Preview
Geometric Optics: Basics
Thin lens ray diagrams — real and virtual images
Hold a magnifying glass over a page and the print swells — slide the same lens a little farther from the page and the world flips upside down. One piece of glass, two opposite behaviors, and the whole story fits on a single line of algebra: 1/f = 1/do + 1/di. This lab hands you that line as a living ray diagram. An orange arrow stands on the optical axis; three principal rays leave its tip — one running parallel to the axis, one aimed through the near focal point, one straight through the lens center — and wherever they re-meet, an image is born. Your three levers are the object distance, the focal length (positive for a converging lens, negative for a diverging one), and the object height. The drama lives at one special spot: walk the object toward the focal point and the image distance runs away — 37.5 cm, then 240 cm, then off the chart entirely, because an object sitting exactly at the focal plane sends its rays out parallel, and parallel rays only meet at infinity. Push one step closer and the same lens stops projecting altogether and becomes the magnifier on your desk: a virtual, upright, enlarged image floating on the object's own side, where no screen can ever catch it. Flip the focal length negative and the lens turns diverging — now it refuses to make a real image at any distance, always upright, always shrunk, always virtual. Three presets stage the three classic cases; a live data panel prices every image to the tenth of a centimeter. By the end, the thin-lens equation stops being a formula and becomes a map you can read at a glance.
What you'll be able to do
- Close the thin-lens equation and magnification on the panel's own rows at the shipped default state (do = 25 cm, f = 15 cm, ho = 3 cm): 1/di = 1/15 − 1/25 = 2/75 gives di = 37.5 cm on the di row, m = −37.5/25 = −1.50× on the m row (negative → inverted, per Eq. 2.22 and the sign conventions), and hi = −1.5 × 3 = −4.50 cm on the hi row — a real, inverted image on the opposite side, 'di is positive if the image is on the side opposite the object' (UP3 §2.4 verbatim), badge reading Real · Inverted
- Locate and interpret the pole at do = f: as Object Distance walks 25 → 16 → 15 cm at f = 15 cm the di row balloons 37.5 → 240.0 → '∞' — 'As the object approaches the focal plane, the image distance diverges to positive infinity… an object at the focal plane produces parallel rays that form an image at infinity' (UP3 §2.4 verbatim) — and below the pole the same lens is a simple magnifier: do = 8 cm gives di = −120/7 → '−17.1', m = +15/7 → '2.14×', a case 2 image, 'virtual, on the same side of the lens as the object, and upright' (UP3 §2.4; CP2e 'To use a convex lens as a magnifier, the object must be closer to the converging lens than its focal length'), with OpenStax Example 2.4b (do = 5, f = 10 → di = −10.0, m = +2.00) as the digit anchor of the same branch
- Prove the diverging lens's single case: with f < 0 and any do > 0, 1/di = 1/f − 1/do is the sum of two negatives, so di < 0 always (virtual, same side), m = −di/do > 0 always (upright), and |di| < do gives |m| < 1 always (reduced) — CP2e Table 25.3 case 3 'formed for any object by a negative focal length or diverging lens' and UP3 Figure 2.25(b) 'the image distance is negative for all positive object distances'; live at the diverging preset (do = 20, f = −15): di = −60/7 → '−8.6', m = +3/7 → '0.43×', badge Virtual · Upright, machine-certified by the diverging-upright-reduced bounds gate
Formulas
Make a prediction
Default state: a converging lens with f = 15 cm, the object at do = 25 cm — the panel reads a real, inverted image at di = 37.5 cm with m = −1.50×. Now drag Object Distance down to 8 cm, INSIDE the focal length (do < f). What happens to the image?
No grading here — commit to a guess, then scroll down and test it yourself.
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Your prediction
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Answer: The image flips to the object's own side and turns virtual, upright, and magnified: 1/di = 1/15 − 1/8 = −7/120 gives di = −17.1 cm and m = +2.14× — the magnifying-glass configuration
The thin-lens equation does not shut off inside f — it changes sign. With do = 8 cm and f = 15 cm: 1/di = 1/15 − 1/8 = (8 − 15)/120 = −7/120, so di = −120/7 = −17.14 cm → the panel's '−17.1', and m = −di/do = +15/7 = +2.14× with hi = +6.43 cm. The negative image distance is the whole answer: the emergent rays still diverge after the lens, but their backward extensions meet on the OBJECT's side — a virtual, upright, magnified image (CP2e Table 25.3 case 2; 'To use a convex lens as a magnifier, the object must be closer to the converging lens than its focal length', verified; UP3 Example 2.4b prices the same branch at do = 5, f = 10 → di = −10.0, m = +2.00, and the sim's own quiz Q1 keys it: 'this is exactly how a magnifying glass works'). Option A is the targeted misconception: it confuses 'no REAL image' with 'no image at all' — the rays really do fail to converge, but the lens's job is to redirect the bundle, and a bundle that merely APPEARS to originate from a point is an image too, just not one a screen can catch. Option C half-remembers the truth: approaching the focal point from OUTSIDE does grow the real image without bound (di runs 37.5 → 240 → ∞ as do runs 25 → 16 → 15) — but at do = f that image ceases to exist as a finite object (the rays exit parallel, 'an image at infinity'), and below f the real branch is gone entirely. The lab's machine gate brackets the magnifier branch at exactly 15/7.
Quiz (0/3)
An object is 30cm from a converging lens with f=10cm. Where is the image?
What is the magnification for the setup above?
Why does a magnifying glass work? (d_o < f)
You can now
- Close the thin-lens equation and magnification on the panel's own rows at the shipped default state (do = 25 cm, f = 15 cm, ho = 3 cm): 1/di = 1/15 − 1/25 = 2/75 gives di = 37.5 cm on the di row, m = −37.5/25 = −1.50× on the m row (negative → inverted, per Eq. 2.22 and the sign conventions), and hi = −1.5 × 3 = −4.50 cm on the hi row — a real, inverted image on the opposite side, 'di is positive if the image is on the side opposite the object' (UP3 §2.4 verbatim), badge reading Real · Inverted
- Locate and interpret the pole at do = f: as Object Distance walks 25 → 16 → 15 cm at f = 15 cm the di row balloons 37.5 → 240.0 → '∞' — 'As the object approaches the focal plane, the image distance diverges to positive infinity… an object at the focal plane produces parallel rays that form an image at infinity' (UP3 §2.4 verbatim) — and below the pole the same lens is a simple magnifier: do = 8 cm gives di = −120/7 → '−17.1', m = +15/7 → '2.14×', a case 2 image, 'virtual, on the same side of the lens as the object, and upright' (UP3 §2.4; CP2e 'To use a convex lens as a magnifier, the object must be closer to the converging lens than its focal length'), with OpenStax Example 2.4b (do = 5, f = 10 → di = −10.0, m = +2.00) as the digit anchor of the same branch
- Prove the diverging lens's single case: with f < 0 and any do > 0, 1/di = 1/f − 1/do is the sum of two negatives, so di < 0 always (virtual, same side), m = −di/do > 0 always (upright), and |di| < do gives |m| < 1 always (reduced) — CP2e Table 25.3 case 3 'formed for any object by a negative focal length or diverging lens' and UP3 Figure 2.25(b) 'the image distance is negative for all positive object distances'; live at the diverging preset (do = 20, f = −15): di = −60/7 → '−8.6', m = +3/7 → '0.43×', badge Virtual · Upright, machine-certified by the diverging-upright-reduced bounds gate