Concave Lens Ray Diagram: Always Virtual

If a convex (converging) lens is a helpful tool — it can make images that are bigger, smaller, inverted, upright, real, or virtual depending on where you put the object — a concave (diverging) lens is, at first glance, maddeningly boring. No matter where you put the object, the image is always virtual, always upright, and always smaller. Always. Move the object from 5 cm to 5 km; the image stays virtual, upright, and diminished. That sounds useless, but it turns out to be one of the most important lenses in optics precisely because of its predictability.

Let's understand why, draw the ray diagram, compute the numbers, and see why we actually need diverging lenses.

What makes a lens "diverging"

A converging lens is thick in the middle and thin at the edges (like a lentil — "lens" comes from the Latin for lentil). It bends parallel rays inward to a focal point.

A diverging lens is thin in the middle and thick at the edges. It bends parallel rays outward — it makes them spread apart, or diverge. Parallel rays entering the lens come out as if they're emanating from a point on the incoming side of the lens. That point is the focal point $F$, and because the focal point is on the same side as the incoming light (not the other side like a converging lens), we assign it a negative focal length.

So $f = -15$ cm means the focal point is 15 cm in front of the lens (on the object side), not behind it.

The thin-lens equation still works

The same thin-lens equation from converging lenses applies here:

$\dfrac{1}{f} = \dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$

The only difference is that $f$ is negative. Let's plug in our numbers: $d_{o} = 20$ cm, $f = -15$ cm.

$\dfrac{1}{-15} = \dfrac{1}{20} + \dfrac{1}{d_{i}}$

$\dfrac{1}{d_{i}} = \dfrac{1}{-15} - \dfrac{1}{20} = \dfrac{-4}{60} - \dfrac{3}{60} = \dfrac{-7}{60}$

$d_{i} = \dfrac{60}{-7} \approx -8.57 \text{ cm}$

The image distance is negative. In our sign convention, that means the image is on the same side as the object — it's a virtual image. No screen can catch it; you can only see it by looking through the lens.

Magnification:

$m = -\dfrac{d_{i}}{d_{o}} = -\dfrac{-8.57}{20} = +0.43$

Positive $m$ → the image is upright. $|m| < 1$ → the image is diminished (43% of the object's height).

Why the image is always virtual (proof)

This is worth understanding, not just memorizing. Rearrange the thin-lens equation:

$\dfrac{1}{d_{i}} = \dfrac{1}{f} - \dfrac{1}{d_{o}}$

For a diverging lens, $f < 0$, so $1/f$ is negative. And $d_{o} > 0$ always (real object), so $1/d_{o}$ is positive. We're subtracting a positive number from an already-negative number:

$\dfrac{1}{d_{i}} = \underbrace{\dfrac{1}{f}}_{\text{negative}} - \underbrace{\dfrac{1}{d_{o}}}_{\text{positive}} = \text{even more negative}$

No matter what positive $d_{o}$ you choose, $1/d_{i}$ remains negative, which means $d_{i}$ is always negative. The image is always virtual. There is no object position that can produce a real image from a single diverging lens. That's a rigorous algebraic proof — not just "it always happens to be virtual," but "it's mathematically impossible for it to be real."

Similarly, $m = -d_{i}/d_{o}$, and since $d_{i} < 0$ and $d_{o} > 0$, $m$ is always positive (always upright). And since $|d_{i}| < d_{o}$ (you can verify this from the equation — the virtual image is always closer to the lens than the object is), $|m| < 1$ (always diminished).

The three principal rays for a diverging lens

The ray rules are similar to a converging lens, but with the focal-point logic reversed because $F$ is on the object side:

Ray 1 — parallel → appears to come from F. A ray from the object tip, parallel to the axis, hits the lens and bends outward. If you trace the outgoing ray backward (dashed line), it appears to come from the focal point $F$ on the near side. The actual refracted ray diverges away from the axis.

Ray 2 — through center, undeflected. Same as always — a ray through the optical center of any thin lens passes straight through with no deflection.

