Answers · Reflection in mirrors

Images in a plane mirror — answer key

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

1. AQ13 · Practice

Model answer. With no parallax, the search pin stands where the image is, so the second column gives the image’s distance behind the mirror. In each row it is within 0.1 cm of the object pin’s distance in front, and the small differences go both ways, as measuring errors do. So the results support the conclusion that the image is as far behind the mirror as the object is in front. The image is also the same size as the object and virtual, but this table measures only distances, so it cannot show those.

A full answer includes

I chose: as far behind the mirror as the object is in front.
I saw that in each row the two distances differ by 0.1 cm or less.
I did not choose a true statement that these results do not measure, such as the size of the image.

Also accept

as far behind the mirror as the object is in front

Watch for

“The image is the same size as the object, at every distance from the mirror.” — That is true, but these results measure only distances, not sizes, so they cannot support it.
“The image is virtual, so it cannot be formed on a screen behind the mirror.” — That is true, but nothing in this table tests a screen. The results only compare two distances.
“The image is always a little nearer the mirror than the object is.” — The search pin is nearer in one row and further in another, by only 0.1 cm: that is measuring error. Within it, the two distances are equal.

2. AQ14 · Practice

Model answer. (a) The image is as far behind the mirror as the vase is in front: 45 cm. (b) The vase is 45 cm in front of the mirror and the image 45 cm behind it, so they are 45 cm + 45 cm = 90 cm apart.

Checked automatically

(a) Distance of the image behind the mirror: 45 cm (within ±0.5 cm)
- Common wrong answer 22.5 : The image is as far behind the mirror as the vase is in front, not half as far.
- Common wrong answer 90 : That is the distance from the vase to its image. Part (a) asks for the distance behind the mirror.
(b) Distance from the vase to its image: 90 cm (within ±0.5 cm)
- Common wrong answer 45 : That is the distance from the mirror to the image; add the 45 cm from the vase to the mirror.

A full answer includes

(a) I wrote 45 cm.
(a) I did not halve the distance.
(b) Method: I added the distance in front and the distance behind: 45 cm + 45 cm.
(b) Answer: I wrote 90 cm.

Also accept

(a) 45 cm (b) 90 cm

Watch for

“The image is 22.5 cm behind the mirror, halfway.” — The image is as far behind the mirror as the object is in front, not half as far.
“The image is 45 cm from the vase.” — That is its distance from the mirror. Add the 45 cm from the vase to the mirror.

3. AQ15 · Practice

Model answer. The reflected rays only seem to come from the image behind the mirror; no light actually goes there. An image that light only seems to come from is virtual, and it cannot be formed on a screen.

A full answer includes

I chose: light only seems to come from it.
I linked the word to where the light goes, not to how bright or how big the image looks.
I did not choose “fainter than the object”.

Also accept

light only seems to come from it

Watch for

“It is fainter than the object, because the mirror absorbs some light.” — Virtual means that no light actually reaches the image; it says nothing about brightness.
“It lies on the surface of the mirror, not behind it.” — The image is behind the mirror. It is virtual because light only seems to come from there.
“It looks smaller than the object, because it is further away.” — The image is the same size as the object. Virtual is about where the light goes, not about size.

4. AQ16 · Practice

Model answer. Her image is always as far behind the mirror as she is in front, so the distance to her image is twice her distance to the mirror. At first she was 5.6 m ÷ 2 = 2.8 m from the mirror; at the end, 3.4 m ÷ 2 = 1.7 m. She walked 2.8 m − 1.7 m = 1.1 m. Her image came 1.1 m towards the mirror too, which is why the gap shrank by 2.2 m.

Checked automatically

Distance Leila walked: 1.1 m (within ±0.05 m)
- Common wrong answer 2.2 : That is how much the gap to her image shrank. Her image moves too, so she walked half of that.
- Common wrong answer 1.7 : That is her last distance from the mirror. Take it from her first distance from the mirror.
- Common wrong answer 2.8 : That is her first distance from the mirror. Take away her last distance from the mirror.

A full answer includes

Method: I halved 5.6 m to get her first distance from the mirror, 2.8 m.
Method: I halved 3.4 m to get her last distance from the mirror, 1.7 m.
Method: I took one from the other: 2.8 m − 1.7 m.
Answer: I wrote 1.1 m.
I did not give 2.2 m: that is how much the gap to her image shrank.

