Convex Lens: Image at Every Object Position

Problem 146 focused on one specific setup — an object beyond 2F. That's like photographing a scene from across the room. But a converging lens is far more versatile. By changing nothing except how close the object is to the lens, you can make the image do wildly different things: shrink, grow, flip, even jump from one side of the lens to the other and become something a screen can't catch.

This problem is a grand tour of all five regimes. We'll walk the object from far away all the way into the focal point and beyond, and at each stop we'll note what the image looks like, where it is, and why.

Why object position matters so much

In the thin-lens equation $\dfrac{1}{f} = \dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$, the focal length $f$ is fixed (it's a property of the lens). The only input you can change is $d_{o}$, the object distance. Everything about the image — its position $d_{i}$, its magnification $m = -d_{i}/d_{o}$, whether it's real or virtual — falls out of that single number. So sweeping $d_{o}$ from infinity down to zero is a complete exploration of everything a converging lens can do.

Think of $d_{o}$ as a dial. As you turn it, the image responds.

Regime 1: Object at infinity ($d_{o} \to \infty$)

When the object is very far away, its light arrives as essentially parallel rays. A converging lens bends parallel rays so they meet at the focal point on the far side. So

$d_{i} = f = 10 \text{ cm}$

The image forms right at $F'$. It's a tiny, bright point (or, if the object is an extended source like the sun, a tiny bright disk). The magnification $m \to 0$: the image is infinitely small compared to the real object, which makes sense — the sun's image through a magnifying glass is a dot, not a 1.4-million-km disk.

This is the regime your camera is in when you focus on distant mountains. The sensor sits right at the focal distance of the lens.

Regime 2: Object beyond 2F ($d_{o} > 2f$)

We covered this case in detail in Problem 146. Example with $d_{o} = 30$ cm:

$d_{i} = \dfrac{1}{1/10 - 1/30} = \dfrac{1}{2/30} = 15 \text{ cm}$

$m = -\dfrac{15}{30} = -0.5$

Image: real (positive $d_{i}$), inverted (negative $m$), diminished ($|m| < 1$), located between $F'$ and $2F'$ on the far side.

As the object moves closer to 2F, $d_{i}$ grows and $|m|$ approaches 1.

Regime 3: Object at exactly 2F ($d_{o} = 2f = 20$ cm)

$d_{i} = \dfrac{1}{1/10 - 1/20} = \dfrac{1}{1/20} = 20 \text{ cm}$

$m = -\dfrac{20}{20} = -1$

This is the crossover point. The image is the same size as the object ($|m| = 1$), inverted, and sits at exactly $2F'$ on the far side. It's a perfect, life-size, inverted copy.

This is a symmetric situation: object at $2F$, image at $2F'$. The lens is equidistant from both. Optically, there's nothing special about this distance except that the magnification passes through unity. Some textbooks make a big deal of it; what's really important is just that it's the boundary between "diminished" and "magnified."

Regime 4: Object between F and 2F ($f < d_{o} < 2f$)

Example with $d_{o} = 15$ cm:

$d_{i} = \dfrac{1}{1/10 - 1/15} = \dfrac{1}{1/30} = 30 \text{ cm}$

$m = -\dfrac{30}{15} = -2$

Now the image is magnified ($|m| = 2$, so twice the size), still inverted, still real, but located far beyond $2F'$. As the object approaches $F$ from the right, $d_{i}$ and $|m|$ both shoot toward infinity: the image gets enormous and flies off toward the horizon.

This is the projector regime. A slide projector places the film slightly beyond $F$ so that the image thrown onto a distant wall is huge and real (and inverted, which is why you load the slide upside-down).

Regime 5: Object at F ($d_{o} = f = 10$ cm)

$\dfrac{1}{d_{i}} = \dfrac{1}{10} - \dfrac{1}{10} = 0 \quad \Rightarrow \quad d_{i} = \infty$

The image goes to infinity. The three principal rays, after passing through the lens, emerge parallel — they never converge. There is no image on any finite screen.

