En anglais

Where curved mirrors put images

Science · Waves, Light & Sound · ages 19-20

Name ______________________ Date ____________

Which ray construction locates a virtual image formed by a convex mirror?
- Intersect backward extensions of two reflected rays
- Intersect two incident rays before they reach the mirror
- Intersect two reflected rays in front of the mirror
A concave spherical mirror has radius of curvature 20 cm. In the paraxial approximation, what is its focal length?
- 5 cm, equal to one quarter of the radius
- 20 cm, equal to the whole radius
- 10 cm, equal to half the radius
Using the lesson's sign convention, di = -6 cm and m = +2 describe a virtual, upright image. True or false?
Circle one: True False
Compared with a plane mirror of the same size, what makes a convex mirror useful for viewing nearby traffic?
- A narrower field with enlarged, upright virtual images
- A wider field with enlarged images on a real screen
- A wider field with reduced, upright virtual images
Use cm and x positive in front. A paraxial concave mirror has f = 6, do = 18, h = 0.3. On paper, draw reflected lines through (0,0.3),(6,0) and through (0,0),(18,-0.3). Locate their intersection and check 1/di = 1/6 - 1/18. Which result agrees?
- (x,y) = (9,-0.15); di = +9 cm
- (x,y) = (9,+0.15); di = +9 cm
- (x,y) = (-9,+0.15); di = -9 cm
Use cm and x positive in front. A paraxial convex mirror has f = -12, do = 24, h = 0.3. Draw reflected lines through (0,0.3),(-12,0) and through (0,0),(24,-0.3). Locate their backward intersection and check 1/di = -1/12 - 1/24. Which result agrees?
- (x,y) = (-8,+0.1); di = -8 cm
- (x,y) = (-8,-0.1); di = -8 cm
- (x,y) = (+8,-0.1); di = +8 cm
Which ideal mirror removes spherical aberration for a broad incoming beam parallel to its optical axis?
- A wider spherical mirror with the same curvature
- A convex spherical mirror that spreads the beam
- An ideal paraboloid aligned with the axial beam
Case A: an axial parallel beam hits a spherical mirror; edge and central rays focus at different distances. Case B: a tilted parallel beam hits an ideal paraboloid and gives a comet-like blur. Which labels distinguish the two cases?
- A: coma; B: spherical aberration
- A: spherical aberration; B: coma
- A: spherical aberration; B: chromatic aberration

Answer key

For grown-ups. Fold this page away before handing over the rest.

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Intersect backward extensions of two reflected rays · The reflected rays diverge. Extend them backward behind the convex mirror; their extensions meet at the virtual image point.
10 cm, equal to half the radius · For a concave spherical mirror, f = R/2 in the paraxial approximation. Thus f = 20/2 = 10 cm.
True · Negative image distance means virtual and behind the mirror. Positive magnification means upright. Its magnitude of 2 also means enlarged.
A wider field with reduced, upright virtual images · For real objects, a convex mirror forms reduced, upright, virtual images. It shows a wider field than an equal-sized plane mirror.
(x,y) = (9,-0.15); di = +9 cm · The forward reflected lines intersect at (9,-0.15), giving a real, inverted image 9 cm in front. Independently, 1/di = 1/6 - 1/18 = 1/9 cm^-1, so di = +9 cm.
(x,y) = (-8,+0.1); di = -8 cm · The backward extensions intersect at (-8,+0.1), locating a virtual, upright image 8 cm behind. The equation gives 1/di = -1/12 - 1/24 = -1/8 cm^-1, hence di = -8 cm.
An ideal paraboloid aligned with the axial beam · An ideal paraboloid focuses an axial parallel beam at one point. This removes spherical aberration for that beam, but does not remove every off-axis aberration.
A: spherical aberration; B: coma · A is spherical aberration: different ray heights give different focal distances even for an axial beam. B is coma: a tilted beam gives a comet-like blur, which an ideal paraboloid can still produce.