Must we always use absolute value for lens magnification?

In summary, the discussion revolves around the concept of magnification and height of image in physics. The usual definition of magnification includes a negative sign, which helps determine whether the image is inverted or erect. The height of an image is determined by multiplying the magnification with the height of the object. However, it is important to consider the context and the question being asked to determine whether the negative sign should be included in the answer. Ultimately, it is best to clarify with the instructor to ensure the correct answer.
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Jonforall
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My brother's physics teacher says that magnification and height of image are always positive. Is she right?
 
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  • #2
Jonforall said:
My brother's physics teacher says that magnification and height of image are always positive. Is she right?
It depends on the context. It does not make sense to have a negative magnification if you look at it strictly as a factor by which an image is reduced or enlarged. However the usual definition of the magnification is a signed quantity ##m = -(i/o)## where ##i## and ##o## are, respectively, the distance of the image and object from the lens. The convention is that if the magnification thus defined is a negative number (both ##i## and ##o## are positive), the image is inverted and real; if the magnification is a positive number (##i## is negative), the image is erect and virtual. This convention helps sort out whether images are real or virtual and inverted or erect and comes in handy when one considers the formation of images by multiple lenses. The height of an image is the magnification multiplied by the height of the object. Thus, the height of the object carries the sign of the magnification.
 
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  • #3
kuruman said:
It depends on the context. It does not make sense to have a negative magnification if you look at it strictly as a factor by which an image is reduced or enlarged. However the usual definition of the magnification is a signed quantity ##m = -(i/o)## where ##i## and ##o## are, respectively, the distance of the image and object from the lens. The convention is that if the magnification thus defined is a negative number (both ##i## and ##o## are positive), the image is inverted and real; if the magnification is a positive number (##i## is negative), the image is erect and virtual. This convention helps sort out whether images are real or virtual and inverted or erect and comes in handy when one considers the formation of images by multiple lenses. The height of an image is the magnification multiplied by the height of the object. Thus, the height of the object carries the sign of the magnification.
So should we always use absolute value in answering every problem that tells us to find height of image and its magnification?
 
  • #4
Jonforall said:
So should we always use absolute value in answering every problem that tells us to find height of image and its magnification?
Let me repeat: There is no "always" because it depends on the context. If the question gives you the focal length and the position of the object and asks for the magnification, I would include the negative sign. If instead the question asks "by what factor is the image smaller than the object?", I would not include the negative sign. The height of the image is usually positive although sometimes people who write a question asking for a height expect a negative sign to indicate that the image is inverted. This becomes important when you want to find the image formed by two of lenses in which the image formed by the first lens becomes the object for the second lens. Dropping the sign will get you in trouble. The upshot of all this is that only your instructor knows for sure what answer is expected, so ask.
 
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kuruman said:
Let me repeat: There is no "always" because it depends on the context. If the question gives you the focal length and the position of the object and asks for the magnification, I would include the negative sign. If instead the question asks "by what factor is the image smaller than the object?", I would not include the negative sign. The height of the image is usually positive although sometimes people who write a question asking for a height expect a negative sign to indicate that the image is inverted. This becomes important when you want to find the image formed by two of lenses in which the image formed by the first lens becomes the object for the second lens. Dropping the sign will get you in trouble. The upshot of all this is that only your instructor knows for sure what answer is expected, so ask.
Thanks
 

1. What is lens magnification?

Lens magnification refers to the ratio of an object's apparent size to its actual size when viewed through a lens. It is commonly used to describe the size of an image produced by a lens.

2. Why do we use absolute value for lens magnification?

We use absolute value for lens magnification because it allows us to accurately measure the size of an image without regard to its direction. This is important because images can be either upright or inverted, and using absolute value ensures that we get a positive number for the magnification.

3. Are there any situations where we don't need to use absolute value for lens magnification?

Yes, there are situations where we may not need to use absolute value for lens magnification. For example, if we are only concerned with the relative size of two objects in an image, we can use the negative sign to indicate which object is larger or smaller.

4. Can we use absolute value for all types of lenses?

Yes, absolute value can be used for all types of lenses, including convex and concave lenses. It is a mathematical concept that applies to any type of magnification, regardless of the lens being used.

5. Is there a relationship between lens magnification and focal length?

Yes, there is a relationship between lens magnification and focal length. As focal length increases, lens magnification also increases. This means that a lens with a longer focal length will produce a larger image compared to one with a shorter focal length.

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