符号付きレンズおよびミラー像
導入済み| English | 意味 |
|---|---|
| virtual image/ˈvɜːtʃuːəl ˈɪmɪdʒ/ | 虚像 |
| transverse magnification/trænsˈvɜːs ˌmæɡnɪfɪˈkeɪʃn/ | 横倍率 |
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?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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.
このケースに当てはまる答えはどれか？
明示された符号則に基づき、近軸薄レンズおよび球面鏡の像を位置づける
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.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
The order follows the stated process; check each stage before the next.