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My textbook states that the linear magnification of concave mirrors can either be equal to one, less than one, or greater than one. It seems plausible enough considering the fact that concave mirrors can form images of the same size, diminished images, or enlarged images. However, upon further thought, I'm confused about the fact that the linear magnification m, can be equal to one.

If the object is placed at the centre of curvature, the image formed is real, inverted and of the same size. Hence, h2 (the size of the image) would be negative (as it is inverted), and h1 (the size of the object) would be positive (as it is assumed to be erect). Therefore, m = (h2)/(h1) = (-h1/h1) = -1.

This seems to contradict the fact that m can be equal to one.

Can someone please correct me if I'm wrong?

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4 Answers 4

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We know that magnification of spherical mirrors is $$m=-\frac{v}{u}$$ and also $$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$$ Substituting $v$ in the magnification formula, we get $$m=\frac{f}{f-u}$$ Now you can clearly see that if $m=1$, then $u=0$, but as there will be no image, so we take $u$ very close to zero and therefore you get an image of $m\approx1$, though practically not possible.

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When the object is placed very close to the mirror$^1$ so that $u \rightarrow 0$, then the magnification $M\rightarrow +1$. There is no contradiction. In this case, $$M=\frac{h_i}{h_o} = 1$$

$^1$This also assumes that radius of curvature is sufficiently large in comparison to the object. An approximate comparison would be in the limit as the concave mirror approached a plane mirror, and the magnification of all plane mirrors is $+1$.

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  • $\begingroup$ By close to the mirror, do you mean the object is placed between the principal focus and pole of the mirror? $\endgroup$
    – Twilight
    Jul 13, 2021 at 7:29
  • $\begingroup$ I mean at the tangent plane of the mirror. $\endgroup$
    – joseph h
    Jul 13, 2021 at 7:29
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Under Gaussian conditions, for an object in the tangent plane of the mirror himself, object and image are identical : magnification of +1.

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  • $\begingroup$ I see. But how is the magnification of +1 derived? $\endgroup$
    – Twilight
    Jul 13, 2021 at 7:25
  • $\begingroup$ The easiest way, to avoid the limit in the formula, is to use Newton's formula: $\gamma=-\frac{f}{\bar{FA}}=1$ if $\bar{FA}=-\bar{FS}=-f$ $\endgroup$
    – Vincent Fraticelli
    Jul 13, 2021 at 7:40
  • $\begingroup$ I think you prefer a proof by the mirror conjugation relations, but in reality, to prove these relations, you use the obvious fact in Gauss's approximation that a point of the mirror is its own image ? $\endgroup$
    – Vincent Fraticelli
    Jul 13, 2021 at 10:51
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You concern is due to the fact that there are many conventions which are used in optics.

Real is positive sign convention
In this convention, distances from optical components to real objects, images, and foci are taken as positive, whereas distances to virtual points are taken as negative.
So if the text book is using this convention the magnification would be $+1$ as both the object and the image are real.
So you need to read the introductory part of the section on mirrors and lenses in your textbook to check on the convention used.

(New) Cartesian sign convention
In this convention the magnification is negative if the image is inverted relative to the object so in the example that you have given the magnification is $-1$. .
I suspect that this is the one that you are usually using?

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