Also accept

1.1 m

Watch for

“She walked 2.2 m, the change in the distance to her image.” — Her image moves towards the mirror as she does, so the gap shrinks twice as fast as she walks: she walked half of 2.2 m.
“She walked 1.7 m.” — That is her distance from the mirror at the end. Find her distance at the start too.

5. AQ17 · Practice

Model answer. (a) Light spreads out from the lamp in all directions, and the part that hits the mirror is reflected: each ray leaves with its angle of reflection equal to its angle of incidence. The reflected rays spread out from the mirror. (b) Traced back, the reflected rays all meet at one point behind the mirror, as far behind it as the lamp is in front. Your eye takes the light to have come in straight lines, so it sees the lamp at that point: the image.

A full answer includes

(a) I wrote that the light reflects off the mirror.
(a) I wrote that each ray’s angle of reflection equals its angle of incidence.
(a) I wrote that the reflected rays spread out from the mirror.
(b) I wrote that the reflected rays seem to come from one point behind the mirror.
(b) I wrote that the eye takes the rays to have come in straight lines.
(b) I did not write that light goes through the mirror.

Also accept

(a) reflected, i = r for each ray (b) rays seem to come from a point behind the mirror

Watch for

“Light goes through the mirror to the image behind it.” — No light goes behind the mirror; the rays only seem to come from there.

6. AQ18 · Practice

Model answer. The paper stays blank. The image of the flame is virtual: the reflected rays only seem to come from behind the mirror, and no light from the candle actually goes there, so none falls on the paper, in any light.

A full answer includes

I chose: nothing, the paper stays blank.
I explained that the image is virtual.
I explained that no light from the candle reaches the paper behind the mirror.
I did not say that a picture forms there in a dark room.

Also accept

nothing; the image is virtual

Watch for

“A sharp picture of the flame, the same size, because that is where the image is.” — Only a real image, which light actually reaches, can be caught on paper. A plane mirror’s image is virtual.
“A dim picture, from light that leaks through the mirror.” — The silvered mirror reflects the light; none of it reaches the paper behind.
“A picture of the flame, but only in a dark room.” — Darkness helps you see a faint real image, but here no light reaches the paper at all: the image is virtual.

7. AQ19 · Practice

Model answer. O is 3 squares in front of the mirror. Its image is on the line through O at 90° to the mirror, 3 squares behind it: that is Q. P is on the mirror, R is twice as far behind as O is in front, and S is off to one side.

A full answer includes

I chose Q.
I checked that Q is on the line through O at 90° to the mirror.
I checked that Q is 3 squares behind the mirror, as O is 3 squares in front.
I did not choose R, which is 6 squares behind the mirror.

Also accept

Watch for

“P, because the image is on the mirror.” — The image is behind the mirror, as far behind as the object is in front.
“R, because the image is 6 squares from O.” — Q is the point 6 squares from O: 3 in front of the mirror and 3 behind. R is 6 squares behind the mirror.
“S, because the image is moved to one side.” — The image is on the line through O at 90° to the mirror, directly opposite O.

8. AQ20 · Practice

Options as numbered on the pupil sheet: 1 — At each point where a ray meets the mirror, draw the normal.; 2 — Extend the two reflected rays back behind the mirror with dashed lines.; 3 — Draw two rays from O to the mirror.; 4 — Mark the image where the dashed lines cross.; 5 — Draw each reflected ray with its angle of reflection equal to its angle of incidence..

Model answer. Draw two rays from O to the mirror. At each point where a ray meets the mirror, draw the normal, then draw the reflected ray with its angle of reflection equal to its angle of incidence. Extend both reflected rays back behind the mirror with dashed lines: the image is where they cross.

A full answer includes

First: I drew two rays from O to the mirror.
Second: I drew the normal where each ray meets the mirror.
Third: I drew each reflected ray with r = i.
Fourth: I extended the reflected rays back behind the mirror, dashed.
Last: I marked the image where the dashed lines cross.
I did not mark the image where the rays meet the mirror.

