A Diverging Lens: Always Virtual, Always Reduced
1 · Predict
A diverging lens has a negative focal length. Does it ever form a real, enlarged image, the way a converging lens can?
- No — a diverging lens always forms a smaller, upright virtual image, no matter the object distance.
- It forms a real image when the object is far enough away.
- It always forms a real, enlarged image.
2 · Set Up
- Open the diverging-virtual 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 and record the image distance and magnification.
3 · Collect Data
| Object distance d_o (cm) | Image distance d_i (cm) | Magnification m |
|---|---|---|
| 20 | ||
| 30 | ||
| 40 |
Plot magnification m (y-axis) against object distance d_o (x-axis) for your three trials. Does m ever exceed 1?
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.
- Explain why, no matter how large d_o gets, d_i stays negative (virtual) and m stays positive and less than 1 for a diverging lens — unlike a converging lens's behavior in the converging-real experiment.
5 · Extend
- A door peephole uses a diverging lens (or a similar wide-angle design) to show a reduced, wide field-of-view image of whoever's outside. Explain why a reduced image is exactly what you'd want for a peephole's purpose.
- As d_o grows very large (object very far away), what does d_i approach? Compare it to the lens's focal length, and explain physically why a diverging lens's virtual image can never move past its own focal point.
The Physics Behind This Experiment
Diverging Lens Imaging
A diverging lens has negative focal length. Substituting f < 0 into 1/f = 1/d_o + 1/d_i always yields negative d_i (virtual) and 0 < m < 1 (reduced, upright) for any positive object distance.