Physics · 3 Waves

3.2.1 Reflection of light

Before you start

Light travels in straight lines, and a ray is drawn as a straight line with an arrow showing which way the light goes.
You see an object when light from it enters your eye.
Measuring and drawing angles with a protractor, including a line at $90 °$ to another line.
Angles on a straight line add up to $180 °$ .

Syllabus objectives covered (4)

Define and use the terms normal, angle of incidence and angle of reflection
Describe the formation of an optical image by a plane mirror and give its characteristics, i.e. same size, same distance from mirror, virtual
State that for reflection, the angle of incidence is equal to the angle of reflection; recall and use this relationship
Use simple constructions, measurements and calculations for reflection by plane mirrors

Part 1 of 7

The normal and the two angles

The ray that arrives at a mirror is the incident ray. The ray that leaves it is the reflected ray. The point where the incident ray meets the mirror is the point of incidence.
Before you can say anything about angles, you need a line to measure them from. That line is the normal: a line drawn at $90 °$ to the surface at the point where the ray hits it. Draw it dashed, so that it cannot be mistaken for a ray.
The angle of incidence, $i$ , is the angle between the incident ray and the normal.
The angle of reflection, $r$ , is the angle between the reflected ray and the normal.
ray diagram — not drawn yet
A horizontal plane mirror with hatching on its back. A dashed vertical normal at the point of incidence. An incident ray from the upper left with an arrow pointing towards the point, and a reflected ray to the upper right with an arrow pointing away. The angles i and r are marked between each ray and the normal; the angle between the incident ray and the mirror surface is also marked and labelled 90° − i.
Both angles are measured from the normal, not from the mirror.
The trap here is the mirror surface. Both angles are measured from the normal, never from the mirror. If you are given the angle between a ray and the mirror, subtract it from $90 °$ first, because the normal and the mirror are at $90 °$ to each other.
$i = 90 ° - (angle between the ray and the mirror)$

Part 2 of 7

The law of reflection

For every reflection: the angle of incidence is equal to the angle of reflection.
$i = r$
When you are asked to state it, write it in words. The equation $i = r$ only means something once you have said what $i$ and $r$ are.
Using it is one step. A ray arrives at $40 °$ to the normal, so it leaves at $40 °$ to the normal, on the other side of the normal.
A ray that travels along the normal has $i = 0 °$ , so $r = 0 °$ too. It reflects straight back along its own path.
Two results follow with no new physics. The incident and reflected rays sit on opposite sides of the normal, so the angle between them is the sum of the two angles.
$angle between the rays = i + r = 2 i$
The reflected ray also makes the same angle with the mirror as the incident ray does, because each of those angles is $90 ° - i$ .

Part 3 of 7

Testing the law with a ray box

You need a ray box, a plane mirror, a sheet of paper, a ruler, a sharp pencil and a protractor. This experiment is where the measurements in this topic come from.
Draw a straight line on the paper. Stand the mirror on it, with its reflecting surface along the line.
Use the protractor to draw a normal at $90 °$ to the line. From the point where the normal meets the mirror, draw lines at $20 °$ , $30 °$ , $40 °$ , $50 °$ and $60 °$ to the normal. These are your angles of incidence.
Shine a narrow ray from the ray box along the first line, so that it hits the mirror exactly where the normal meets it.
Mark the reflected ray with two small crosses, well apart. Remove the mirror and join the crosses back to the point of incidence with a ruler. Two crosses far apart fix the direction of the ray more accurately than two close together.
Measure the angle between this line and the normal: that is $r$ . Repeat for each angle of incidence and record $i$ and $r$ in a table.
You should find $r = i$ every time, to within the accuracy of your drawing. A narrow ray and a darkened room make the ray easier to mark precisely.

Part 4 of 7

How a plane mirror forms an image

You see an object because light from it enters your eye. Light leaves each point on an object in many directions.
Take one point on the object and call it O. Some of the rays from O hit the mirror. Each one reflects with $i = r$ , and the reflected rays spread out (diverge) as they travel away from the mirror.
Some of the reflected rays enter your eye. Your eye and brain take it that light has travelled in a straight line, so they trace each ray straight back.
Traced back behind the mirror, the reflected rays all meet at one point, I. That point is the image of O, and it is where you see O.
ray diagram — not drawn yet
A vertical plane mirror with hatching on its back. A point object O on the left, in front of the mirror. Two rays leave O, reflect off the mirror at different points, and enter an eye on the left. Each reflected ray is extended behind the mirror as a dashed line with no arrow; the dashed lines meet at the image I on the right. A dashed line joins O to I, crossing the mirror at 90°, and the two equal distances from the mirror are marked d.
The reflected rays only appear to come from I. Nothing behind the mirror is a real ray.
No light travels behind the mirror. The lines you draw there are construction lines that show where the rays appear to come from, so draw them dashed and without arrows.
Every point on the object forms its own image point in this way. Together, those image points make up the whole image.

