KAHIBARO
Table of Contents
Image formation by a thin lens
A thin lens bends light rays so that an object on one side can form an image on the other side, or sometimes on the same side. The lens equation gives a simple relation between the object distance, the image distance, and the focal length.
For a thin lens, the basic formula is
$$
\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}
$$
where $f$ is the focal length of the lens, $d_o$ is the object distance, and $d_i$ is the image distance.
This equation is one of the central tools of geometrical optics.
For a thin lens,
$$
\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}
$$
You must use a consistent sign convention when inserting values.
Meaning of the quantities
The object distance $d_o$ is the distance from the lens to the object. The image distance $d_i$ is the distance from the lens to the image. The focal length $f$ depends on the type of lens.
A converging lens has positive focal length, $f > 0$.
A diverging lens has negative focal length, $f < 0$.
If the image distance is positive, the image is real and forms on the opposite side of the lens from the object. If the image distance is negative, the image is virtual and appears on the same side as the object.
Sign convention
A common sign convention for thin lenses is shown below.
| Quantity | Positive when | Negative when |
|---|---|---|
| $f$ | converging lens | diverging lens |
| $d_o$ | object is on incoming light side | rarely negative in basic problems |
| $d_i$ | real image, opposite side of lens | virtual image, same side as object |
For most beginner problems, the object is real, so $d_o > 0$.
Typical beginner sign rules:
A converging lens has $f>0$.
A diverging lens has $f<0$.
A real image has $d_i>0$.
A virtual image has $d_i<0$.
Rearranging the lens equation
Sometimes you know two of the three quantities and solve for the third.
To solve for image distance,
$$
\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}
$$
so
$$
d_i = \frac{1}{\frac{1}{f} - \frac{1}{d_o}}
$$
To solve for object distance,
$$
\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}
$$
Connection to image position
The lens equation tells where the image forms. It does not by itself tell the size or orientation of the image. That is handled with magnification, which belongs to a separate topic, but the position found from the lens equation is the starting point.
A positive $d_i$ means you could place a screen there and catch the image. A negative $d_i$ means no screen can catch it, because the image is virtual.
Converging lens cases
For a converging lens, the result depends strongly on where the object is placed compared with the focal length.
If the object is farther than one focal length from the lens, the image is real, so $d_i > 0$.
If the object is exactly at the focal point, then the outgoing rays are parallel and the image is effectively at infinity.
If the object is closer than one focal length, the image is virtual, so $d_i < 0$.
Diverging lens cases
For a diverging lens, a real object produces a virtual image. That means the image distance is usually negative. The image appears on the same side as the object and is closer to the lens than the focal point in appearance.
Special cases
There are a few important limiting situations.
If $d_o \to \infty$, then $\frac{1}{d_o} \to 0$, so
$$
\frac{1}{f} = \frac{1}{d_i}
\quad \Rightarrow \quad
d_i = f
$$
This means light from a very distant object forms an image at the focal plane.
If $d_o = f$ for a converging lens, then
$$
\frac{1}{d_i} = \frac{1}{f} - \frac{1}{f} = 0
$$
so $d_i \to \infty$.
Important special results:
If $d_o \to \infty$, then $d_i = f$.
If a converging lens has $d_o = f$, the image is at infinity.
Example calculation
Suppose a converging lens has focal length
$$
f = 10\ \text{cm}
$$
and the object is placed at
$$
d_o = 30\ \text{cm}
$$
Use the lens equation:
$$
\frac{1}{10} = \frac{1}{30} + \frac{1}{d_i}
$$
Then
$$
\frac{1}{d_i} = \frac{1}{10} - \frac{1}{30}
= \frac{3}{30} - \frac{1}{30}
= \frac{2}{30}
= \frac{1}{15}
$$
So
$$
d_i = 15\ \text{cm}
$$
The positive answer means the image is real and forms $15\ \text{cm}$ from the lens on the opposite side from the object.
Example with a diverging lens
Take a diverging lens with
$$
f = -12\ \text{cm}, \qquad d_o = 24\ \text{cm}
$$
Then
$$
\frac{1}{-12} = \frac{1}{24} + \frac{1}{d_i}
$$
So
$$
\frac{1}{d_i} = -\frac{1}{12} - \frac{1}{24}
= -\frac{2}{24} - \frac{1}{24}
= -\frac{3}{24}
= -\frac{1}{8}
$$
Thus
$$
d_i = -8\ \text{cm}
$$
The image distance is negative, so the image is virtual and on the same side as the object.
Practical interpretation
The lens equation is useful because it predicts where an image appears without tracing many rays. In cameras, microscopes, telescopes, and eyeglasses, the location of the image is crucial. Once $d_i$ is known, one can then study image size and orientation using magnification.
Common mistakes
A very common mistake is forgetting the sign of the focal length for a diverging lens. Another common mistake is using distances measured from the wrong point. In thin lens problems, distances are measured from the lens itself.
Another mistake is treating a virtual image as if it had positive image distance. If the image appears on the same side as the object, then $d_i$ must be negative in the standard convention used here.
Common errors to avoid:
Use $f<0$ for diverging lenses.
Use $d_i<0$ for virtual images.
Measure $d_o$ and $d_i$ from the lens.
Keep all distances in the same unit.
Summary
The thin lens equation is
$$
\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}
$$
It connects the geometry of an object and its image through the focal length of the lens. With the correct sign convention, it works for both converging and diverging lenses, and for both real and virtual images.