Also accept

rays; normals; reflected rays; extend back; mark the crossing

Watch for

“The image is where the rays meet the mirror.” — The rays meet the mirror at two different points. The image is where the reflected rays, extended back, cross behind the mirror.

9. AQ21 · Practice

Model answer. The image in a plane mirror is the same size as the object, so it is 14 cm tall. It is as far behind the mirror as the candle is in front, 36 cm. It is virtual, because light only seems to come from it. Row C.

A full answer includes

I chose row C.
I checked the height: the same as the candle, 14 cm.
I checked the position: 36 cm behind the mirror.
I checked the kind: virtual.
I did not choose row B: 72 cm is the distance from the candle to its image.

Also accept

Row C

Watch for

“Row A.” — The image only looks smaller because it is further from your eye than the mirror is. It is the same size as the candle: 14 cm.
“Row B.” — 72 cm is the distance from the candle to its image. The image is 36 cm behind the mirror.
“Row D.” — No light goes behind the mirror, so the image cannot be formed on a screen: it is virtual, not real.

10. AQ22 · Practice

Model answer. The image of the light is 1.1 m behind the mirror, at the same height as the light. The ray reaches the camera along the straight line from the image to the camera, which covers 1.1 m + 2.4 m = 3.5 m across and drops 0.90 m. It reaches the mirror after 1.1 m of the 3.5 m, so it has dropped 0.90 m × 1.1 ÷ 3.5 = 0.28 m (to 2 significant figures). Check: the incident ray drops 0.28 m in 1.1 m across, and the reflected ray drops the other 0.62 m in 2.4 m; both drop about 0.26 m for each metre across, so they make the same angle with the normal.

Checked automatically

Distance below the height of the light: 0.28 m (2 significant figures)
- Common wrong answer 0.45 : Halfway works only when the light and the camera are equally far from the mirror. Join the camera to the image of the light, 1.1 m behind the mirror.
- Common wrong answer 0.62 : That is how far the point is above the camera. Measure down from the height of the light.
- Common wrong answer 0.41 : Divide by the whole distance across from the image to the camera, 1.1 m + 2.4 m, not by 2.4 m.

A full answer includes

Method: I placed the image of the light 1.1 m behind the mirror, at the same height as the light.
Method: I used the straight line from the image to the camera, which covers 1.1 m + 2.4 m = 3.5 m across.
Method: I saw that it reaches the mirror after 1.1 m of the 3.5 m.
Method: I multiplied: 0.90 m × 1.1 ÷ 3.5.
Answer: I wrote 0.28 m.
I did not take the halfway point, 0.45 m.

Also accept

0.28 m

Watch for

“The ray hits the mirror halfway down, 0.45 m below the light.” — Halfway is right only when the light and the camera are the same distance from the mirror. Join the camera to the image of the light instead.
“The ray hits the mirror 0.62 m below the light.” — 0.62 m is how far that point is above the camera. Measure down from the height of the light.

11. AQ23 · Practice

Model answer. (a) The horizontal ray is at 42° to the surface of the mirror, so i = 90° − 42° = 48°. (b) r = i = 48°, so the angle between the rays is 2 × 48° = 96°. (c) The reflected ray leaves at 42° to the mirror, on the other side of the normal, and the mirror is at 42° to the horizontal, so the reflected ray is at 42° + 42° = 84° to the horizontal. It is 6° from the vertical, so it slants across the tube and hits the side instead of reaching the bottom mirror straight. With the mirror at 45°, it would be 45° + 45° = 90°: straight down.

Checked automatically

(a) Angle of incidence (°): 48 (within ±0.5)
- Common wrong answer 42 : That is the angle to the surface; measure from the normal: 90° − 42°.
- Common wrong answer 45 : That is for a mirror at 45°; this one is at 42°.
(b) Angle between the rays (°): 96 (within ±0.5)
- Common wrong answer 84 : 84° is 180° − 96°, the angle the ray turns through. The angle between the rays is the two angles either side of the normal: 48° + 48°.
- Common wrong answer 90 : That is for a mirror at 45°; here i = 48°.

