Answers · Reflection in mirrors

The law of reflection — answer key

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

1. AQ1 · Practice

Options as numbered on the pupil sheet: 1 — The incident ray; 2 — The reflected ray; 3 — The normal; 4 — The mirror.

Model answer. Line 1 is the plane mirror: its back is hatched. Line 2 is the normal: it is dashed, because it is a construction line, and it is at 90° to the mirror where the ray hits it. Line 3 is the incident ray, with its arrow pointing towards the mirror, and line 4 is the reflected ray, with its arrow pointing away from it.

A full answer includes

I matched line 1 to the mirror.
I matched line 2 to the normal.
I matched line 3 to the incident ray.
I matched line 4 to the reflected ray.
I did not match the dashed line to a ray: a dashed line is a construction line.

Also accept

1 mirror; 2 normal; 3 incident ray; 4 reflected ray

Watch for

“Line 1 is the normal, because it is at 90° to line 2.” — The normal is drawn at 90° to the mirror, and it is dashed. Line 1, with its back hatched, is the mirror itself.
“Line 4 is the incident ray, because it is on the right.” — Follow the arrows: the incident ray points towards the mirror and the reflected ray away from it.

2. AQ2 · Practice

Model answer. The angle of incidence is the angle between the incident ray and the normal. The incident ray is line 3, with its arrow pointing towards the mirror, and the normal is the dashed line 2, so the angle of incidence is x. Angle y, between the normal and the reflected ray, is the angle of reflection; w and z are angles to the surface of the mirror.

A full answer includes

I chose x.
I used the angle between the incident ray (line 3) and the normal (line 2).
I did not choose w: it is measured from the surface of the mirror.

Also accept

Watch for

“w, because it is the angle between the incoming ray and the mirror.” — The angles of incidence and reflection are measured from the normal, not from the surface of the mirror.
“y, because it is the other angle next to the normal.” — y is between the normal and the reflected ray: it is the angle of reflection.
“z, because it is the angle at the mirror on the side the light goes to.” — z is between the reflected ray and the surface of the mirror. The angle of incidence belongs to the incident ray and is measured from the normal.

3. AQ3 · Practice

Model answer. The normal is a line drawn at right angles (90°) to the surface of the mirror, at the point where the ray hits it. It is a construction line, drawn dashed, and both the angle of incidence and the angle of reflection are measured from it.

A full answer includes

I chose the line at 90° to the mirror’s surface.
I checked that it is drawn at the point where the ray hits the mirror.
I did not choose the line at 45°: halfway between is not the normal.

Also accept

a line at 90° to the mirror where the ray hits it
a line perpendicular to the mirror where the ray hits it

Watch for

“The normal is a line along the mirror’s surface.” — The normal is at 90° to the mirror, not along it.
“The normal is a line at 45° to the mirror’s surface.” — The normal makes a right angle with the mirror: 90°, not 45°.
“The normal is the path of the ray towards the mirror.” — That is the incident ray. The normal is a construction line at 90° to the mirror; it is not a ray at all.

4. AQ4 · Practice

Model answer. The incident ray crosses the scale at 124° and the reflected ray at 57°. The straight edge of the protractor lies along the mirror, so its 90° mark is on the normal, and each angle from the normal is the difference between a scale reading and 90°. Angle of incidence: 124° − 90° = 34°. Angle of reflection: 90° − 57° = 33°. Row C. The two angles differ by 1° because a ray is marked and a protractor is read only to about 1°, not because the law fails.

A full answer includes

I chose row C.
I read the incident ray’s crossing as 124° and the reflected ray’s as 57°.
I took 90° from the incident ray’s reading: 124° − 90° = 34°.
I took the reflected ray’s reading from 90°: 90° − 57° = 33°.
I did not choose row A: those are scale readings, not angles from the normal.

Also accept

Row C

Watch for

“Row A.” — These are the scale readings. Each angle is measured from the normal, which is on the 90° mark, so take the difference between each reading and 90°.
“Row B.” — These are the angles between each ray and the mirror, which lies along the straight edge of the protractor. Measure from the 90° mark instead.
“Row D.” — Only the incident ray’s reading was turned into an angle from the normal. The reflected ray’s 57° must also be taken from 90°.

5. AQ5 · Practice

Model answer. The law of reflection: the angle of incidence is equal to the angle of reflection, i = r. Both angles are measured from the normal, the line at 90° to the mirror where the ray hits it.

