Signed lens and mirror images
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| virtual image/ˈvɜːtʃuːəl ˈɪmɪdʒ/ | 虚像 | xū xiàng |
| transverse magnification | 横向放大率 | héng xiàng fàng dà lǜ |
A decision before an answer
- A negative image distance is a geometric result that locates a virtual image, not a failed calculation to replace by its absolute value.
- Your goal: Locate paraxial thin-lens and spherical-mirror images with a declared sign convention.
Declare distances and signs
- Use the real-is-positive convention for this unit: a real object sends diverging incident rays into the element, so s>0; an already converging incident beam can represent a virtual object, s<0. A real image has s′>0, while a virtual image has s′<0. The paraxial equation is 1/f=1/s+1/s′ and transverse magnification M=−s′/s.
- A converging lens has f>0 and a diverging lens f<0. For left-to-right light and a real object at x<0 relative to a lens at x=0, a real image lies at x=s′>0; a negative s′ puts its virtual image on the input side. State the convention before substitution, as other signed-coordinate conventions give different-looking equations.
A converging lens has f=12 cm and a real object s=8 cm. Which image result is correct?
1/s′=1/12−1/8=−1/24; M=−(−24)/8=+3, virtual and upright.
Trace a lens image
- For a converging thin lens, a ray parallel to the axis exits through the far focal point, and a ray through the ideal optical centre continues undeviated. Their intersection identifies a real image; backwards extensions can locate a virtual image. The paraxial approximation uses rays close to the axis and neglects thickness and aberrations.
- With s>f>0, s′ is positive and M is negative: the image is real and inverted. With 0<s<f, s′ is negative and M is positive: the image is virtual, upright and enlarged. At s=f, emerging rays are parallel and no finite image plane exists. A diverging lens with a real object produces an upright reduced virtual image.
A concave mirror has f=10 cm and real object s=30 cm. The image is:
s′=15 cm and M=−1/2; a positive real mirror image is in front of the mirror.
Apply reflected-image geometry
- For a spherical mirror in the same real-is-positive convention, concave f>0 and convex f<0, with f=R/2 in the paraxial limit. A real reflected image is in front of the mirror on the incoming-light side; a virtual image is behind it. Thus a positive image distance has a different physical side for a mirror than for a transmitting lens.
- A parallel ray reflects through the concave focus; a ray aimed through the centre of curvature retraces its path. Convex reflected rays diverge as if from the focus behind the mirror. Use the same equation and M=−s′/s with signed values, rather than silently combining mirror geometry with a lens coordinate diagram.
A converging lens at x=0 has f=10 cm and a real object at x=−30 cm. 1/s′=1/10−1/30=1/15, so s′=15 cm and M=−1/2. A second f=10 cm lens at x=40 cm receives rays diverging from x=15: s₂=25 cm, s₂′=50/3 cm and final x=170/3 cm≈56.67 cm. M₂=−2/3, so total M=+1/3. Separately a convex mirror f=−12 cm with s=24 cm gives s′=−8 cm and M=+1/3, an upright virtual image behind the mirror.
A diverging lens has f=−15 cm and s=30 cm. Signed image distance is ____ cm.
1/s′=−1/15−1/30=−1/10; M=+1/3.
Locate the next element’s object
- For a sequence of separated lenses, first calculate the actual image coordinate from the first lens. Relative to the next lens, decide whether the incoming rays diverge from a point before it or are still converging toward a point beyond it; these are real and virtual objects respectively. Only then assign its signed object distance and solve again.
- Total transverse magnification is the product of the individual signed magnifications for aligned paraxial elements. Element separation is not automatically the second object distance. Geometric real/virtual character depends on convergence, not on whether the final image is magnified. These ideal results do not model wave-optical resolution or spherical/chromatic aberration.
Keep signs until the geometry is interpreted; a positive mirror image lies on the incident side, while a positive transmitting-lens image lies on the outgoing side.
Which answer fits this case?
Locate paraxial thin-lens and spherical-mirror images with a declared sign convention
For a converging thin lens with a real object exactly at its front focal point, the ideal output rays are parallel and the finite-image formula has no finite solution.
1/s′=1/f−1/f=0; the ideal image is at infinity.
Keep the distinctions
- virtual image 虚像 — An apparent image located by backwards ray extensions rather than actual outgoing-ray convergence.
- transverse magnification 横向放大率 — Signed image-height to object-height ratio, −s′/s in the declared convention.
- Locate paraxial thin-lens and spherical-mirror images with a declared sign convention.
- Use signed magnification to classify orientation and real or virtual character.
- Track successive image locations as the next element’s object without resetting geometry.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.