Part 5 of 7

The three characteristics of the image

Same size. The image is the same size as the object.
Same distance from the mirror. The image is as far behind the mirror as the object is in front of it. The line that joins a point on the object to its image point crosses the mirror at $90 °$ .
Virtual. A virtual image is one that light only appears to come from. No light actually passes through it, so it cannot be formed on a screen.
Compare a real image, which forms where rays actually meet. A real image can be caught on a screen placed at that position. A screen placed behind a plane mirror stays dark, because no light reaches it.
You may also read that the image is upright and laterally inverted (left and right appear swapped). Both are true, but the syllabus names three characteristics: same size, same distance from the mirror, virtual. Learn those three as a set.
Do not write 'imaginary' for virtual. You really do see the image. 'Virtual' says how it is formed, not whether it is there.

Part 6 of 7

Constructing a ray diagram

A construction gives you a mirror, an object O and an eye E, and asks you to show how E sees the image of O. Work in this order, with a sharp pencil and a ruler.
Step 1, find the image. From O, draw a line at $90 °$ to the mirror and continue it behind the mirror. Measure the distance from O to the mirror, and mark the image I the same distance behind it.
Step 2, the line of sight. Draw a straight line from I to E. The point where it crosses the mirror is P, the point where the ray reflects. Make the part from I to P dashed.
Step 3, the real ray. Draw a solid line from O to P, and make the part from P to E solid. Put arrows on both, pointing from O towards E. Light travels from the object to the eye, never from the eye to the object.
Step 4, check. Draw the normal at P and measure both angles with a protractor. If $i$ and $r$ differ by more than a degree or two, your image point or your line is out.
Why Step 2 works: any reflected ray that reaches E must appear to come from I, so it lies on the straight line from I to E. You do not have to find the angle by trial; the image fixes it for you.
For an object with size, such as an arrow, find the image of its top and the image of its bottom separately, then join them.

Part 7 of 7

Calculations with plane mirrors

Distances. If the object is a distance $d$ in front of the mirror, the image is $d$ behind it. So the object and its image are $2 d$ apart.
$object-to-image distance = 2 d$
If you walk towards a mirror, your image moves towards the mirror by the same distance. The gap between you and your image therefore shrinks by twice the distance you walked.
Two mirrors. Follow the ray one mirror at a time. At each mirror, draw the normal, find $i$ , and use $r = i$ . Then use angles on a straight line, or angles in a triangle, to find the angle of incidence at the next mirror.
A periscope has two parallel mirrors. The light meets each mirror at $45 °$ to its surface, so $i = 45 °$ and the angle between the incident and reflected rays is $90 °$ : the light turns through a right angle. After both mirrors it travels in its original direction, at a different height.
A turning mirror. Keep the incident ray fixed and turn the mirror through an angle $θ$ . The normal is fixed at $90 °$ to the mirror, so it turns through $θ$ too, and $i$ changes by $θ$ .
The angle between the incident and reflected rays is $2 i$ , so it changes by $2 θ$ . The incident ray has not moved, so all of that change is the reflected ray turning.
$angle turned by the reflected ray = 2 θ$

Worked examples

Example 1. A ray of light strikes a plane mirror. The angle between the ray and the surface of the mirror is $34 °$ . Determine (a) the angle of incidence, (b) the angle of reflection and (c) the angle between the incident ray and the reflected ray.

Show the working

The $34 °$ is measured from the mirror, not from the normal, so it is not the angle of incidence. Convert it first.
The normal is at $90 °$ to the mirror, so the angle between the ray and the normal is what is left of the $90 °$ .
$i = 90 ° - 34 ° = 56 °$
The law of reflection gives the angle of reflection directly.
$r = i = 56 °$
The two rays are on opposite sides of the normal, so the angle between them is the sum.
$i + r = 56 ° + 56 ° = 112 °$
Check another way. The reflected ray also makes $34 °$ with the mirror, and the three angles along the mirror line must make a straight line: $34 ° + 112 ° + 34 ° = 180 °$ .

