# Why the real image is always inverted while virtual image is always erect?

While studying optics and the ray diagrams, I observed that the real images were always inverted and virtual were always erect. I have read several articles but haven't got any convincing answer. Someone please help !

• There is no result of the sort, at least in general. Could you at least specific which optical systems you're studying? Thin lens? Thick lens? Spherical mirrors? Commented Jun 20, 2022 at 18:12
• Not any specifically, I just observed it in all the optical systems. Commented Jun 20, 2022 at 18:29

$$\frac{1}{p'}-\frac{1}{p}=\frac{1}{f'} \quad\text{with}\quad \begin{cases} p=\overline{OA} & \text{object}\\ p'=\overline{OA'} & \text{image} \end{cases}$$
$$\gamma=\frac{p'}{p}$$
If you want to get a real image ($$p'>0$$) but not inverted ($$\gamma>0$$), then the object has to be virtual ($$p>0$$). The first equation shows that it is possible. For example for a convergent lens ($$f'>0$$), it can work with $$p' (both object and image on the right-hand side, the image closer to the lens than the object).