A full answer includes

Gaps (1) and (3): I wrote incidence and reflection, in either order.
Gap (2): I wrote equal.
Gap (4): I wrote normal.
I did not write mirror or surface in gap (4).

Also accept

incidence; equal; reflection; normal
reflection; equal; incidence; normal

Watch for

“The angle of incidence and the angle of reflection add up to 90°.” — They are equal. It is an angle to the normal and the angle to the surface that add up to 90°.
“Both angles are measured from the mirror.” — Both are measured from the normal.

6. AQ6 · Practice

Model answer. By the law of reflection, the angle of incidence equals the angle of reflection. The reflected ray is at 28° to the normal, so the angle of reflection is 28°, and the angle of incidence is 28° too. The incident ray is 90° − 28° = 62° from the surface of the glass, but the angle of incidence is measured from the normal.

Checked automatically

Angle of incidence (°): 28 (within ±0.5)
- Common wrong answer 62 : That is the angle to the glass. The angle of incidence is measured from the normal.
- Common wrong answer 56 : That is the angle between the two rays, i + r. The angle of incidence alone equals r.

A full answer includes

I used the law of reflection: i = r.
I wrote 28°.
I did not give 62°: that is the angle to the glass.

Also accept

28°

Watch for

“The angle of incidence is 62°.” — That is the angle between the incident ray and the glass. The angle of incidence is measured from the normal: 28°.
“The angle of incidence is 56°.” — That is the angle between the incident ray and the reflected ray, i + r.

7. AQ7 · Practice

Model answer. A ray at 90° to the surface of the mirror travels along the normal, so its angle of incidence is 0°. By the law of reflection, the angle of reflection is also 0°, so the reflected ray leaves along the normal: it goes straight back the way it came.

A full answer includes

I chose: it goes back along the normal, the way it came.
I saw that a ray at 90° to the surface travels along the normal.
I took the angle of incidence as 0°.
I used r = i, so r = 0°.
I did not take the angle of incidence to be 90°.

Also accept

back along the normal, the way it came

Watch for

“It travels along the surface of the mirror.” — That would need an angle of incidence of 90°. A ray that hits the mirror straight on, at 90° to the surface, travels along the normal, so i = 0°, and r = 0° too.
“It leaves at 45° to the normal, on the far side.” — The 90° to the surface does not split into two angles of 45°. Along the normal, i = 0°, so r = 0°.
“It spreads out in all directions from the mirror.” — A rough surface scatters light in all directions. A plane mirror is smooth, so one ray reflects as one ray, by r = i: along the normal, straight back.

8. AQ8 · Practice

Model answer. (a) In the row with an angle of incidence of 42°, the angle of reflection is 48°: 6° out, far more than in any other row. It does not fit the pattern, so it is anomalous, most likely a ray marked or read wrongly. The pupil should repeat that reading, or, if it cannot be repeated, leave it out of the conclusion. (b) In the other four rows the two angles differ by 1° or less, and as the angle of incidence goes up from 12° to 72°, the angle of reflection goes up with it by about the same amount. Within the accuracy of the measurements, the angle of reflection equals the angle of incidence: i = r. (c) Differences of 1° come from the measuring, not from the light. The ray is about a millimetre wide, so the crosses marking it are not placed exactly, and a protractor can only be read to about 1°.

A full answer includes

(a) I identified the row with 42° and 48° as the anomalous result.
(a) I wrote that the pupil should repeat that reading, or leave it out of the conclusion.
(a) I did not call a row anomalous when its two angles differ by only 1°.
(b) I wrote that the angle of reflection equals the angle of incidence.
(b) I wrote that in the other rows the two angles differ by 1° or less.
(c) I gave any one of: the ray is about 1 mm wide; the crosses are not placed exactly; a protractor reads only to about 1°.
(c) I did not write that the law of reflection fails or that the mirror is faulty.

Also accept

(a) 42° and 48°: repeat it or leave it out (b) i = r to within 1° (c) a measuring error, such as a wide ray, the crosses, or a protractor read to 1°

Watch for

“The 42° row shows that the law of reflection does not always work.” — One row out of five that does not fit is far more likely to be a mistake in marking or reading the ray. The other four agree with i = r to within 1°, so repeat that reading.
“The angle of reflection is sometimes bigger and sometimes smaller, so there is no rule.” — Apart from the anomalous row, the differences are only 1°, about as small as a protractor can measure, and they go both ways; i = r fits all those readings.
“The mirror must be faulty.” — A plane mirror obeys i = r; differences of a degree come from marking and measuring the rays.