This is not a failure; it's exactly what a flashlight or spotlight does. Put the bulb at $F$; the lens sends out a parallel beam. It's also the reverse of Regime 1: light from infinity focuses at $F$, so light from $F$ goes to infinity.

Regime 6: Object inside F ($d_{o} < f$)

Example with $d_{o} = 5$ cm:

$d_{i} = \dfrac{1}{1/10 - 1/5} = \dfrac{1}{-1/10} = -10 \text{ cm}$

$m = -\dfrac{-10}{5} = +2$

For the first time, $d_{i}$ is negative. That means the image is on the same side as the object — behind the lens from the viewer's perspective. Light rays are diverging after the lens as if they're coming from that point, but they never actually meet there. If you put a screen at that location, you'd see nothing. This is a virtual image.

The magnification is positive: the image is upright (same orientation as the object) and magnified ($|m| = 2$).

This is the magnifying glass. Hold a lens close to a page (object inside $F$), look through it, and you see a larger, upright, virtual image floating behind the lens. Every hand-held magnifier works in this regime.

Summary of all regimes

• $d_{o} \to \infty$: image at $F'$, infinitely diminished, real, inverted
• $d_{o} > 2f$: image between $F'$ and $2F'$, diminished, real, inverted
• $d_{o} = 2f$: image at $2F'$, same size, real, inverted
• $f < d_{o} < 2f$: image beyond $2F'$, magnified, real, inverted
• $d_{o} = f$: image at infinity, no finite image
• $d_{o} < f$: image on same side, magnified, virtual, upright

This progression — from diminished-real through same-size-real to magnified-real, then "infinity break," then magnified-virtual — is the single most important conceptual picture in geometric optics for converging lenses. If you understand this sweep, you understand the lens.

The graphical relationship

There's a beautiful way to see all of this at once: plot $d_{i}$ vs $d_{o}$. The thin-lens equation rearranges to

$d_{i} = \dfrac{f\,d_{o}}{d_{o} - f}$

This is a hyperbola with a vertical asymptote at $d_{o} = f$ (where the image blows up to infinity) and a horizontal asymptote at $d_{i} = f$ (when the object is at infinity, the image settles at $f$). The function is positive (real image) for $d_{o} > f$ and negative (virtual image) for $d_{o} < f$. The crossover point $(2f, 2f)$ sits on the diagonal $d_{i} = d_{o}$.

Real-world applications at each regime

• Infinity → beyond 2F (camera): Object far away, small real image on sensor. Every camera, telescope objective, human eye.
• Beyond 2F → at 2F (photocopier at 1:1): Equal-size reproduction. Photocopiers and scanners use a lens at this distance for 1:1 document imaging.
• Between F and 2F (projector): Object slightly past $F$, big real image far away. Movie projectors, overhead projectors, stage lighting.
• At F (collimator/spotlight): Object at focal point, parallel beam output. Flashlights, laser collimators, infinity-focus telescopes.
• Inside F (magnifier): Object closer than $F$, large virtual image. Hand magnifiers, jeweler's loupe, simple microscope eyepiece.

Common mistakes

• Thinking all real images are inverted: True for a single thin converging lens — but not for multi-lens systems. Two inversions restore uprightness (compound microscope, astronomical telescope with an erecting lens).
• Confusing virtual with "not there": A virtual image is not visible on a screen, but it's absolutely visible to your eye. The virtual image of your face in a bathroom mirror is something you look at every day.
• Plotting $d_{i}$ vs $d_{o}$ and ignoring the vertical asymptote at $f$: Many students think $d_{i}$ smoothly transitions from positive to negative as $d_{o}$ crosses $f$. It doesn't — it blows up to $+\infty$ from the right and comes back from $-\infty$ on the left. That discontinuity is where the image type changes.

Try it in the visualization

Drag the object distance slider continuously from 60 cm down to 3 cm. Watch four things as you do: (1) the image arrow flips from inverted to upright when you cross $F$; (2) the image size passes through same-size at $2F$; (3) the image arrow dashes (virtual) inside $F$; (4) the formula panel updates live. Turn on the $d_{i}$-vs-$d_{o}$ plot to see the hyperbolic curve with the vertical asymptote at $d_{o} = f$.