# Magnification, when is it negative?

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1. Jun 7, 2017

### izMuted

1. The problem statement, all variables and given/known data
In magnification, I keep on confusing the signs. From what I understand currently, magnification is positive when the image is erect. An image is only erect when it is a virtual image, therefore virtual images = positive magnification. Vice versa, magnification is negative when the image is inverted, therefore a real image.

However using the equation m = v/u, m is negative when v is negative. And v is only negative when the image is on the same side of the lens as the object. Therefore meaning a virtual image (According to physicsclassroom "The negative value for image distance indicates that the image is a virtual image located on the object's side of the lens."). This contradicts the statement above where virtual images = positive magnification.

Where am I going wrong?

2. Relevant equations
m = v/u

3. The attempt at a solution

2. Jun 7, 2017

### Staff: Mentor

That's only true for a single lens. A multi-element optical system can have positive magnification without the image being a virtual image. Consider a system which uses a relay lens. The objective generates an intermediate image which the relay lens then images, resulting in an upright final image of positive magnification. No virtual image necessary.

Assuming u is the distance between the lens and the object, then sign convention dictates that it should be negative. If you imagine the lens to be at the origin of a graph, then the object is to the left of 0, which is the negative direction. A virtual image would also be to the left and would also have a negative value for distance. The negatives would cancel out and the magnification would be positive.

3. Jun 7, 2017