Lens problems chain the thin lens equation with the size and kind of image, then move the object and ask what changes. A ray diagram from Lens Ray Diagrams is the quickest check. Each problem here is one short scene and a chain of questions. Answer them in order: each later question uses what you found before, so a slip early on carries through. Keep full precision between steps and round only at the end.
Key idea
$$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}, \qquad m = -\frac{d_i}{d_o}.$$ Object beyond $f$: real, inverted image on the far side ($d_i > 0$). Object inside $f$: virtual, upright, enlarged image on the same side ($d_i < 0$).
Worked example: A candle and a screen
A $2\ \text{cm}$ flame is $20\ \text{cm}$ from a lens with $f = 12\ \text{cm}$. Find (1) the image distance and (2) the image height.
- (1) $\dfrac{1}{d_i} = \dfrac{1}{12} - \dfrac{1}{20} = \dfrac{1}{30}$, so $d_i = 30\ \text{cm}$.
- (2) $m = -30/20 = -1.5$: the image is $3\ \text{cm}$ tall and upside down.
Answer: $30\ \text{cm}$, $3\ \text{cm}$ (inverted)
Common mistake
Getting a negative $d_i$ and calling it a mistake. A negative image distance means a virtual image on the object's side, as in a magnifying glass. Check with a ray diagram.