Can a Concave Lens Form a Real Image, and What Is Its Magnification?

A concave (diverging) lens normally makes the rays from a real object spread out, so the image is virtual. A real image is possible, though, if the rays arriving at the lens are already converging. The lens then changes where those rays meet.

There is no single numerical magnification without distances or a focal length. The signed thin-lens equation gives both the image location and its magnification once those values are known.

When can a concave lens make a real image?

Use the Cartesian sign convention for light traveling from left to right: the lens is at coordinate zero, and distances to the right are positive. The focal length of a concave lens is negative. The thin-lens equation is

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f},$$

where $u$ is the object coordinate and $v$ is the image coordinate. A real image lies on the outgoing side of the lens, so $v>0$.

For an ordinary real object to the left, $u<0$. Since $f<0$, the equation gives $1/v=1/f+1/u<0$, so $v<0$: the image is virtual. To get a real image, the incoming rays must instead be converging toward a point to the right of the lens. That point acts as a virtual object, so $u>0$.

For the image to be real, the equation requires $1/v=1/f+1/u>0$. Since $f<0$, this means $0<u<|f|$. In physical terms, without the lens the rays would meet at a point closer to the lens than the magnitude of its focal length. The concave lens pushes their meeting point farther away, but the rays still meet.

Magnification with signed distances

In this Cartesian convention, the transverse magnification is

$$m=\frac{y_i}{y_o}=\frac{v}{u},$$

where $y_o$ and $y_i$ are the object and image heights. A positive value means the image has the same orientation as the virtual object; a negative value means it is inverted. The image size relative to the object is $|m|$.

The sign convention matters. In a convention that uses positive distances for real objects and real images, the virtual-object distance is negative, and the magnification formula is often written $m=-q/p$. These are equivalent when signs are applied consistently.

Worked example

Suppose a concave lens has focal length $f=-10\,\text{cm}$. The incident rays are converging toward a point $6\,\text{cm}$ to the right of the lens, so $u=+6\,\text{cm}$. Solve for $v$:

$$\begin{aligned} \frac{1}{v}-\frac{1}{6}&=-\frac{1}{10}\\ \frac{1}{v}&=\frac{1}{6}-\frac{1}{10}=\frac{1}{15}\\ v&=+15\,\text{cm}. \end{aligned}$$

Because $v$ is positive, the rays meet $15\,\text{cm}$ to the right of the lens and form a real image. The magnification is

$$m=\frac{v}{u}=\frac{15}{6}=2.5.$$

The image is 2.5 times the height of the virtual object and has positive magnification, so it is upright relative to that virtual object. The virtual object is a point toward which the incoming rays are heading, not necessarily a physical object at that location.

Common mistakes

• Assuming a concave lens always makes only virtual images. That is true for a single concave lens acting on a real object, but not when the incident rays are converging.
• Treating the virtual object as a real object on the far side. It describes the direction of the incoming rays; no light has yet converged there before passing through the lens.
• Mixing sign conventions. Choose one convention and keep the signs of $u$, $v$, and $f$ consistent in both the lens equation and magnification formula.
• Calling positive magnification “always upright” without specifying the reference. Here, orientation is relative to the virtual object represented by the incoming ray bundle.