Answer: (a) $56 °$ (b) $56 °$ (c) $112 °$

Example 2. A girl stands $1.8$ m in front of a vertical plane mirror. (a) Calculate the distance between the girl and her image. (b) She walks $0.5$ m towards the mirror. Calculate the new distance between the girl and her image, and state by how much it has changed.

Show the working

The image is the same distance behind the mirror as she is in front of it, so it is $1.8$ m behind the mirror.
The distance from her to her image is the two distances added.
$1.8 + 1.8 = 3.6 m$
Find her new distance to the mirror first. After walking, she is $1.8 - 0.5 = 1.3$ m from the mirror, so her image is now $1.3$ m behind it.
$1.3 + 1.3 = 2.6 m$
The change is $3.6 - 2.6 = 1.0$ m. That is twice the $0.5$ m she walked, because she moved $0.5$ m closer to the mirror and her image moved $0.5$ m closer to it as well.
Notice what did not change: the size of her image. It is still the same size as she is.

Answer: (a) $3.6$ m (b) $2.6$ m, a decrease of $1.0$ m.

Example 3. On squared paper, a plane mirror is drawn as a straight line. A small object O is $3.0$ cm in front of the mirror, and N is the point on the mirror closest to O. An eye E is $6.0$ cm in front of the mirror, and the point on the mirror closest to E is $8.0$ cm from N. Construct the ray from O that reflects into E, and find the distance from N to the point P where it reflects.

Show the working

Step 1, the image. Draw the line from O to N, at $90 °$ to the mirror, and continue it behind. Mark I on it $3.0$ cm behind the mirror, because the image is as far behind the mirror as the object is in front.
ray diagram — not drawn yet
A vertical plane mirror drawn on a square grid. Object O 3.0 cm in front of the mirror, with N the foot of the perpendicular from O; image I 3.0 cm behind the mirror on the same perpendicular. Eye E 6.0 cm in front of the mirror, with the foot of its perpendicular 8.0 cm along the mirror from N. A dashed line from I to the mirror at P, 2.7 cm from N; solid rays from O to P and from P to E with arrows towards E; a dashed normal at P with equal angles i and r marked.
The line from the image to the eye fixes where the ray reflects.
Step 2, the line of sight. Join I to E with a straight line. It crosses the mirror at P, the point of reflection. Make the part from I to P dashed.
Step 3, the ray. Draw solid lines from O to P and from P to E, with arrows pointing towards E.
To find P by calculation, look at the line from I to E. It goes $3.0 + 6.0 = 9.0$ cm across, at $90 °$ to the mirror, and $8.0$ cm along the mirror. On a straight line, the distance along grows in proportion to the distance across, and the line reaches the mirror after $3.0$ cm of the $9.0$ cm across.
$N P = \frac{3.0}{9.0} \times 8.0 = 2.7 cm$
Check with a protractor at P. The angle of incidence and the angle of reflection both measure about $42 °$ . They are equal, as the law of reflection requires.

Answer: The image I is $3.0$ cm behind the mirror on the line through O at $90 °$ to it. The ray reflects at P, $2.7$ cm along the mirror from N, with $i = r$ , both about $42 °$ .

Example 4. A ray of light strikes a plane mirror with an angle of incidence of $30 °$ . The incident ray stays fixed and the mirror is turned through $12 °$ , so that the angle of incidence increases. Calculate the new angle between the incident and reflected rays, and the angle through which the reflected ray turns.

Show the working

The normal is fixed at $90 °$ to the mirror, so when the mirror turns through $12 °$ , the normal turns through $12 °$ as well.
The question says the angle of incidence increases, so it goes up by $12 °$ .
$i = 30 ° + 12 ° = 42 °$
Before the turn, the angle between the rays was $2 i$ .
$2 \times 30 ° = 60 °$
After the turn it is $2 i$ again, with the new $i$ .
$2 \times 42 ° = 84 °$
The incident ray has not moved, so the whole change belongs to the reflected ray.
$84 ° - 60 ° = 24 °$
That is $2 \times 12 °$ : the reflected ray turns through twice the angle the mirror turns.

Answer: The angle between the rays becomes $84 °$ , and the reflected ray turns through $24 °$ .

Where people go wrong

Mistake: Measuring the angle of incidence from the mirror surface.

Instead: Measure it from the normal. If you are given the angle to the mirror, subtract it from $90 °$ .

