Lateral magnification in mirrors, simple issue

  • Thread starter Addez123
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  • #1
Addez123
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Homework Statement:
An object at point p reflects in a convex or concave mirror at origin.
Given that m = 0.4 (lateral magnification), p = +30
The image is inverted. Find i (distance from orgin to reflected image).
Relevant Equations:
m = -i/p
The equation gives us that
i = -mp = -0.4 * 30 = -12

The 'correct' answer is +12.
Did I miss something here?
 

Answers and Replies

  • #2
Antarres
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Well, you have to be careful with signs. Also it is important whether the mirror is convex or concave, that's not the same thing. Assuming it is convex, if your object is between the focus and the pole of the mirror, you will get an imaginary image, that is not inverted. If it is behind the focus, you get a real image and it's inverted. Magnification should always be defined as ## m = i/p##, where we take ##i## to be negative if the image is imaginary, and positive if it is real.

In case I'm using a term that's not used in English, what I mean by real image - it's image created at intersection of real rays that come from the object. Imaginary image is image created by extensions of rays(the real rays diverge but their extensions converge and the intersection is called imaginary image).

So in your case since you're told the image is inverted, you conclude that it is real, because imaginary image with spherical mirror is never inverted. And then you can get the right coordinate ##i## of the image.
 
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  • #3
Addez123
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I agree with what you're saying except the formula in my book is m = -i/p, should I've used m = i/p then I'd get wrong answer on previous questions.
m and p are given with + or - signs infront of them, all I do is put them into the equation.

What you're saying is basically that if the image is inverted then m = -i/p isn't valid. But that formula is for 'lateral magnification produced by a spherical mirror' which is exactly what I'm calculating.
 
  • #4
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No, I'm saying that that formula depends on how you calculate those coordinates. So my formula basically uses all those lengths without signs and attaches signs according to whether the image is imaginary or real. That's the standard thing I've found in optics books. Your formula probably just uses a different convention, but I can't compare since I have no access to your literature.

That's why I gave you the principle, so you can understand if there's a difference in conventions.

Edit: Also there can be a confusion arising from whether ##i## and ##p## are heights of image and object, or their distances from the pole of the mirror. By similarity of triangles those quotients should be the same, but the sign convention can play a role there depending on how you calculate heights etc. In my definition I used horizontal distances, not heights.

For example, if you say that non inverted height of an object/image is positive, and inverted one is negative, then in this case, where you have inverted image, magnification would have that extra minus sign(in general it is defined as ratio of absolute heights of image vs object). So you need to take into account how you defined those lengths you measure in optics, that's the only issue here.
 
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