but when there are multiple lenses that doesn’t hold true.

Thank you, it is very thoughtful of you to upload those pictures, can you show me an example of a multiple lens system forming an erect real image and vice versa?

 

[jtbell]

The purpose of my question was to get informed of any cases at all (if they exist)- virtual objects or real- to be forming a real erect OR virtual inverted image. Is that possible in a single lens/mirror system?

A virtual object requires another lens or mirror, in order to produce the virtual object. Do you mean a single lens/mirror in addition to that one?

If yes, then the example on the page that I linked to, is a real, erect image. You can verify this by using the thin-lens equation with a suitable value of u > 0. Hint: try both u > f and u < f, i.e. object outside and inside the focal length.

If no, then the object must be real (u < 0). You should be able to enumerate all four possible combinations of object outside and inside the focal length, with a converging and a diverging lens, and determine whether any of them produce a real erect or virtual inverted image.

 

[YoungPhysicist]

Thank you, it is very thoughtful of you to upload those pictures, can you show me an example of a multiple lens system forming an erect real image and vice versa?

Compound microscopes are a great example of an inverted virtual image:
https://goo.gl/images/75NFeM