Both definitions use the normal. A ray at $20 °$ to the mirror has an angle of incidence of $70 °$ , not $20 °$ .

Mistake: Saying that the image is on the surface of the mirror.

Instead: The image is behind the mirror, as far behind it as the object is in front.

The reflected rays, traced back, meet behind the mirror. That is the point they appear to come from.

Mistake: Thinking that a screen placed behind the mirror would show the image.

Instead: A virtual image cannot be formed on a screen, because no light passes through it.

The light reflects at the mirror. None of it reaches the space behind the mirror, so a screen there stays dark.

Mistake: Drawing the lines behind the mirror as solid rays with arrows, or putting arrows from the eye towards the object.

Instead: Lines behind the mirror are dashed construction lines with no arrows. Arrows on real rays point from the object, via the mirror, to the eye.

Light does not travel behind the mirror, and it travels into the eye, not out of it.

Mistake: Believing that your image gets bigger as you walk towards the mirror.

Instead: The image is always the same size as you. It looks bigger only because it is closer to your eye.

Any object looks larger when it is nearer to you, without changing its size. Your image is no different.

Check yourself

Define the normal.
Answer
A line at $90 °$ to the surface of the mirror at the point where the ray strikes it.
A ray strikes a plane mirror at $20 °$ to its surface. State the angle of reflection.
Answer
$70 °$ . The angle of incidence is $90 ° - 20 ° = 70 °$ , and the angle of reflection is equal to it.
A ray strikes a plane mirror along the normal. Describe what happens to it.
Answer
The angle of incidence is $0 °$ , so the angle of reflection is $0 °$ : the ray reflects straight back along its own path.
State the three characteristics of the image formed by a plane mirror.
Answer
Same size as the object; the same distance behind the mirror as the object is in front; virtual.
Explain why the image in a plane mirror is described as virtual.
Answer
The reflected rays only appear to come from the image. No light actually passes through it, so it cannot be formed on a screen.
A candle is $25$ cm in front of a plane mirror. It is moved $10$ cm further away from the mirror. Calculate the new distance between the candle and its image.
Answer
The candle is now $35$ cm from the mirror and its image is $35$ cm behind it, so they are $70$ cm apart.
The angle between an incident ray and its reflected ray is $130 °$ . Calculate the angle of incidence and the angle between the reflected ray and the mirror.
Answer
The angle between the rays is $2 i$ , so $i = 65 °$ . The reflected ray makes $90 ° - 65 ° = 25 °$ with the mirror.
In a periscope, light meets the first mirror at $45 °$ to its surface. State the angle of incidence, and calculate the angle between the incident and reflected rays.
Answer
The angle of incidence is $90 ° - 45 ° = 45 °$ . The angle between the rays is $2 \times 45 ° = 90 °$ , so the light turns through a right angle.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A ray of light from a ray box strikes a plane mirror at point P. The angle between the ray and the surface of the mirror is $28 °$ .

(a) Define the angle of incidence. 1 mark No calculator

Mark scheme

B1 the angle between the incident ray and the normal (at P); do not credit 'the angle between the ray and the mirror'

(b) Calculate the angle of incidence at P. 1 mark

Mark scheme

B1 62°

(c) State the law of reflection, and use it to give the angle of reflection at P. 2 marks No calculator

Mark scheme

B1 angle of incidence = angle of reflection (in words, or $i = r$ with $i$ and $r$ identified)
B1 62°; ecf from (b)

(d) Calculate the angle between the incident ray and the reflected ray. 2 marks

Mark scheme

C1 $62 ° + 62 °$ , or $2 \times 62 °$ , or $180 ° - 2 \times 28 °$ seen
A1 124°; ecf from (b)

Question 2. A boy stands $2.4$ m in front of a large vertical plane mirror and looks at his image.