Ray 3 — aimed at F′ → emerges parallel. If you draw a ray from the object tip aimed at $F'$ (the point a distance $|f|$ on the far side of the lens), when this ray hits the lens it exits parallel to the principal axis. (This is the reverse-symmetry rule: for a converging lens, a ray through the near $F$ exits parallel; for a diverging lens, a ray aimed at the far $F'$ exits parallel.)

After the three rays leave the lens, they're all diverging — they never meet on the far side. But if you extend each of them backward (dashed lines), the backward extensions meet at a single point on the near side of the lens. That's the virtual image. It's where the rays appear to come from, even though they don't actually pass through that point.

More worked examples

Example 2: $d_{o} = 10$ cm, $f = -15$ cm.

$\dfrac{1}{d_{i}} = \dfrac{1}{-15} - \dfrac{1}{10} = \dfrac{-2}{30} - \dfrac{3}{30} = \dfrac{-5}{30}$

$d_{i} = -6 \text{ cm}, \quad m = -\dfrac{-6}{10} = +0.6$

Virtual, upright, 60% of original size.

Example 3: $d_{o} = 100$ cm, $f = -15$ cm.

$\dfrac{1}{d_{i}} = \dfrac{1}{-15} - \dfrac{1}{100} = -0.0667 - 0.01 = -0.0767$

$d_{i} \approx -13.04 \text{ cm}, \quad m \approx +0.13$

Object very far away: virtual image very close to $F$ (approaching $-15$ cm from the right), very small. This makes sense — for an infinitely distant object, the virtual image would be exactly at $F$, just like a real image forms at $F$ for a converging lens.

Example 4: $d_{o} = 5$ cm, $f = -15$ cm.

$\dfrac{1}{d_{i}} = \dfrac{1}{-15} - \dfrac{1}{5} = -0.0667 - 0.2 = -0.2667$

$d_{i} = -3.75 \text{ cm}, \quad m = +0.75$

Object close to the lens: virtual image even closer, fairly large (75% of original). The closer the object, the closer to 100% the magnification gets — but it never reaches or exceeds 1.

Where diverging lenses are actually used

Given that a diverging lens always produces a smaller virtual image, why does it exist? Because in combination with other optical elements, its diverging power is essential:

• Correcting myopia (nearsightedness): The myopic eye's lens is too strong — it converges light too aggressively, forming the image in front of the retina. A diverging eyeglass lens in front of the eye spreads the rays slightly before they enter, compensating for the over-convergence. The focal length of the corrective lens is chosen so that the combined (eye + glasses) system focuses exactly on the retina.

• Telephoto camera lenses: A telephoto lens isn't just a single big converging lens. It's a converging element (long focal length, good magnification) followed by a diverging element that brings the focal point closer to the sensor. This lets you have a long effective focal length in a physically shorter barrel — without the diverging element, a 400 mm telephoto would need to be 400 mm long.

• Laser beam expanders: Two lenses — a diverging one first to spread the beam, then a converging one to make it parallel again but wider. The ratio of the focal lengths determines the expansion factor.

• Galilean telescope: Galileo's original telescope design used a converging objective and a diverging eyepiece. The result is a compact, upright-image telescope — no need for an inverting prism. Opera glasses still use this design.

Common mistakes

• Drawing the focal point on the wrong side. For a diverging lens, $F$ is on the object side (the same side the light comes from). Many students reflexively put $F$ behind the lens because that's what they learned for converging lenses.
• Using a positive $f$ for a diverging lens. The sign convention is not optional. If the lens is diverging, $f < 0$. If you accidentally use $f = +15$ instead of $-15$, you'll get a real, inverted image — which physically cannot happen with a diverging lens.
• Thinking the virtual image is "behind" the lens. It's not — it's on the same side as the object, between the object and the lens. Saying "behind" is ambiguous and usually wrong.

Try it in the visualization

Notice how the image (dashed arrow, always upright) hugs close to the lens and never grows bigger than the object. Drag the object distance from 5 cm to 60 cm — the image barely moves, staying pinned between the lens and $F$. Toggle the backward ray extensions to see where the three diverging rays appear to originate. Turn on the magnification graph to watch $m$ creep from ~0.75 (close objects) asymptotically toward 0 (far objects) — it never reaches 1.