Where is the virtual image plane when the vergence is 4 diopters?

Light has a way of bending reality in the blur between what we see and where we think it comes from. In visual optics, that blurred line is often described with a few tidy ideas: where rays bend, where they appear to originate, and how that “image” sits relative to the lens. Here’s a neat little puzzle that scratches that surface: refracted light waves diverge from a lens in air, and the question is, where is the virtual image plane? The multiple-choice options look like a brain teaser, but the path to the answer is pretty friendly if you keep the concepts straight.

What the numbers actually mean

Let’s set the scene in plain terms. A lens sits in air. Light leaving the lens to the right has been bent, and at a certain plane, 15 cm to the right of the lens, you measure a vergence of 4 diopters (D). Vergence is just a way to quantify how strongly the light rays are diverging (or converging) as they head toward a point. Positive numbers usually point you toward convergence or toward a real focal point; negative ones toward a virtual point.

Two quick reminders help:

• Vergence in diopters is the reciprocal of distance, with sign conventions kept in mind.

• A virtual image appears to come from a point on the same side of the lens as the incoming light (the left side in our setup), so its distance from the lens is considered negative in the usual Cartesian sign convention.

The punchy trick to solve this

The plane you’re looking at is a certain distance from the lens: s = 15 cm = 0.15 m to the right. The light leaving the lens behaves as if it’s diverging from a virtual image located somewhere on the left side of the lens, at a distance d_left (positive for magnitude, but the actual distance from the lens is negative in the sign convention).

The key relation (using the plane’s vergence) is:

V_plane = 1 / (s - v_image)

Here, v_image is the image distance from the lens with the sign convention: negative if the image is on the left (virtual), positive if on the right (real, if it existed). Since the image is virtual and on the left, v_image is negative. If the virtual image sits a distance d_left to the left of the lens, then v_image = -d_left.

Plugging in what we know:

• V_plane = 4 D

• s = 0.15 m

• v_image = -d_left (where d_left is the magnitude we want)

So:

4 = 1 / (0.15 - (-d_left)) = 1 / (0.15 + d_left)

Solve for d_left:

0.15 + d_left = 1 / 4 = 0.25

d_left = 0.25 − 0.15 = 0.10 m

That gives d_left = 0.10 m, i.e., 10 cm. Since this is a virtual image on the left, the virtual image plane sits 10 cm to the left of the lens.

The final answer: 10 cm to the left of the lens.

Why this makes sense, in plain language

• A vergence of 4 D corresponds to a light bundle that behaves as if it comes from a point 25 cm away from the plane where you measure it. But that “plane” is not the lens itself—it’s 15 cm off to the right. If the virtual image were 25 cm to the left of the lens, the plane to the right would see a different vergence.

• When you carry the geometry from the lens to the measurement plane, you add the left-distance of the image to the right-distance to the plane. In our numbers, 15 cm to the right plus 10 cm to the left yields 25 cm. The math checks out: a 25 cm separation from the measurement plane gives a 4 D vergence, matching the given data.

• So the image isn’t perched 25 cm away on the left—it’s closer, at 10 cm to the left. The short, neat picture is that the plane sees the rays as if they come from a point 25 cm away, and the actual virtual image that produces that appearance sits 10 cm to the left.

A quick sanity check you can do in your head

• If the virtual image were right at the lens (0 cm left), the distance from the lens to the measurement plane would be 15 cm, and the vergence would be 1/0.15 ≈ 6.7 D — not our 4 D.

• If the image were 25 cm to the left, the distance from the plane to the image would be 0.15 + 0.25 = 0.40 m, giving 1/0.40 = 2.5 D — again not matching.

• Our 10 cm left sits exactly at the sweet spot that makes 0.15 + 0.10 = 0.25 m, and 1/0.25 = 4 D. The numbers click into place.

Why this matters beyond the quiz

You’ll recognize this as the kind of thinking that comes up a lot when you’re dealing with lenses, virtual images, and how the eye or a camera interprets light. It’s not just about plugging numbers; it’s about building a mental model:

• Vergence is a handy shorthand for angular thoughts about where light is heading.

• The position of a virtual image isn’t just a single point; it’s tied to where you measure the rays’ behavior elsewhere in the optical path.

• Sign conventions matter. A virtual image on the left is negative in the standard lens-sign world, but that negative sign is just a reminder of direction, not a mystery.

Analogies for real life

Think of the lens as a storyteller, and the light rays as the audience. The “virtual image plane” is where the crowd imagines the story is coming from. If you tilt your head and look from a different distance, the same story looks like it’s coming from a different spot. The math you use to track that is the same math you’d use to plan a camera focus or to simulate how a pair of glasses shapes your view.

A quick note on language and flavor

In conversations about light, you’ll hear terms that look like abstract math at first. The trick is to stay with the intuition: vergence tells you how strongly rays are diverging or converging; the image plane is the phantom source seen by the eye or sensor; and distances are your bridges between what you see and where it came from. If a diagram helps, sketch it: a lens, a rightward measurement plane, a suspected leftward image position, and the numbers 0.15 m and 0.10 m doing the dance in your notebook.

Where this leads next, in a broader sense

If you enjoyed this little problem, you’ll find that similar tricks pop up in other optical setups: how diopters relate to focal lengths, how eyeglass prescriptions translate to real-world lens arrangements, and how cameras render virtual depth. The thread tying it all together is the idea that light isn’t just a bundle of rays; it’s a geometry puzzle that your brain is trained to solve.

Wrap-up, with a touch of curiosity

So the virtual image plane sits 10 cm to the left of the lens. It’s a tidy little result that showcases how a measured vergence at a distance can reveal a hidden position of a virtual source. And once you’ve got that, you’re no longer just chasing multiple-choice answers—you’re gaining a lens into how light behaves in everyday life, from eyeglasses to smartphone cameras.

If you’re ever curious about applying this kind of reasoning to a real-world setup—say, tweaking a magnifier, evaluating a contact-lens scenario, or framing a shot with a tricky focal arrangement—hang onto that mental model: vergence, sign conventions, and the distance between the plane you’re measuring and the image you infer. With those in your pocket, a good chunk of optical puzzles becomes pleasantly approachable, even enjoyable.