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According to my textbook, it says that according to the thin lens equation, object distance and image distance vary inversely. However, when I do a converging lens simulation, as I increase the distance the object is from the lens, the distance the image is from the lens also increases. Why is this? Is there a sign convention I am not aware of? enter image description here

enter image description here

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It is an easy calculation:

$$ -\frac{1}{a}+\frac{1}{a'}=\frac{1}{f'} $$

$$ \frac{1}{a'}=\frac{1}{f'}+\frac{1}{a} $$

$$ \frac{1}{a'}=\frac{a}{af'}+\frac{f'}{af'} $$

$$ \frac{1}{a'}=\frac{a+f'}{af'}$$

$$ a'=\frac{af'}{a+f'}$$

So it is not inversely proportional, as it is not $\propto \frac{1}{a}$ only.

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  • $\begingroup$ why is 1/a negative? $\endgroup$ – 54284User Jul 10 '18 at 8:59
  • $\begingroup$ Isn't it the lens equation? $\endgroup$ – FGSUZ Jul 10 '18 at 9:00
  • $\begingroup$ I thought there's no negative in the lens equation $\endgroup$ – 54284User Jul 10 '18 at 9:08
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    $\begingroup$ @54284User It all depends on which one of the many sign conventions is being used. $\endgroup$ – Farcher Jul 10 '18 at 11:07

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