A full answer includes

(a) Method: I took the angle to the surface from 90°: 90° − 42°.
(a) Answer: I wrote 48°.
(a) I did not give 42°: that is the angle to the surface.
(b) Method: I added i and r: 48° + 48°.
(b) Answer: I wrote 96°.
(c) I wrote no.
(c) I found that the reflected ray is at 42° + 42° = 84° to the horizontal.
(c) I wrote that it is 6° from the vertical.
(c) I wrote that it slants towards the side of the tube.

Also accept

(a) 48° (b) 96° (c) no: 84° to the horizontal, 6° off vertical

Watch for

“The angle of incidence is 42°.” — The 42° is the angle to the surface. The angle of incidence is measured from the normal: 90° − 42°.
“Any mirror inside the tube sends the light straight down.” — Only a mirror at 45° turns a horizontal ray through exactly 90°.

12. AQ24 · Practice

Model answer. The ray to the left goes 2.8 cm across while it goes 5.5 cm towards the mirror, so on an exact drawing its angle of incidence is about 27°, and the reflected ray leaves at 27° on the other side of the normal. The ray to the right has an angle of incidence of about 16°. Extended back, the two reflected rays cross 5.5 cm behind the mirror, on the normal through O: the image is as far behind the mirror as the object is in front. A careful drawing comes within about 2° and 0.3 cm of these values.

Checked automatically

(a) Angle of incidence, measured (°): 27 (within ±2)
- Common wrong answer 63 : That is the angle to the mirror; measure from the normal.
(b) Distance behind the mirror where the lines cross: 5.5 cm (within ±0.3 cm)
- Common wrong answer 0 : Extend the reflected rays further back, behind the mirror.
- Common wrong answer 11.0 : That is the distance from O to the crossing point; measure from the mirror line.

A full answer includes

(a) I measured from the normal, not from the mirror.
(a) I wrote about 27°, between 25° and 29°.
(a) I did not give about 63°: that is the angle to the mirror.
(b) I measured along the normal through O, from the mirror line to the crossing.
(b) I wrote about 5.5 cm, between 5.2 cm and 5.8 cm.

Also accept

(a) 27° (b) 5.5 cm

Watch for

“The angle of incidence is 63°.” — That is the angle between the ray and the mirror. The angle of incidence is measured from the normal: 90° − 63° = 27°.
“The dashed lines cross on the mirror.” — Extend them further: they cross behind the mirror, as far behind it as O is in front.

13. AT2 · Apply

Model answer. (a) The chart is 0.6 m further from the mirror than the patient: 2.7 m + 0.6 m = 3.3 m. (b) The image of the chart is 3.3 m behind the mirror, and the patient is 2.7 m in front of it, so the image is 2.7 m + 3.3 m = 6.0 m from the patient: a small room gives the 6 m reading distance an eye test needs. (c) Mark the image of the letter on the line through it at 90° to the mirror, 3.3 m (to scale) behind the mirror. Draw a straight line from the eye to that image; where it crosses the mirror is where the ray reflects. Draw the ray from the letter to that point and from there to the eye, with arrows, and the part behind the mirror dashed.

A full answer includes

(a) Method: I added the patient’s distance and the extra 0.6 m: 2.7 m + 0.6 m.
(a) Answer: I wrote 3.3 m.
(b) Method: I placed the image 3.3 m behind the mirror.
(b) Method: I added the patient’s 2.7 m in front of the mirror: 2.7 m + 3.3 m.
(b) Answer: I wrote 6.0 m.
(b) I did not give 3.3 m: that is the image’s distance behind the mirror.
(c) I marked the image of the letter on the normal through the letter, as far behind the mirror as the letter is in front.
(c) I joined the eye to that image with a straight line.
(c) I took the point where that line crosses the mirror as the point where the ray reflects.
(c) I drew the ray from the letter to that point, then on to the eye.
(c) I drew the part behind the mirror dashed, because no light goes there.

Also accept

(a) 3.3 m (b) 6.0 m (c) join the eye to the image; the ray reflects where that line crosses the mirror

Watch for

“The image of the chart is 3.3 m from the patient.” — That is its distance behind the mirror; add the 2.7 m from the patient to the mirror.
“The ray reflects at the point of the mirror straight in front of the letter.” — The ray reaches the eye as if it came from the image, so it reflects where the line from the eye to the image crosses the mirror.