Preview

Geometric Optics — Lenses & Mirrors

Q: An object is 30 cm from a convex lens with f = 20 cm. Where does the image form?

You can work it out this way: apply 1/f = 1/v + 1/u with u = 30 and f = 20.

Q: Under what condition does a convex lens produce a virtual image?

You can work it out this way: compare the object distance to the focal length.

Q: An object is placed at 2f from a convex lens. Calculate the magnification.

You can work it out this way: find image distance first, then use m = -v/u.

Trace rays through lenses and discover image formation

Every microscope and telescope you have ever used is two lenses playing catch. The first lens — the objective — takes the object and throws a real image into the empty space inside the barrel. That image hangs there, made of nothing but converging light, and then the second lens — the eyepiece — catches it and magnifies it again. The astonishing part is where the hand-off happens. In this lab's default state, the first lens (f₁ = 10 cm, object at 12 cm) is already trying to form its image 60 cm down the axis — but the second lens sits only 25 cm away. The rays have not finished converging when they hit it. Your instinct says the system should stall: no meeting point, no image, nothing to catch. The physics disagrees. The lens equation does not care whether the object is a physical arrow or a would-be image that never got to form: it reads the still-converging bundle as a virtual object with a negative object distance, do₂ = 25 − 60 = −35 cm, and quietly returns a real final image 4.4 cm behind the eyepiece. This is the engine inside every compound microscope — the objective's image often forms beyond the eyepiece, and the instrument works anyway. You get four live controls: both focal lengths, the object distance, and the separation between the lenses. Push the object toward the focal point and watch the image distance run away to infinity — the same equation, blowing up exactly where the textbook says it must. Then tour the three instruments: a telescope whose angular reach is −5× while its panel magnification reads −0.200, a microscope that magnifies −1.667× when the famous f₁/f₂ rule says −1, and a camera stack whose concave second element flips the image back upright. Two lenses, one equation applied twice — and a hand-off that works even when the catch happens before the ball arrives.

What you'll be able to do

Apply the thin-lens equation twice along a two-lens chain and read the panel quantitatively: at the shipped defaults (f₁ = 10, f₂ = 5, do₁ = 12, sep = 25 cm) lens 1 gives di₁ = (1/10 − 1/12)⁻¹ = 60.0 cm and m₁ = −5.000; the intermediate image falls 35 cm BEYOND lens 2, so lens 2 receives a virtual object do₂ = 25 − 60 = −35 cm and still delivers a real final image di₂ = (1/5 + 1/35)⁻¹ = 4.375 cm with m₂ = +0.125 and mTot = −0.625 — 'this first image serves as the object for the second lens' (OpenStax §2.8 verbatim)
Locate and interpret the chain's two singularities: as do₁ → f₁ the image distance diverges to positive infinity because 'an object at the focal plane produces parallel rays that form an image at infinity' (OpenStax §2.4 verbatim) — on the panel the di₁ row flips to '∞' at do₁ = 10.0 cm — and below f₁ the same lens is a simple magnifier (do₁ = 5 cm → di₁ = −10.0 cm, m₁ = +2.000, matching OpenStax Example 2.4b digit for digit); the second pole sits at do₂ = 0 (sep = di₁), where the would-be object lands inside lens 2 itself
Distinguish linear from angular magnification and bound the f₁/f₂ rule: total LINEAR magnification is the product mTot = m₁·m₂ of each stage, while M = −fobj/feye (OpenStax Eq. 2.40) is the telescope's ANGULAR magnification derived only for the afocal configuration (focal planes superposed) — at the microscope preset mTot = −1.667 ≠ −f₁/f₂ = −1, and at the telescope preset the panel's mTot = −0.200 coexists with the instrument's angular −5×, the two numbers never interchangeable

Formulas

\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

Thin Lens Equation

m = -\frac{v}{u}

Magnification

v > 0 \Rightarrow \text{Real image},\quad v < 0 \Rightarrow \text{Virtual image}

Image type by sign of image distance

Make a prediction

Default state: f₁ = 10 cm, f₂ = 5 cm, object at do₁ = 12 cm, lenses separated by only 25 cm. The panel shows lens 1 forming its intermediate image at di₁ = 60.0 cm — 35 cm BEYOND lens 2, so the converging rays are intercepted before they can ever meet. What does lens 2 do?

No grading here — commit to a guess, then scroll down and test it yourself.