9. AQ9 · Practice

Model answer. (a) By the law of reflection, the ray leaves the first mirror at 33° to its normal, which is 90° − 33° = 57° to the first mirror’s surface. The second mirror is at 90° to the first, so its normal runs parallel to the first mirror’s surface. The ray therefore meets the second mirror at 57° to its normal: the angle of incidence there is 57°. (b) The ray reflects at 57° and leaves parallel to the ray that arrived at the first mirror, travelling back the opposite way, a little to one side of it. Corner mirrors like these send light back towards where it came from.

Checked automatically

(a) Angle of incidence at the second mirror (°): 57 (within ±0.5)
- Common wrong answer 33 : The second mirror is at 90° to the first, so its normal is not parallel to the first one. Find the ray’s angle to the first mirror’s surface.
- Common wrong answer 66 : That is i + r at the first mirror. At the second mirror, measure from its own normal.

A full answer includes

(a) Method: I found that the ray leaves the first mirror at 90° − 33° = 57° to its surface.
(a) Method: I used that the second mirror’s normal runs along the first mirror’s surface.
(a) Answer: I wrote 57°.
(a) I did not give 33°: the two mirrors’ normals are at 90° to each other, not parallel.
(b) I wrote that the ray leaves parallel to the ray that arrived.
(b) I wrote that it travels back the opposite way.

Also accept

(a) 57° (b) parallel to the arriving ray, going back the opposite way

Watch for

“The angle of incidence at the second mirror is 33°, the same as at the first.” — That would be true for parallel mirrors. Here the second mirror is at 90° to the first, so its normal lies along the first mirror’s surface: 90° − 33° = 57°.
“The angle of incidence at the second mirror is 66°, i + r at the first.” — i + r is the angle between the two rays at the first mirror. At the second mirror, measure from its own normal.
“The ray leaves the second mirror at 90° to the ray that arrived.” — Each mirror turns the ray. After both, it travels parallel to the arriving ray, back the way it came.

10. AQ10 · Practice

Model answer. By the law of reflection, r = i = 38°. The angle between the incident ray and the reflected ray is made of the two angles either side of the normal: i + r = 38° + 38° = 76°.

Checked automatically

Angle between the rays (°): 76 (within ±0.5)
- Common wrong answer 38 : That is only the angle of reflection; add the angle of incidence.
- Common wrong answer 104 : That is 180° − 76°. The angle between the rays is the two angles either side of the normal.
- Common wrong answer 52 : That is the angle between one ray and the surface. Add the angles either side of the normal.

A full answer includes

Method: I used r = i = 38°.
Method: I added the angles on both sides of the normal: 38° + 38°.
Answer: I wrote 76°.
I did not give 38°: that is only one of the two angles.

Also accept

76°

Watch for

“The angle between the rays is 38°.” — That is only the angle of reflection. The angle between the rays includes both angles either side of the normal.
“The angle between the rays is 104°.” — 104° is 180° − 76°, the angle the ray turns through. The angle between the rays is the two angles either side of the normal: 38° + 38° = 76°.
“The angle between the rays is 52°.” — 52° is the angle between one ray and the surface. Add the two angles either side of the normal.

11. AQ11 · Practice

Model answer. (a) The ray box does not move, but the normal turns with the mirror. The angle of incidence went from 15° to 27°, so the mirror turned through 27° − 15° = 12°. (b) Before the turn, the angle between the rays was 2 × 15° = 30°; after it, 2 × 27° = 54°. The incident ray did not move, so the reflected ray turned through 54° − 30° = 24°: twice the angle the mirror turned through.

Checked automatically

(a) Angle through which the mirror turned (°): 12 (within ±0.5)
- Common wrong answer 24 : That is how far the reflected ray turned. The mirror’s turn is the change in the angle of incidence.
- Common wrong answer 42 : Find the change in the angle of incidence, not the sum of the two angles.
(b) Angle through which the reflected ray turned (°): 24 (within ±0.5)
- Common wrong answer 12 : That is the mirror’s turn. Find the angle between the rays before and after.
- Common wrong answer 54 : That is the angle between the rays after the turn. Take away the angle between them before it.
- Common wrong answer 30 : That is the angle between the rays before the turn. Find how much it changed.

