Learn how to position a +2 D lens to collimate light from a minus lens in visual optics.

Outline (quick skeleton)

• Hook: why this two-lens setup is a neat little puzzle in visual optics

• The setup in plain terms: a minus lens leaves a virtual image 15 cm to the left

• Step-by-step reasoning: turning diopters into a focal length, then back to lens spacing

• The key calculation: place the +2 D lens so its distance to the virtual image equals its focal length

• Result and a quick sanity check

• Real-world flavor: where you’d see a setup like this, beyond the math

• Tips and common gotchas for similar two-lens problems

• Parting thought: you’ve got this kind of geometry in your toolkit

A little puzzle with a familiar vibe

If you’ve ever messed with lenses, you know the vibe: the way light bends, the way images shift, and how a simple distance can change everything. Here’s a clean little scenario that feels almost like a riddle: after a minus lens in air, the light is diverging as it heads toward image space. The catch? The virtual image sits 15 cm to the left of that minus lens. Now, you add a second lens—this one with a positive power of +2 diopters. The goal is to position that second lens so the light leaves in a perfectly parallel beam, i.e., collimated. The multiple-choice options ask, essentially, “How far to the right of the first lens should the second lens sit to make this happen?” The correct answer is 35 cm.

What the setup means in plain terms

Let me unpack the scene a bit, so the geometry doesn’t get tangled:

• The first lens is a concave (minus) lens. It makes a virtual image rather than a real one, and that virtual image sits 15 cm to the left of the lens.

• Light emerging from a concave lens tends to spread out as if it came from that virtual image. If you could place your eye where the light converges, you’d see the image at that leftward position.

• The second lens is a plus lens with a focal length derived from its power. A power of +2 diopters corresponds to a focal length of 1/2 meter, or 50 cm (since f in meters is 1/P in diopters).

Let’s connect the dots to find where to place the second lens

Here’s the clean, mechanical way to reason it, without any mysticism:

• For the second lens to convert the diverging beam into parallel (collimated) rays, that second lens must be located at a distance from the virtual image equal to its focal length.

• The virtual image is 15 cm to the left of the first lens. If we put the second lens to the right of the first lens, the distance from the second lens to that virtual image is the sum of two pieces: the gap between the lenses plus that 15 cm on the left.

• The second lens has f2 = 50 cm. So we want: distance from L2 to the virtual image = 50 cm.

Mathematical trace (in plain numbers)

• Let x be the distance from the first lens to the second lens (to the right). Then the distance from the second lens to the virtual image is x + 15 cm.

• Set x + 15 cm = 50 cm (the focal length of the +2 D lens).

• Solve for x: x = 50 − 15 = 35 cm.

So, the second lens should sit 35 cm to the right of the first lens. That places the two lenses just right to coax that diverging beam into a neat, parallel stream—exactly what “collimate” means in this context.

A quick mental check

• The unit check lines up: x comes out in centimeters, the focal length was in centimeters (50 cm). No funny business with meters versus centimeters here.

• If the second lens were any closer, the rays would still be diverging after the second lens; if it were farther, they’d still be diverging but with different acceleration. By landing at 35 cm, the second lens brings those rays to a bright, parallel line—perfect for situations that require sharp, well-formed beams.

Why this matters beyond the numbers

Two-lens configurations are everywhere in optics, from simple magnifying setups to more elaborate diagnostic tools. Here’s why this little exercise matters:

• It reinforces sign conventions and how virtual objects behave. A diverging beam from a minus lens behaves as if it’s coming from a point on the same side as the object, which is why the virtual image sits to the left.

• It shows how a positive lens can “neutralize” divergence when placed properly, turning a spread into a straight line. That’s the core idea behind beam shaping and even some basic laser alignment tricks.

• In practical terms, you’ll see this kind of thinking in choosing lens placements for ophthalmic devices, diagnostic instruments, or even simple projection setups where you want a crisp, collimated output.

A few related ideas to chew on

• Focal length and diopters: you don’t always hold a ruler up to the problem. The rule of thumb is f (in meters) = 1 / P (in diopters). It’s a tiny formula, but it unlocks quick intuition about how “strong” a lens is and where it acts in a system.

• Virtual vs real images: with a minus lens, the image can be virtual and on the left side—this is a common source of confusion if you skip the sign convention. Drawing a quick ray diagram helps a lot.

• Real-world analogies: think of a diverging beam like a sprinkler head that’s spraying water outward. A positive lens placed at just the right distance acts like a nozzle that refracts that spray into a neat, straight stream—still water, just reframed.

Common traps and how to sidestep them

• Mixing up the sign conventions: keep straight that the virtual image is on the same side as the object for a minus lens. If you flip this, the distance math will feel off.

• Forgetting the lens-to-image distance applies to the second lens as well: the second lens needs to “see” the virtual image at its focal length. It’s not enough to know the second lens’s focal length in isolation.

• Assuming the answer depends on real images: this setup relies on a virtual image acting as the object for the second lens. If you mistakenly chase a real-image route, the numbers won’t line up.

A few practical tips you can try on your own

• Sketch it first: a quick diagram with the first lens, the virtual image location, and the proposed position of the second lens. Label distances in centimeters; it makes the algebra feel tactile.

• Use a simple ruler and a straight line of sight: imagine a beam leaving the first lens toward the right, then check where that line would appear to originate from on the left.

• Do a quick sanity check by working backward: suppose the second lens is at 35 cm from the first lens; verify the distance to the virtual image is 50 cm. If that checks out, you’re aligned.

A little more context for the curious mind

If you’ve ever pondered how eyeglasses are tuned for people with different refractive quirks, you’ve touched on the same physics. Lenses don’t just bend light in a vacuum; they shape a whole path—one that can be orchestrated with a precise hand and a patient mind. It’s a bit of practical physics that you can see in a pocket-size format if you’ve got a spare pair of lenses or a tiny lab setup.

Wrapping it up with the bottom line

In this scenario, the second lens—the +2 D one—belongs 35 cm to the right of the first, so that the distance from the virtual image (which sits 15 cm to the left of the first lens) to the second lens equals the second lens’s focal length (50 cm). That placement makes the light emerging from the second lens become parallel, i.e., collimated.

If you want to keep sharpening your intuition for these two-lens problems, try varying the numbers a bit: change the position of the virtual image, alter the second lens’s power, or switch the order of lenses to see how the geometry shifts. The core idea—collimating light by tying a lens’s focal length to the separation between optical elements—stays the same. And that’s the kind of mental toolkit you’ll find handy across the world of visual optics, whether you’re drafting a quick diagram for a project, adjusting a diagnostic instrument, or just exploring how light behaves when it meets glass.