The Memory Trick
💡 1/f = 1/do + 1/di
The thin lens equation locates the image formed by any lens: 1/f = 1/do + 1/di, where f is the focal length (positive for a converging lens, negative for a diverging lens), do is the object distance, and di is the image distance (positive for a real image on the opposite side from the object, negative for a virtual image on the same side as the object). Magnification is m = −di/do, where a negative m indicates an inverted image.
Why It Works
The sign conventions built into this equation do a lot of work automatically: solving for di and getting a positive value tells you the image is real (light rays actually converge there); a negative value tells you it's virtual (rays only appear to diverge from there). Combined with the magnification sign, you get a complete description of the image — real/virtual, upright/inverted, enlarged/reduced — from just two calculated numbers.
Step by Step
Applying the Thin Lens Equation
1
Interpreting the sign of di
A positive di means a real image forms on the opposite side of the lens from the object (light rays actually converge there, and it could be projected onto a screen). A negative di means a virtual image on the SAME side as the object (rays only appear to diverge from there; it cannot be projected).
A camera lens produces a real image (positive di) on the sensor; a magnifying glass held close to an object produces a virtual image (negative di) that you look 'into' but couldn't project onto a screen.
2
Interpreting the sign and magnitude of m
A negative m indicates an inverted image; a positive m indicates an upright image. |m| > 1 means the image is enlarged; |m| < 1 means it's reduced.
A camera typically produces a small, inverted real image (negative m, |m|<1) on its sensor, which is why early cameras needed to flip the resulting photograph right-side up.
3
Converging vs. diverging lens sign convention
A converging (convex) lens has a positive focal length f; a diverging (concave) lens has a negative focal length f — get this sign wrong, and every subsequent calculation will be incorrect.
Always double-check whether the problem specifies a converging or diverging lens before assigning the sign of f in your calculation.
🏥 Worked Example
An object is placed 30 cm from a converging lens with a focal length of 10 cm. Find the image distance and magnification, and describe the resulting image.
1
Apply the thin lens equation: 1/f = 1/do + 1/di → 1/10 = 1/30 + 1/di → 1/di = 1/10 − 1/30 = 3/30 − 1/30 = 2/30.
2
Solve for di: di = 30/2 = 15 cm (positive, meaning a real image).
3
Find magnification: m = −di/do = −15/30 = −0.5 — negative (inverted image) and |m|<1 (reduced). The image is real, inverted, and smaller than the original object.
📌 Exam Application
Exams test correctly applying the thin lens equation and magnification formula, and correctly interpreting the resulting signs to describe whether an image is real/virtual, upright/inverted, and enlarged/reduced.
⚠️ Most Common Thin Lens Equation Mistakes
The most common trap is misapplying the sign convention for f (converging vs. diverging) or misinterpreting the resulting sign of di or m — always work out the sign meaning systematically rather than guessing based on intuition, since these conventions aren't always intuitive at first.
✓ Quick Self-Test
1) Write the thin lens equation. 1/f = 1/do + 1/di. 2) Write the magnification formula for a thin lens. m = −di/do. 3) What does a positive value of di indicate about the image? It's a real image, forming on the opposite side of the lens from the object. 4) What does a negative value of m indicate about the image? It's inverted. 5) What sign convention applies to the focal length f of a converging lens versus a diverging lens? Positive for converging (convex); negative for diverging (concave).