Sign convention for spherical lenses

Every numerical on convex and concave lenses hides a trap: the lens formula 1v−1u=1f\frac{1}{v}-\frac{1}{u}=\frac{1}{f} only gives the right answer if you first attach the correct plus or minus sign to every distance you substitute. Get one sign wrong and a perfectly real, inverted image on your paper turns into a virtual, upright one — the arithmetic looks fine, the physics is backwards. The sign convention for spherical lenses is the fixed rulebook that removes this ambiguity: measure every distance from the optical centre, treat the direction the incident light travels as positive, and heights above the axis as positive. Apply it consistently and the same two formulas — the lens formula and m=h′/h=v/um = h'/h = v/u — work for every combination of convex or concave lens and real or virtual image, with the final sign of vv and mm telling you exactly what kind of image you got.

The core insight

Distances measured in the direction of incident light (and heights above the axis) are positive, distances against it are negative, so a convex lens always has f>0f>0, a concave lens always has f<0f<0, and the object distance uu is almost always negative.

The core idea

All distances for a spherical lens are measured from its optical centre, and the reference direction is the direction the incident light travels (conventionally left to right).

  • Distances measured in the direction of incident light are positive; distances measured against it are negative.
  • Heights measured upward from the principal axis are positive; heights measured downward are negative.
  • The object is conventionally drawn to the left of the lens, so the object distance uu is almost always negative.
  • A convex (converging) lens has focal length ff positive; a concave (diverging) lens has focal length ff negative — this is fixed by the lens type alone, never by where the object sits.

This convention is what lets one pair of formulas cover every case:

1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

where: uu is the object distance, vv is the image distance, and ff is the focal length, all measured from the optical centre with the sign rules above.

m=h′h=vum = \frac{h'}{h} = \frac{v}{u}

where: hh is the object height, h′h' is the image height, and mm is the magnification — a pure ratio, so its sign alone tells you the image's orientation. Note the lens formula is the mirror image of the mirror formula: 1v−1u=1f\frac{1}{v}-\frac{1}{u}=\frac{1}{f} for lenses versus 1v+1u=1f\frac{1}{v}+\frac{1}{u}=\frac{1}{f} for mirrors. For a mirror the reflected light returns on the same side as the incident light; for a lens the refracted, image-forming light usually passes through to the other side, so the minus sign appears instead. This one rulebook, applied without exception, is also what lets a single formula predict image position and nature for every combination you will meet in Class 10 — a convex lens with a distant object, a convex lens with an object closer than the focal length, or a concave lens anywhere at all — without needing a separate case-by-case formula for each situation the way an untrained ray diagram might suggest. Once you solve for vv or mm, read the sign back into physics:

  • Positive vv → real image, on the side opposite the object.
  • Negative vv → virtual image, on the same side as the object.
  • Positive mm → image is virtual and erect; negative mm → image is real and inverted.
Figure
Sign convention axes for a lens showing positive distances to the right of the optical centre and positive heights above the principal axis
Sign convention at the optical centre O principal axis (vertical) / lens O incident light direction (+) right of O: + distances left of O: − distances heights up: + heights down: − object usually placed here (u negative)
Derivation AFinding the image distance for a convex lens

The goal is to show, with a real convex-lens numerical, exactly how assigning signs before substitution turns the lens formula 1v−1u=1f\frac{1}{v}-\frac{1}{u}=\frac{1}{f} into a correct, unambiguous answer, using an object placed 15 cm in front of a convex lens of focal length 10 cm.

Step 1. Assign the sign to uu. The object sits to the left of the lens, against the direction of incident light, so by convention u=−15 cmu = -15\ \text{cm}.

Step 2. Assign the sign to ff. The lens is convex (converging), so its focal length is positive by the type-of-lens rule: f=+10 cmf = +10\ \text{cm}.

