Magnification, when is it negative?

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SUMMARY

Magnification in optics is defined by the equation m = v/u, where m represents magnification, v is the image distance, and u is the object distance. Positive magnification indicates an erect image, which is typically a virtual image, while negative magnification signifies an inverted image, usually a real image. However, in multi-element optical systems, positive magnification can occur without a virtual image, as demonstrated by systems utilizing relay lenses. The sign convention dictates that a virtual image has a negative image distance, leading to the cancellation of negatives and resulting in positive magnification.

PREREQUISITES
  • Understanding of optical systems and lens behavior
  • Familiarity with the magnification equation m = v/u
  • Knowledge of sign conventions in optics
  • Basic principles of virtual and real images
NEXT STEPS
  • Study multi-element optical systems and their magnification properties
  • Explore the role of relay lenses in image formation
  • Review sign conventions in optics for various lens types
  • Investigate the differences between virtual and real images in detail
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding the complexities of image formation and magnification in optical systems.

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.
 

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