A Lens as a Magnifying Glass
1 · Predict
An object sits closer to a converging lens than its focal length. Does the lens still form a real image, or something different?
- It forms an enlarged, upright virtual image on the same side as the object — a magnifying glass.
- It still forms a real image, just farther away.
- No image forms at all inside the focal length.
2 · Set Up
- Open the magnifier preset and press Reset. The lens's focal length is fixed at 10 cm.
- Enable the image-distance and magnification readouts.
- Set the object distance for each trial (always less than the focal length) and record the image distance and magnification.
3 · Collect Data
| Object distance d_o (cm) | Image distance d_i (cm) | Magnification m |
|---|---|---|
| 4 | ||
| 6 | ||
| 8 |
Plot magnification m (y-axis) against object distance d_o (x-axis) for your three trials. Does m grow as d_o shrinks toward zero?
4 · Analyze
- For one trial, compute d_i from 1/f = 1/d_o + 1/d_i using f = 10 cm, then m = −d_i/d_o. Compare both to the table.
- Every trial gives negative d_i and positive m greater than 1. Explain what a negative image distance means (virtual image, same side as the object) and why m is positive here instead of negative.
5 · Extend
- Your data shows moving the object closer to the lens (smaller d_o, always under f) increases the magnification. Explain why a magnifying glass held very close to a small object shows it bigger than holding it farther away (but still inside f).
- What would happen to the image (and magnification) if the object were placed exactly at the focal length, d_o = f? Compare to the focal-point experiment's result.
The Physics Behind This Experiment
Virtual Image Magnification
When an object sits inside a converging lens's focal length, the image distance d_i comes out negative (virtual, same side as the object), and the magnification m = −d_i/d_o comes out positive and greater than 1 — an upright, enlarged image, the basis of a magnifying glass.