Step 3. Substitute into the lens formula and solve for 1v\frac{1}{v}. 1v=1f+1u=110+1−15=330−230=130\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{-15} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30}, so v=+30 cmv = +30\ \text{cm}.

Step 4. Interpret the sign of vv. Since vv is positive, the image forms on the side opposite the object, so it is a real image, 30 cm from the optical centre — exactly the sign-convention rule "positive vv means real image" being confirmed by the algebra, not just quoted.

Worked example

A 4 cm tall object is placed 20 cm in front of a concave lens of focal length 15 cm. Find the image distance and magnification, and state the nature and orientation of the image.

Step 1. Assign signs to the given quantities. The object is to the left of the lens, so u=−20 cmu = -20\ \text{cm}. The lens is concave, so f=−15 cmf = -15\ \text{cm}. The object stands upright above the axis, so h=+4 cmh = +4\ \text{cm}.

Step 2. Apply the lens formula. 1v=1f+1u=1−15+1−20=−460−360=−760\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-15} + \frac{1}{-20} = -\frac{4}{60} - \frac{3}{60} = -\frac{7}{60}, giving v≈−8.6 cmv \approx -8.6\ \text{cm}.

Step 3. Interpret vv. Negative vv means the image lies on the same side as the object — a virtual image, as expected whenever a concave lens forms an image of a real object.

Step 4. Find the magnification. m=vu=−8.6−20=+0.43m = \frac{v}{u} = \frac{-8.6}{-20} = +0.43.

Step 5. Interpret mm and find the image height. Positive mm confirms the image is virtual and erect, and ∣m∣<1|m|<1 means it is diminished: h′=m×h=0.43×4≈1.7 cmh' = m \times h = 0.43 \times 4 \approx 1.7\ \text{cm}, upright and smaller than the 4 cm object.

The image is virtual, erect, diminished, about 1.7 cm tall, and located 8.6 cm from the optical centre on the same side as the object.

Common traps

  • Keeping ff positive for a concave lens "because it still works like a normal lens" — the sign of ff is fixed by lens type alone: convex is always ++, concave is always −-.
  • Writing uu as positive because the object is placed in front of the lens "the usual way" — since the object is almost always to the left, against incident light, uu is negative.
  • Swapping the lens formula 1v−1u=1f\frac{1}{v}-\frac{1}{u}=\frac{1}{f} for the mirror formula 1v+1u=1f\frac{1}{v}+\frac{1}{u}=\frac{1}{f} — the two look alike but use different signs of the middle term.
  • Computing a numeric vv or mm correctly but forgetting to translate the sign back into physical meaning (real/virtual, erect/inverted) in the final answer.

Ask the tutor: Still mixing up which sign goes with concave versus convex — ask your AI tutor to walk through two or three more sign-convention numericals with you, step by step.

Why is the object distance u almost always negative if the object is just sitting in the normal position

Because "normal position" means the object is placed to the left of the lens, against the direction the incident light travels toward the lens, and the sign convention makes any distance measured against that direction negative. It is not that objects are special — it is a direct consequence of always drawing the object on the side the light comes from. If you ever place a virtual object to the right of the lens (rare at this level), uu would become positive instead.

Why can the focal length f be positive for one lens and negative for another, when u follows one fixed rule

Object distance uu follows a positional rule because the object's placement (left of the lens) barely changes across problems. Focal length instead reflects a real physical difference between lens types: a convex lens converges light to a real focus on the transmission side, giving f>0f>0, while a concave lens diverges light so rays only appear to meet on the incidence side, giving f<0f<0. So ff's sign is tied to what the lens is, not where anything is placed.

If I get a negative magnification, does that always mean the image is upside down

Yes, for the sign convention used with m=h′/h=v/um = h'/h = v/u: a negative mm always pairs with a real, inverted image, and a positive mm always pairs with a virtual, erect image, no exceptions within this convention. It is worth checking that your vv and uu signs are correct first, though, since an arithmetic slip on either one flips the sign of mm and would make you misread the image's orientation.

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