Is It Possible to Create a Virtual Image by Adjusting Object and Lens Positions?

[lampshader]

Homework Statement

Can you move the object and the lens in such a way that the image is virtual? If so, relate where the object is with respect to the focal length of the lens

Homework Equations

1/f =1/p +1/q

or

f = (p + q)^-1

The Attempt at a Solution

What I can do, is add a convex lens into my lab environment. I can use the source of the light and shine it directly at the center of the convex lens. What happens is that the light is bent in some way (can't descibe it), but I know that the light does in fact go through the convex lens, so, by defintion the image is virtual.

I know that it is possible to get a virtual image, but how would I relate the objects positon?

I know that the mathematical phrase 'with respect' is a d/dx. But this is an algebra based course, so having to use leibniz (however its spelt) is confusing.

Could someone help with this? Thanks in advance!

 

[ideasrule]

I don't get the question. A convex lens ALWAYS produces a virtual image. The question asks us to "relate where the object is", but relate it with what? Where the image is?

 

[lampshader]

Yes, the object is the source of the light beam, my lab environment does not have a scale, so I don't know whehter the convex lens is inches, meters, or even how many feet it may be away from the object (light beam).

I do have some data.

the object (light) has the coordinates:

x = 23.2
y = 87.25

and the convex lens has:

x = 242.0
f = 50.0

What formula do I use to relate where the object is with respect to the focal lenght?

ideasrule, I appreciate your time with my problem.

 

[collinsmark]

I don't get the question. A convex lens ALWAYS produces a virtual image. The question asks us to "relate where the object is", but relate it with what? Where the image is?

Actually, ideasrule, I think you mean "concave" lens (or a convex "mirror"). A concave lens can only produce virtual images (cannot produce real images).

Can you move the object and the lens in such a way that the image is virtual? If so, relate where the object is with respect to the focal length of the lens

[...snip...]

What I can do, is add a convex lens into my lab environment. I can use the source of the light and shine it directly at the center of the convex lens. What happens is that the light is bent in some way (can't descibe it), but I know that the light does in fact go through the convex lens, so, by defintion the image is virtual.

Hello lampshader. And sorry. I disagree with your definition. Just because light travels through the lens doesn't make the image virtual.

Camera lenses produce real images on the film plane (well, replace 'film' with CCD or CMOS sensors these days). What kind of compound lens (overall concave or overall convex) is used on cameras?

A magnifying glass is a simple example of a convex lens. If you vary the height of the magnifying glass above a table, is it ever possible to focus in a real image on the table, of an actual light source on the ceiling? (i.e. looking at the table, could you adjust the glass's height in order to see an image of the light source; such that the magnifying glass is acting like a camera lens?)

Now, what happens when you use the magnifying glass normally, such was when reading the newspaper, or fine print. What kind of image are you looking at in that case (real or virtual)?

I know that it is possible to get a virtual image, but how would I relate the objects positon?

I know that the mathematical phrase 'with respect' is a d/dx. But this is an algebra based course, so having to use leibniz (however its spelt) is confusing.

Could someone help with this? Thanks in advance!

I have a suspicion that you are over-thinking this question. When the question asks, "relate where the object is with respect to the focal length of the lens," I believe it's merely asking you to specify if the actual object (not the image) is between the focal length and the lens, or on the far side of the focal length (relative to the lens).