# Homework Help: Optical lens - Virtual Object

1. Sep 8, 2012

### jaumzaum

I was solving the following problem, and I've got confused at the second question

http://img407.imageshack.us/img407/7921/43545856.png [Broken]

1) An object is put at the center of curvature C of a convergent lens. Where is the image? Calculate by gauss equation.

It's obvious that the image is in C', and is inverted and has the same size of the object.
By 1/f = 1/s + 1/s', we get 1/s'= 2/R-1/R => s' = +R (so is in the right of the lens)

2) The object now is put at C'. Where is the image? Calculate by gauss equation.

I would say it's also obvious it would be in C, inverted and equal to the object.
Now I have the following property: A lens, if turned changes nothing, s' and i stays the same. A convergent lens if seen from right to left is also convergent and with the same focus.

So I can use 1/R + 1/s' = 2/R, if I consider the light rays are coming from right to left, and I get s' = +R (R left the lens)

But now is the question:

If I consider a virtual object, and the light rays coming from left to right, we have f = +R/2 and s = -R, so we get

1/-R + 1/s' = 1/f => s' = R/3 right the lens. Why I got this wrong result? Can't I consider virtual objects by negative values?

[]'s
John

Last edited by a moderator: May 6, 2017
2. Sep 8, 2012

### TSny

f $\cong$ R/2 for a converging mirror. For a lens, f depends on the index of refraction of the lens, the index of refraction of the surrounding medium, and the radii of curvature of the two faces of the lens.

It doesn't matter which side of a lens you place a "real object". In either case, the object distance would be considered positive. Object distances are negative for "virtual objects".

If light rays from the object are diverging from the object as they reach the lens, then the object is a "real object" and s is positive (no matter which side of the lens the object is located). This is the "usual" case.

If the light rays that are approaching a lens are converging as they approach the lens, then you are dealing with a "virtual object" (no matter which side of the lens the rays are approaching from). The object distance will be the negative of the distance from the lens to the place where the rays would converge to if the lens were not there.

A virtual object is often the result of the light having already passed through a previous lens before passing through the lens that you are dealing with.