# Combined lens & mirror system

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1. Apr 17, 2017

### horsedeg

1. The problem statement, all variables and given/known data
The object in the figure below is midway between the lens and the mirror, which are separated by a distance d = 24.9 cm. The magnitude of the mirror's radius of curvature is 20.4 cm, and the lens has a focal length of -16.5 cm.

(a) Considering only the light that leaves the object and travels first toward the mirror, locate the final image formed by this system.
(b) Is this image real or virtual?
(c) Is it upright or inverted?
2. Relevant equations
1/p + 1/q = 1/f

3. The attempt at a solution
Considering only the light that goes to the mirror, I plugged in the focal point of the mirror (using the radius) as well as the object distance (p) to the formula above and this would give me the virtual image (q) created by the light going into the mirror. This gives me a distance of 56.44 to the left of the mirror. Ideally this would become the real object for the lens, but I don't see how it makes sense in this case. Wouldn't the light be going through the lens on the way from the mirror to the location of the virtual image? Is there something I'm misunderstanding? In addition, as a result the object location that would be used for the lens in the equation would supposedly be negative as well and I'm not sure why. I guess it makes sense because if someone was looking at the system from the left through the lens to look at the object, the projected object would be behind the lens which would mean it is negative for a real object. Just doesn't make sense to me when considering the previous issue though.

2. Apr 17, 2017

### stevendaryl

Staff Emeritus
Even though both lenses and mirrors use the equation $\frac{1}{p} + \frac{1}{q} = \frac{1}{f}$, the quantity $q$ is measured differently in the two cases. For a lens, positive $q$ means on the opposite side of the object, while negative $q$ means on the same side. For a mirror, positive $q$ means on the same side as the object, while negative $q$ means on the opposite side.

So in your case, for the mirror, you get $-56.44$. Since that's negative, it's on the opposite side of the mirror from the object. So this image is to the right of the mirror, not the left.

But your post has another error: you say "Wouldn't the light be going through the lens on the way from the mirror to the location of the virtual image?".

If your calculation had been correct, and the image due to the mirror had been to the left of the lens, then that would have been a case of a "virtual object", one with a negative value of $p$. The formula relating p, q and f still works in that case.