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Audit the default state row by row: di₁ = 60.0 cm sits 35 cm beyond lens 2 (sep = 25) — find the pink Image 1 arrow standing past the cyan lens while its rays are intercepted mid-convergence. do₂ = −35 cm (virtual object), di₂ = 4.4 cm, m₂ = +0.125, mTot = −0.625. The system your instinct said would stall is the factory default.
Walk into the pole: drag Object Distance from 12 down toward f₁ = 10 and watch the di₁ row balloon — at 10.0 it flips to '∞' (object at the focal plane: the emergent rays go parallel and the image forms at infinity). Keep going to 5 cm: di₁ reads −10.0 and m₁ flips to +2.000 — the same lens is now a simple magnifier, virtual and upright. These are the two branches of one hyperbola; the panel just showed you both, plus the asymptote between them.
Tour the three instruments: Simple Telescope (di₁ = 56.3, mTot = −0.200 — yet its angular magnification is −5×: f₁/f₂ is an angular law for objects at infinity, never the panel's linear number), Compound Microscope (mTot = −1.667 where the f₁/f₂ rule predicts −1 — the misconception card's own live counter), Camera Lens System (concave f₂ = −10: m₂ = +2.000, a diverging stage re-erecting the chain, Galileo's arrangement).
Stretch the barrel: sweep Lens Separation 10 → 59 cm and read the mTot row deepening −0.455 → −4.167 as the virtual object do₂ = sep − 60 slides toward zero. Stop before 60: at sep = 60 the would-be object lands inside lens 2 itself (do₂ = 0) and the equation has no answer — the chain's second pole, approached but never crossed.

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Your prediction

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Answer: Lens 2 reads the still-converging bundle as a virtual object (do₂ = 25 − 60 = −35 cm) and the thin-lens equation answers unchanged: a real final image at di₂ = 4.4 cm behind lens 2, total magnification −0.625

The thin-lens equation is a local accountant: it only asks what the rays arriving at a lens are doing, not whether they have finished their previous job. At lens 2 the bundle is still converging toward a point 35 cm past it, so that would-be image acts as a VIRTUAL OBJECT with do₂ = sep − di₁ = −35 cm — and the equation handles a negative object distance exactly as written: 1/di₂ = 1/f₂ − 1/do₂ = 1/5 + 1/35 = 8/35, giving di₂ = 4.375 → the panel's 4.4 cm, with m₂ = +0.125 and mTot = m₁·m₂ = −5 × 0.125 = −0.625 (the lab's second analytic gate anchors this entire chain on the engine's own arithmetic). OpenStax §2.8 states the chain rule this lab is built on — 'this first image serves as the object for the second lens' — and the sim's own quiz Q2 keys the same answer: 'the point of convergence acts as a virtual object for Lens 2 (do₂ < 0)'. Option A is the targeted misconception: convergence does not need to complete before the next element — compound microscopes run in exactly this regime by design (the objective's image frequently falls beyond the eyepiece). Option C violates the single-ray-bundle fact: the light arriving at lens 2 IS lens 1's light, already bent; there is one bundle on the axis, not two, and Goldberg & McDermott's documented finding — that students miss 'the uniqueness of the relationship among the components of the optical system' — is precisely the assumption C smuggles in.

Quiz (0/3)

An object is 30 cm from a convex lens with f = 20 cm. Where does the image form?

Under what condition does a convex lens produce a virtual image?

An object is placed at 2f from a convex lens. Calculate the magnification.

You can now

Apply the thin-lens equation twice along a two-lens chain and read the panel quantitatively: at the shipped defaults (f₁ = 10, f₂ = 5, do₁ = 12, sep = 25 cm) lens 1 gives di₁ = (1/10 − 1/12)⁻¹ = 60.0 cm and m₁ = −5.000; the intermediate image falls 35 cm BEYOND lens 2, so lens 2 receives a virtual object do₂ = 25 − 60 = −35 cm and still delivers a real final image di₂ = (1/5 + 1/35)⁻¹ = 4.375 cm with m₂ = +0.125 and mTot = −0.625 — 'this first image serves as the object for the second lens' (OpenStax §2.8 verbatim)
Locate and interpret the chain's two singularities: as do₁ → f₁ the image distance diverges to positive infinity because 'an object at the focal plane produces parallel rays that form an image at infinity' (OpenStax §2.4 verbatim) — on the panel the di₁ row flips to '∞' at do₁ = 10.0 cm — and below f₁ the same lens is a simple magnifier (do₁ = 5 cm → di₁ = −10.0 cm, m₁ = +2.000, matching OpenStax Example 2.4b digit for digit); the second pole sits at do₂ = 0 (sep = di₁), where the would-be object lands inside lens 2 itself
Distinguish linear from angular magnification and bound the f₁/f₂ rule: total LINEAR magnification is the product mTot = m₁·m₂ of each stage, while M = −fobj/feye (OpenStax Eq. 2.40) is the telescope's ANGULAR magnification derived only for the afocal configuration (focal planes superposed) — at the microscope preset mTot = −1.667 ≠ −f₁/f₂ = −1, and at the telescope preset the panel's mTot = −0.200 coexists with the instrument's angular −5×, the two numbers never interchangeable