Answers · Reflection in mirrors

Reflection in mirrors — final check key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. AK1 · Final check

Model answer. (a) The normal is a line at right angles (90°) to the surface of the mirror, at the point where the ray hits it. (b) The angle of incidence is the angle between the incident ray and the normal. (c) The reflected ray is 53° from the surface, so the angle of reflection is 90° − 53° = 37°. By the law of reflection, the angle of incidence equals the angle of reflection: 37°. (d) The mirrors are parallel, so their normals are parallel too. The ray leaves the first mirror at 37° to its normal, so it meets the second mirror at 37° to the second normal: the angle of incidence there is 37°. (e) It reflects at 37° and leaves parallel to the ray that hit the first mirror, travelling in the same direction, but shifted to one side. This is how the two parallel mirrors of a periscope work.

A full answer includes

(a) I wrote: a line at 90° to the surface of the mirror.
(a) I wrote that it is drawn at the point where the ray hits the mirror.
(b) I wrote: the angle between the incident ray and the normal.
(b) I did not write “the angle between the incident ray and the mirror”.
(c) Method: I found the angle of reflection: 90° − 53° = 37°.
(c) Method: I used i = r.
(c) Answer: I wrote 37°.
(c) I did not give 53°: that is the angle to the surface.
(d) Method: I used that the two normals are parallel, because the mirrors are.
(d) Answer: I wrote 37°.
(d) I did not give 53°: that is the ray’s angle to the second mirror’s surface.
(e) I wrote that it is parallel to the ray that hit the first mirror.
(e) I wrote that it travels in the same direction as that ray.

Also accept

(a) at 90° to the mirror where the ray hits (b) between the incident ray and the normal (c) 37° (d) 37° (e) parallel to the first ray, in the same direction

Watch for

“The angle of incidence is the angle between the ray and the mirror.” — It is measured from the normal, not from the surface.
“The angle of incidence in (c) is 53°.” — The 53° is the reflected ray’s angle to the surface. Its angle to the normal is 90° − 53° = 37°, and i = r.
“The ray leaving the second mirror goes back the way it came.” — Parallel mirrors facing each other send the ray on in its first direction. Only mirrors at 90° to each other send it back.

2. AK2 · Final check

Model answer. (a) Light spreads out from the object, and each ray that hits the mirror reflects with its angle of reflection equal to its angle of incidence. The reflected rays spread out as if they came from one point behind the mirror, and the eye sees the image there. (b) The image is the same size as the object, as far behind the mirror as the object is in front, and virtual. (c) The object is now 27 cm + 9 cm = 36 cm from the mirror, so its image is 36 cm behind the mirror. The object was 2 × 27 cm = 54 cm from its image and is now 2 × 36 cm = 72 cm from it. The object has not moved, so the image has moved 72 cm − 54 cm = 18 cm, twice as far as the mirror. (d) The incident ray goes 6.0 cm across in 30 cm towards the mirror, 0.20 cm across for each centimetre. By the law of reflection the reflected ray makes the same angle with the normal, so it also goes 0.20 cm across for each centimetre away from the mirror. It has 23 cm − 6.0 cm = 17 cm still to go across, so the eye is 17 cm ÷ 0.20 = 85 cm from the mirror. Check with the image, 30 cm behind the mirror: the straight line from the image through the point on the mirror reaches 23 cm across after 23 ÷ 0.20 = 115 cm, and 115 cm − 30 cm = 85 cm.

A full answer includes

(a) I wrote that each ray from the object reflects with r = i.
(a) I wrote that the reflected rays spread out as if from one point behind the mirror.
(a) I wrote that the eye sees the image at that point.
(b) I wrote: the same size as the object.
(b) I wrote: as far behind the mirror as the object is in front.
(b) I wrote: virtual.
(b) I did not write that the image is on the surface of the mirror.
(b) I did not write that the image is real.
(c) Method: I found the new distance from the object to the mirror: 27 cm + 9 cm = 36 cm.
(c) Method: I found the distance from the object to its image before and after: 2 × 27 cm = 54 cm and 2 × 36 cm = 72 cm.
(c) Method: I took one from the other: 72 cm − 54 cm.
(c) Answer: I wrote 18 cm.
(c) I did not give 9 cm: the image moves twice as far as the mirror.
(d) Method: I found how far the incident ray goes across for each centimetre towards the mirror: 6.0 cm ÷ 30 cm = 0.20.
(d) Method: I used i = r, so the reflected ray goes 0.20 cm across for each centimetre away from the mirror.
(d) Method: I found how far the reflected ray goes across: 23 cm − 6.0 cm = 17 cm.
(d) Method: I divided: 17 cm ÷ 0.20.
(d) Answer: I wrote 85 cm.
(d) I did not give 115 cm: that is the eye’s distance from the image.

Also accept

(a) reflected rays seem to come from a point behind the mirror (b) same size; same distance behind; virtual (c) 18 cm (d) 85 cm

Watch for

“The image is on the surface of the mirror.” — It is as far behind the mirror as the object is in front.
“The image moves 9 cm, the same as the mirror.” — The image stays as far behind the mirror as the object is in front. The object is now 9 cm further from the mirror, and the mirror itself has moved 9 cm, so the image moves 9 cm + 9 cm = 18 cm.
“The eye is 115 cm from the mirror.” — 115 cm is how far the eye is in front of the image. The image is 30 cm behind the mirror, so take 30 cm away.