Answers · Reflection in mirrors

Reflection in mirrors — return questions key

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

1. AR1 · Return

Model answer. (a) The angle of reflection is the angle between the reflected ray and the normal, the line at 90° to the mirror where the ray hits it. (b) The angle of incidence is equal to the angle of reflection. (c) The angle between the beams is i + r, and r = i, so the angle of incidence is 64° ÷ 2 = 32°. That angle is measured from the normal, which is at 90° to the surface, so the incident beam is 90° − 32° = 58° from the surface of the mirror.

A full answer includes

(a) I wrote: the angle between the reflected ray and the normal.
(a) I did not write “the angle between the reflected ray and the mirror”.
(b) I wrote that the angle of incidence equals the angle of reflection.
(b) I did not write that the two angles add up to 90°.
(c) Method: I halved the angle between the beams for the angle of incidence: 64° ÷ 2 = 32°.
(c) Method: I took the angle of incidence from 90°: 90° − 32°.
(c) Answer: I wrote 58°.
(c) I did not give 32°: that is the angle to the normal.

Also accept

(a) between the reflected ray and the normal (b) i = r (c) 58°

Watch for

“The angle of reflection is the angle between the reflected ray and the mirror.” — Both angles in the law are measured from the normal, not from the surface of the mirror.
“The incident beam is at 32° to the surface.” — 32° is the angle of incidence, measured from the normal. The surface is at 90° to the normal: 90° − 32° = 58°.
“The incident beam is at 26° to the surface, 90° − 64°.” — The 64° is between the two beams, i + r, not one beam’s angle to the normal. Halve it first.

2. AR2 · Return

Model answer. (a) The image is not on the surface: it is behind the mirror, as far behind it as the pupil is in front. So the image is twice as far from the pupil as the mirror is. (b) The incident ray goes 0.7 cm across while it travels 2.5 cm towards the mirror. By the law of reflection the reflected ray makes the same angle with the normal, so it goes 0.7 cm across for every 2.5 cm away from the mirror: in 5.0 cm, 1.4 cm. The eye is 0.7 cm + 1.4 cm = 2.1 cm to the right of the normal through O. Check with the image: I is 2.5 cm behind the mirror, and the line from I through the point on the mirror reaches the eye after 2.5 cm + 5.0 cm = 7.5 cm, three times as far, so 3 × 0.7 cm = 2.1 cm across.

A full answer includes

(a) I wrote that the image is behind the mirror, not on its surface.
(a) I wrote that it is as far behind the mirror as the pupil is in front.
(a) I wrote that the image is twice as far from the pupil as the mirror is.
(b) Method: I used i = r, so the reflected ray goes as far across for each centimetre as the incident ray: 0.7 cm for every 2.5 cm.
(b) Method: I found that in 5.0 cm away from the mirror, the reflected ray goes 1.4 cm further across.
(b) Method: I added 0.7 cm + 1.4 cm.
(b) Answer: I wrote 2.1 cm.
(b) I did not give 1.4 cm: that is measured from the point where the ray hits the mirror.

Also accept

(a) the image is behind the mirror, twice as far from the pupil as the mirror (b) 2.1 cm

Watch for

“The image is on the surface of the mirror.” — The image is behind the mirror, as far behind as the object is in front.
“The eye is 1.4 cm to the right of the normal through O.” — 1.4 cm is how far the reflected ray goes across after the mirror. Add the 0.7 cm the incident ray went across first.