Lens Equation (Thin Lens)

The thin lens equation tells you where an image forms for a thin lens:

$$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$

Here $f$ is the focal length, $d_o$ is the object distance, and $d_i$ is the image distance. In intro physics, you use this equation when the lens can be treated as thin, the surrounding medium is usually air, and the rays stay close to the principal axis.

If your answer seems strange, the sign convention is usually the reason. The equation itself stays the same, but the meaning of a positive or negative distance depends on the convention your class uses.

Thin Lens Equation: What Each Symbol Means

For a thin lens in air with paraxial rays, the equation connects object position to image position.

• $f$ is the focal length of the lens.
• $d_o$ is the distance from the lens to the object.
• $d_i$ is the distance from the lens to the image.

In a common introductory sign convention:

• $d_o > 0$ for a real object placed in front of the lens.
• $d_i > 0$ for a real image formed on the opposite side of the lens.
• $d_i < 0$ for a virtual image on the same side as the object.
• $f > 0$ for a converging lens and $f < 0$ for a diverging lens.

If your class uses a different convention, the physics is the same but the signs can change. That is usually where confusion starts.

What The Lens Equation Tells You Physically

The thin lens equation is useful because it shows how image position changes when the object moves.

• If a converging lens has an object far away, the image forms close to the focal plane.
• If the object moves closer to the focal point, the image moves farther away.
• If the object is inside the focal length of a converging lens, the image becomes virtual, so $d_i < 0$ in the convention above.

That last case explains why a magnifying glass can make an enlarged upright virtual image.

Worked Example: Find The Image Distance

Suppose a converging lens has focal length $f = 10\ \mathrm{cm}$, and an object is placed $30\ \mathrm{cm}$ in front of the lens. Using the sign convention above, $f = +10\ \mathrm{cm}$ and $d_o = +30\ \mathrm{cm}$.

Start with the lens equation:

$$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$

Substitute the known values:

$$\frac{1}{10} = \frac{1}{30} + \frac{1}{d_i}$$

Solve for $d_i$:

$$\frac{1}{d_i} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$$ $$d_i = 15\ \mathrm{cm}$$

So the image forms $15\ \mathrm{cm}$ on the far side of the lens. Because $d_i$ is positive, the image is real in this sign convention.

If you also want the image size, use magnification:

$$m = -\frac{d_i}{d_o} = -\frac{15}{30} = -0.5$$

So the image is inverted and half as tall as the object.

Common Mistakes With The Thin Lens Formula

• Mixing sign conventions from different textbooks or teachers.
• Treating all image distances as positive, even for virtual images.
• Forgetting that the thin lens equation is an approximation, not an exact law for every thick or strongly curved lens system.
• Using the focal length of a converging lens as negative, or a diverging lens as positive, without checking the chosen convention.
• Solving for $d_i$ correctly but then misreading what the sign says about real versus virtual images.

When You Can Use The Lens Equation

This equation is used in basic ray optics for cameras, magnifiers, microscopes, telescopes, and simple eye-lens models. It works best when the lens thickness is small compared with the other distances in the problem and the rays stay near the principal axis.

For real optical systems, thickness effects and aberrations can matter. In that case, the thin lens equation is still a useful model, but not the whole story.

Try A Similar Problem

Try your own version with the same lens, but move the object to $8\ \mathrm{cm}$ from the lens. Solve for $d_i$ and check the sign. If you get a negative image distance, that is the signal that the image is virtual rather than real.