What does a negative distance indicate in a combination of lenses?

[synergix]

Homework Statement

As shown in the diagram below (not to scale), a 3.0 cm high object is placed 50.0 cm from thin converging lens A that has a focal length of 10.0 cm. Converging lens B with a focal length of 15.0 cm is placed 25.0 cm to the right of lens A. Find and describe the final image.
http://webct6.nic.bc.ca/webct/RelativeResourceManager/Template/CourseMaterials/CourseContent_2007FA/Assignments/PHY060W_Assignment_09_files/image013.jpg

Homework Equations

1/d_i = 1/f - 1/d_o

The Attempt at a Solution

1/ d_ia = 5/50.0cm - 1/50.0cm= 1/12.5 cm

d_ia=12.5cm

d_ob= 25.0cm - 12.5 cm = 12.5 cm

f_b= 15.0cm

1/d_ib= 1/15.0cm- 1/12.5cm=

12.5/187.5 - 15.0/187.5=

-2.5/ 187.5

d_ib= -75

so now to my question. I am not sure what a negative distance indicates when dealing with a combination of lenses. does it mean the image is virtual? because I know from my diagram that it is on the opposite isde of the lenses from where the light is coming.

 

[jbsteele]

I am assuming that since it is a negative distance that the image is virtual, upright, and magnified. Please let me know if I am correct or not. Thanks

 

[nlightner]

A negative distance in a combination of lenses indicates that the image formed is a virtual image. This means that the light rays appear to be coming from a point behind the lens, rather than actually converging at that point. In this case, the negative distance for the final image indicates that it is located behind the lenses, on the opposite side from the object. This is consistent with the diagram, where the final image is shown on the right side of the lenses. Additionally, a negative distance for the final image also indicates that the image is magnified, since the distance is greater than the distance from the object to the first lens.