How Do You Determine Image Location and Type Using a Spherical Mirror Equation?

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Homework Help Overview

The discussion revolves around determining image location and type using the spherical mirror equation, specifically in the context of concave mirrors. Participants are exploring the relationships between object distance, image distance, and magnification, while grappling with the conventions of sign usage in mirror equations.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand how to apply the spherical mirror equation, particularly in determining whether an image is virtual or real based on given conditions. Questions arise regarding the sign conventions for mirrors and the implications of magnification values.

Discussion Status

The discussion is active, with participants sharing insights about the nature of images formed by concave mirrors. Some guidance has been offered regarding the relationship between image type and orientation, though there is no explicit consensus on the best approach to understanding the underlying principles.

Contextual Notes

Participants express confusion over the varying conventions in different texts regarding the sign of magnification and image distance. There is an acknowledgment that these conventions are agreed upon rather than universally fixed, which adds to the complexity of the problem.

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If you place an object a distance so from a spherical mirror, how do you calculate the image location given that the image is right side up and magnified by two times? I know the equation but I always get confused about what signs to use for lenses vs. mirrors, etc.

I know the magnification is si/so = 2. I don't understand how to determine whether the image should be virtual or real, from the equation 1/so+1/si = 1/f. Maybe if someone can show me how to use this equation in the context of a reflecting mirror I will understand how to do the problem.
Thanks
 
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With a spherical mirror that is concave (the magnifying kind) any image that is "real" will be inverted. Any image that is upright will be virtual.

In this case the magnification is 2.0, and image distance will be a negative value according to convention.
 
Chi Meson said:
With a spherical mirror that is concave (the magnifying kind) any image that is "real" will be inverted. Any image that is upright will be virtual.

In this case the magnification is 2.0, and image distance will be a negative value according to convention.

How do you know that any image that is real is inverted and that any image that is virtual is upright, based on the equation? I know I can just memorize this fact, but is there another way to think about it?
 
It is a fact. The equation is modified to fit the fact. Unfortunately, the equation changes depending on the text. In many cases the magnification equation is: m=si/so=-hi/ho where the negative is thrown into have a "positive magnification" indicate a real image.

Sometimes the negative is left out in order to have a negative magnification indicate an inverted image.

These are only conventions (something people agreed to do). But to repeat: it starts as a fact, due to the simple geometric nature of basic *single* lens or *single* mirror optics, virtual images are always upright, real images are always inverted.
 

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