Magnification (major confusion with the signs)

[s3a]

In magnification, I keep confusing the signs. I don't know if the signs change when comparing converging and diverging lenses (apart from the focal point). I think that:

(a) real image = positive magnification
(b) virtual image = negative magnification
(c) upright/errected = positive magnification
(d) inverted/upside down = negative magnification.

But then when I try to do #12 on page 757 of Harris Benson's University Physics, "The focal length of a diverging lens is -20cm. Locate the object given that the image is (a) virtual, erect and 20% of the size of the object; (b) real, erect, and 150% of the size of the object."

For (a), I see virtual which means negative and erect which means positive so I conclude that the final magnification is negative. By this I mean 0.2 = -q/p which leads me to the correct answer. However for (b), when I attempt to use the same logic, which is that since the image is real/positive and erect/positive, the concluding magnification should be positive as well, so it should be 1.5 = q/p. But the final answer is -6.67cm according to the solution manual I'm looking at and I do get that answer when I make 1.5 = -q/p. I've been breaking my head for several days now with the signs and I still don't get it so I'd greatly appreciate a breakdown of all the situations that change the signs. (By the way - to solve the problems you have two equations, two unknowns)

Thanks in advance.

 

[ehild]

In case of real objects, the real images are upside down and virtual images are upright. Magnification is defined as the height of image / height of the object so its is negative for real images. do is the distance of the object from the lens, and it is positive if the object is in front of the lens. di is the distance of the image, and it is positive if the image is behind the lens. The magnification is hi/ho=-di/do.
You never get real image with a diverging lens from a real object, but you can have one if the object is supplied by an other lens.

For (b), the image is real, so di is positive, the image is behind the lens. It is erect, so the magnification is positive. So m=1,5=-di/do. As di >0 do must be negative, the object is not real.

ehild