(a) State the size of the image, its distance from the mirror, and whether it is real or virtual. 3 marks No calculator

Mark scheme

B1 the same size as the boy
B1 $2.4$ m behind the mirror (accept: the same distance behind the mirror as the boy is in front)
B1 virtual

(b) Explain why the image cannot be formed on a screen placed behind the mirror. 1 mark No calculator

Mark scheme

B1 no light actually passes through the image position / reaches behind the mirror; the rays only appear to come from the image

(c) The boy walks $0.9$ m towards the mirror. Calculate the new distance between the boy and his image. 2 marks

Mark scheme

C1 $2.4 - 0.9 = 1.5$ (m) seen, or $2 \times 1.5$ seen, or $4.8 - 2 \times 0.9$ seen, or any equivalent working
A1 3.0 m; unit required

(d) The boy says: 'My image got bigger as I walked towards the mirror.' Comment on his statement. 2 marks No calculator

Mark scheme

B1 the statement is wrong: the image is (still) the same size as the boy
B1 it only looks bigger because it is now closer to his eyes

Question 3. Make this drawing yourself, full size, on squared paper. A plane mirror is drawn as a straight line. A small object O is $4.0$ cm in front of the mirror, and N is the point on the mirror closest to O. An eye E is $8.0$ cm in front of the mirror, and the point on the mirror closest to E is $9.0$ cm from N.

(a) Draw the image I of O, and a ray from O that reflects off the mirror into E. 3 marks No calculator

Mark scheme

B1 I on the line through O at 90° to the mirror, $4.0$ cm behind the mirror
B1 straight line from I to E, dashed behind the mirror, crossing the mirror at a point P
B1 solid lines from O to P and from P to E, with arrows pointing from O towards E

(b) Determine the distance NP, from N to the point P where the ray reflects. 2 marks

Mark scheme

C1 $\frac{4.0}{12.0} \times 9.0$ seen, or $\frac{N P}{4.0} = \frac{9.0 - N P}{8.0}$ or any equivalent working, or P correctly located on a drawing
A1 3.0 cm (accept 2.9 cm to 3.1 cm from a drawing); unit required

(c) Measure the angle of incidence and the angle of reflection at P. 1 mark No calculator

Mark scheme

B1 both 37° (accept 36° to 38°), and equal to each other

How this is examined

Describing how to show that the angle of incidence equals the angle of reflection: a Describe of an experiment with a ray box, a plane mirror, paper and a protractor Each step is a separate point: draw a normal at $90 °$ to the mirror line; shine a narrow ray at a measured angle of incidence to the point where the normal meets the mirror; mark the reflected ray with two crosses and join them to that point; measure the angle of reflection from the normal; repeat for several angles of incidence and compare $i$ with $r$ .
Completing a ray diagram that shows how an eye sees the image of an object in a plane mirror: a Draw on a diagram of a mirror, an object and an eye: mark the image with a ruler, then show one ray from the object to the eye The image on the line through the object at $90 °$ to the mirror, the same distance behind it. A straight line from the image to the eye, dashed behind the mirror. Solid rays from the object to the mirror and from the mirror to the eye, meeting on the mirror, with arrows pointing towards the eye.
Finding the angles when the angle you are given is measured from the mirror: a Calculate or Determine that gives the angle between a ray and the mirror surface, in words or on a diagram $90 °$ minus the given angle for the angle of incidence; the angle of reflection equal to it; $2 i$ for the angle between the two rays. Each value needs its degree sign.
The three characteristics of a plane-mirror image, and why it is virtual: a State asking for the characteristics of the image, and an Explain asking why it is virtual 'Same size', 'same distance behind the mirror as the object is in front' and 'virtual' are separate points. For why it is virtual: the rays only appear to come from it, no light passes through it, and it cannot be formed on a screen.
The distance between an object and its image, before and after the object moves: a Calculate that gives the object's distance from the mirror, and possibly a distance the object then moves Find the object's distance from the mirror (after any move), then double it. The final value needs its unit.
Defining the normal, the angle of incidence and the angle of reflection: a 1-mark Define, or labelling $i$ and $r$ on a ray diagram Normal: a line at $90 °$ to the mirror at the point where the ray strikes it. Angle of incidence: between the incident ray and the normal. Angle of reflection: between the reflected ray and the normal. The angle between a ray and the mirror is the wrong angle.

Words to know

incident ray: The ray of light that arrives at a surface.
reflected ray: The ray of light that leaves a surface after reflection.
point of incidence: The point where the incident ray meets the surface.
normal: A line at $90 °$ to a surface at the point where a ray strikes it.
angle of incidence: The angle between the incident ray and the normal.
angle of reflection: The angle between the reflected ray and the normal.
law of reflection: The angle of incidence is equal to the angle of reflection.
plane mirror: A flat mirror.
virtual image: An image that light only appears to come from. No light passes through it, so it cannot be formed on a screen.
real image: An image formed where rays of light actually meet. It can be formed on a screen.

All Physics lessons Next: 3.2.2 Refraction of light