1
$\begingroup$

Formula involved in such calculations, similar examples, link associated with d question

$\endgroup$
  • $\begingroup$ Have you done any of your own research? There are many sites on line with equations. $\endgroup$ – Bill N May 3 at 13:22
0
$\begingroup$

With magnification alone? Impossible. You need to know at least something more. Here's, for example, if you also know focal length.

First, one of the standard formulas for magnification is: $m=-\frac{s_i}{s_o}$. Using that and the usual $\frac{1}{f}=\frac{1}{s_i}+\frac{1}{s_o}$:

$$ \frac{1}{f}=\frac{1}{s_o}-\frac{1}{m s_o} $$

$$ s_o = f \left( 1 - \frac{1}{m} \right) $$

For an upright image, $m$ is positive. For convex lens, $f$ is positive. For an object in front of the lens, $s_o$ is positive and for the image behind the lens, $s_i$ is positive.

For a real image, image and object are on the opposite sides so $s_o$ and $s_i$ are either both positive or both negative. For a virtual image, they are on the same side so they have opposite signs.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Shouldn't your very first equation have the ratio of image distance to object distance? $\endgroup$ – Not_Einstein May 3 at 13:23
  • $\begingroup$ @not Yeah, I caught that. Was too quick to post. -.- $\endgroup$ – relatively_random May 3 at 13:25

Not the answer you're looking for? Browse other questions tagged or ask your own question.