# Thins Lens Equation Question

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?

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.

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