Reflection off mirror problem

In summary, when dealing with a spherical mirror that is concave and magnifies an image by two times, the image will be inverted if it is real and upright if it is virtual. This can be determined using the equation 1/so+1/si=1/f, and the convention is for a negative magnification to indicate an inverted image. However, it is important to remember that these conventions are just agreements and the fact remains that virtual images are always upright and real images are always inverted in basic single lens or mirror optics.
  • #1
positron
21
0
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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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 

1. How does reflection work in a mirror?

When light hits a mirror, it bounces off at an angle equal to the angle at which it hit the mirror. This is known as the law of reflection. The smooth surface of the mirror causes the light to reflect in a predictable way, creating a clear image.

2. Why do objects appear reversed in a mirror?

Objects appear reversed in a mirror because of the way that light reflects off of the mirror's surface. When an object is placed in front of a mirror, the light rays that bounce off of it are reflected in the opposite direction, creating a mirror image of the object.

3. Why do some mirrors distort our reflection?

Some mirrors may distort our reflection due to imperfections in the surface of the mirror. This can cause the light rays to reflect at slightly different angles, resulting in a distorted image. Mirrors made from high-quality materials and with smooth surfaces will produce a more accurate reflection.

4. How is the distance between an object and its reflection determined?

The distance between an object and its reflection is determined by the angle at which the light rays reflect off of the mirror's surface. The farther away the object is from the mirror, the larger the angle of reflection will be, resulting in a larger distance between the object and its reflection.

5. Can reflections in a mirror be altered or manipulated?

Yes, reflections in a mirror can be altered or manipulated using various techniques such as tilting the mirror or using multiple mirrors to create a series of reflections. However, the basic principles of reflection will still apply, and the angle of reflection will always be equal to the angle of incidence.

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