How to use the lens formula to find image distance (with a worked example)

How to · Updated 2026-11-10

Put the object distance and focal length into 1/v − 1/u = 1/f with the right signs, solve for v, then get magnification and power. Worked example and checks.

To find where a lens forms an image, write the object distance u as a negative number, give the focal length f a plus sign for a convex lens or a minus sign for a concave one, and solve 1/v = 1/f + 1/u. The lens ray diagram does this for any numbers you type, draws the rays to scale and prints every line of the working with signs.

What you need

• The focal length f of the lens, in cm. A convex (converging) lens is fatter in the middle; a concave (diverging) lens is thinner in the middle.
• The object distance, measured from the centre of the lens to the object, in cm.
• The sign convention. Light travels left to right and every distance starts at the optical centre O. Left of O is negative, right is positive, so an object in front of the lens has a negative u.
• Optional: the object height h, if you also want the height of the image.

Step by step

1. Open the lens tool and press Convex or Concave at the top.

2. Type your numbers into the Object distance (cm) and Focal length (cm) boxes. Type plain sizes; the tool adds the signs. You can also drag the object, use the sliders, or tap a standard position such as At 2F₁.

3. Read the line at the top of The image: real or virtual, bigger or smaller, and where it forms.

4. Look under Lens formula, with signs. It shows the steps you would write in a test:

   1/v = 1/f + 1/u, then v, then m = v/u, then P = 1/f.

   Here v is the image distance, m the magnification (image height divided by object height) and P the power in dioptres, with f in metres.

5. If you need the image height, switch to Class 12, type the object height, and the tool adds h′ = m × h.

What the signs tell you: a positive v means the image is on the far side of the lens, where light really meets, so it is real and can be caught on a screen. A negative v means it sits on the same side as the object and is virtual. A negative m means the image is upside down; a positive m means it is upright.

A worked example

A convex lens has a focal length of 15 cm. A candle stands 25 cm in front of it, and the flame is 4 cm tall. Where is the image, and how big?

• Signs first: u = −25 cm and f = +15 cm.
• 1/v = 1/15 + 1/(−25) = 0.0667 − 0.04 = 0.0267, so v = +37.5 cm.
• m = v/u = 37.5 ÷ (−25) = −1.5.
• Image height: h′ = −1.5 × 4 cm = −6 cm.
• Power: P = 1/0.15 m = +6.67 D.

So a sharp, upside-down flame 6 cm tall appears on a screen 37.5 cm behind the lens. That fits the rule for a convex lens: an object between F₁ and 2F₁ (15 to 30 cm here) gives a real, inverted, enlarged image beyond 2F₂.

Now move the candle to 10 cm, inside the focal length. Then 1/v = 1/15 − 1/10 = −0.0333, so v = −30 cm and m = +3. The image is upright, three times taller and on the candle’s side of the lens. That is how a magnifying glass works.

Last, swap in a concave lens of the same 15 cm, with the object at 30 cm. Now f = −15 cm and 1/v = −1/15 − 1/30 = −0.1, so v = −10 cm and m = +0.33: upright, a third of the size, and virtual. A concave lens gives this kind of image wherever the object stands. Curved mirrors use a near-identical formula with a plus sign, 1/v + 1/u = 1/f, which you can try in the mirror ray diagram; two lenses working together make the instruments in microscope and telescope.

Check your answer

• Look at the picture. The rays in the diagram should meet where your v says. A positive v with only dashed rays on the object’s side means a sign slipped.
• Use the size rule (convex lens). At 2F₁ the image is the same size (m = −1). Between F₁ and 2F₁, |m| is more than 1; beyond 2F₁, less than 1.
• Put v back in. Work out 1/v − 1/u with your numbers. It should give 1/f. For the first example: 1/37.5 + 1/25 = 0.0267 + 0.04 = 0.0667 = 1/15.
• Units. Keep u, v and f all in cm while you solve, then change f to metres only for the power.

Questions people ask

Why is u negative when the distance is clearly 25 cm? The minus sign is a direction, not a size: the side light comes from counts as negative. That keeps one formula working for every case.

What if the object sits exactly at the focus? Then 1/v works out to zero and the image is “at infinity”: the rays leave the lens parallel and never meet. A spotlight puts its bulb at this point to make a parallel beam; a projector puts its slide just outside it.

Some books use 1/f = 1/u + 1/v. Which is right? Both, with different sign rules. That older form treats real distances as positive. The tool, and Class 10 and 12 books in India, use the New Cartesian rule with 1/v − 1/u = 1/f. Pick one convention and stick to it for the whole problem.