Magnification, when is it negative?

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Magnification is positive when the image is erect and virtual, while it is negative for inverted, real images. The confusion arises from the equation m = v/u, where magnification is negative when the image distance (v) is negative, indicating a virtual image on the object's side of the lens. This contradicts the general rule that associates virtual images with positive magnification. However, in multi-element optical systems, positive magnification can occur without a virtual image, as seen with relay lenses. Understanding the sign conventions and the context of the optical system is crucial for correctly interpreting magnification.
izMuted
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Homework Statement


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?

Homework Equations


m = v/u

The Attempt at a Solution

 
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izMuted said:
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.

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.

izMuted said:
However using the equation m = v/u, m is negative when v is negative.

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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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