A full answer includes

(a) Method: I saw that the normal turns with the mirror, so the angle of incidence changes by the mirror’s turn: 27° − 15°.
(a) Answer: I wrote 12°.
(b) Method: I found the angle between the rays before the turn: 2 × 15° = 30°.
(b) Method: I found the angle between the rays after the turn: 2 × 27° = 54°.
(b) Method: I took one from the other: 54° − 30°.
(b) Answer: I wrote 24°.
(b) I did not give 12°: the reflected ray does not turn through the same angle as the mirror.

Also accept

(a) 12° (b) 24°

Watch for

“The reflected ray turns through the same angle as the mirror, 12°.” — Turning the mirror changes both i and r by 12°, so the angle between the rays changes by 24°.
“The reflected ray has turned through 54°.” — 54° is the angle between the rays after the turn. Before it they were 30° apart, so the reflected ray turned through 24°.

12. AQ12 · Practice

Options as numbered on the pupil sheet: 1 — Looking into the mirror, stick two more pins exactly in line with the images of the first two.; 2 — Draw a line for the mirror and a normal at 90° to it, and stand the mirror on the line.; 3 — Draw a line to the point where the normal meets the mirror, and stick two pins on it, well apart.; 4 — Compare the angle of reflection with the angle of incidence.; 5 — Remove the mirror and draw a line through the last two pins, back to the mirror line.; 6 — Measure the angles of incidence and reflection from the normal with a protractor..

Model answer. Draw the mirror line and the normal first, and stand the mirror on the line, so that both angles can be measured from the same normal. A line drawn to the point where the normal meets the mirror is the incident ray, and two pins stuck on it, well apart, mark it. Looking into the mirror from the other side of the normal, stick two more pins so that they and the images of the first two pins all look to be in one line: these two pins mark the reflected ray. Remove the mirror, draw the line through the last two pins back to the mirror line, and measure both angles from the normal with a protractor. If the angle of reflection equals the angle of incidence to within about 1°, the result agrees with the law.

A full answer includes

First: I drew the mirror line and the normal, and stood the mirror on the line.
Second: I drew a line to where the normal meets the mirror and stuck two pins on it.
Third: I stuck two more pins in line with the images of the first two.
Fourth: I removed the mirror and drew the line through the last two pins.
Fifth: I measured both angles from the normal.
Last: I compared the two angles.
I did not stick all four pins before the mirror was standing on its line.

Also accept

mirror line, normal and mirror; incident line and two pins; two pins in line with the images; line through them; measure from the normal; compare

Watch for

“Stick all four pins first, then stand the mirror on its line.” — The last two pins are placed by looking at the images of the first two in the mirror, so the mirror must be standing before they go in.
“Line up the last two pins with the first two pins themselves.” — Line them up with the images of the first two pins, seen in the mirror. The light from the first pins reaches your eye along the reflected ray.
“Measure the angles between the pin lines and the mirror.” — The angles of incidence and reflection are measured from the normal.

13. AT1 · Apply

Model answer. (a) The normal is at 90° to the surface of the mirror, and the mirror is on a vertical wall, so the normal is horizontal: it points straight out from the wall into the room. (b) By the law of reflection, the angle of reflection equals the angle of incidence: 47°. (c) The angle between the rays is i + r, and r = i, so 2i = 136° and i = 136° ÷ 2 = 68°.

A full answer includes

(a) I wrote horizontal: at 90° to the wall.
(a) I did not write that the normal is vertical, along the wall.
(b) I used r = i.
(b) I wrote 47°.
(c) Method: I halved the angle between the rays: 136° ÷ 2.
(c) Answer: I wrote 68°.
(c) I did not give 44°: that is 180° − 136°, not an angle to the normal.

Also accept

(a) horizontal, at 90° to the wall (b) 47° (c) 68°

Watch for

“The normal is vertical, because the mirror is on a vertical wall.” — The normal is at 90° to the surface of the mirror, so it is at 90° to the wall: for a vertical wall, it is horizontal.
“The angle of incidence in (c) is 136°.” — The 136° is between the two rays, i + r. Since r = i, halve it.
“The angle of incidence in (c) is 44°.” — 44° is 180° − 136°. The angle between the rays is made of i and r, one on each side of the normal: halve the 136°.