Lens Combination

1. Nov 14, 2008

Shackleford

21. Two 25cm focal length converging lenses are placed 16.5cm apart. An object is placed 35cm in front of one lens. Where will the final image formed by the second lens be located? What is the total magnification?

HERE IS WHAT I NEED CHECKED. Because the image formed by the first lens is on the opposite side of the second lens from where the light is emanating (right side of second lens), I give the new object distance for the second lens a negative sign. Plugging those numbers into the thin lens equation, I get an image distance of .1848 m. This has a positive sign because it is on the opposite side from where the light is originally emanating.

2. Nov 14, 2008

alphysicist

Hi Shackleford,

Yes, that's right; it will be a virtual object and the object distance for the second lens will be negative.

Yes, a positive image distance for the final lens means here that the final image will be to the right of the second lens.

3. Nov 14, 2008

Shackleford

The book says image distance is positive for a real image and negative for a virtual image. The image distance for the first lens is positive. Wouldn't that make it a real image since the rays do pass through the point? The image distance is also positive for the second lens.

How do you draw the ray diagram to get everything in its proper analytical spot? I think incorporating the second lens is confusing me a bit in this regard.