Physics · 3 Waves

3.2.2 Refraction of light

Before you start

The normal and the angle of incidence from reflection: angles are measured from the normal, not from the surface (3.2.1).
Waves refract because their speed changes when they pass into a different region, as seen with water waves in a ripple tank (3.1).
The wave equation $v = f λ$ (3.1).
Light travels at $3.0 \times 10^{8}$ m/s in a vacuum, and at almost the same speed in air.
Using the sin and $\sin^{- 1}$ keys on a calculator set to degrees.

Syllabus objectives covered (9)

Define and use the terms normal, angle of incidence and angle of refraction
Describe an experiment to show refraction of light by transparent blocks of different shapes
Describe the passage of light through a transparent material (limited to the boundaries between two mediums only)
State the meaning of critical angle
Describe internal reflection and total internal reflection using both experimental and everyday examples
Define refractive index, n, as the ratio of the speeds of a wave in two different regions
Recall and use the equation n = sin i / sin r
Recall and use the equation n = 1 / sin c
Describe the use of optical fibres, particularly in telecommunications

Part 1 of 8

The normal, and the two angles you measure from it

Refraction is the change in direction of light as it crosses a boundary between two transparent materials, such as air and glass. The cause is a change in speed, which the next section explains.
The normal is a line drawn at right angles ( $90 °$ ) to the boundary, at the point where the ray meets it. Draw it as a dashed line, so that nobody mistakes it for a ray.
The angle of incidence, $i$ , is the angle between the incoming (incident) ray and the normal.
The angle of refraction, $r$ , is the angle between the refracted ray and the normal, on the other side of the boundary.
ray diagram — not drawn yet
A horizontal boundary with 'air' written above and 'glass' below. A vertical dashed normal crosses the boundary at one point. An incident ray comes down from the upper left to that point, and a refracted ray continues into the glass on the lower right, closer to the normal than the incident ray. The angle between the incident ray and the normal is labelled i; the angle between the refracted ray and the normal is labelled r. Arrows on both rays point away from the light source.
Both angles are measured from the dashed normal, never from the boundary itself.
The trap is to put the protractor on the surface. If a question says a ray makes $35 °$ with the surface, the angle of incidence is $90 ° - 35 ° = 55 °$ . Check every angle you are given: is it from the normal or from the surface?

Part 2 of 8

Which way the ray bends, and why

Light travels more slowly in glass, water and Perspex than it does in air. Each boundary you meet in this topic is between a material where light is faster and one where it is slower.
Into a slower material (air into glass): the ray bends towards the normal, so $r$ is smaller than $i$ .
Into a faster material (glass into air): the ray bends away from the normal, so the angle in the air is the larger angle.
Along the normal ( $i = 0 °$ ): the ray goes straight through with no change in direction. Its speed still changes. A change of direction only happens when the ray meets the boundary at an angle.
Why a change of speed gives a bend: picture a wavefront arriving at the glass at an angle. One end of the wavefront reaches the glass first and slows down, while the other end is still in air and moving faster. In the same time, the end still in air covers more distance than the end in the glass, so the wavefront swings round and the ray turns towards the normal.
The frequency of the light does not change at the boundary. Since $v = f λ$ , a lower speed with the same frequency means a shorter wavelength in the glass.
Paths of light are reversible. If a ray goes from air into glass along one path, a ray sent backwards along the refracted path comes out along the original incident path. That is why one rule ('slower: towards; faster: away') covers both directions.

Part 3 of 8

Showing refraction with glass blocks

The standard method uses a rectangular glass (or Perspex) block on a sheet of white paper. Draw round the block with a sharp pencil so you can put it back in the same place.
Use a ray box with a single slit to send a narrow ray at one long face, at an angle to it. Mark the path of the incident ray with two crosses, well apart, and do the same for the ray that comes out of the opposite face (the emergent ray).
Remove the block. Rule lines through the crosses to the outline. Join the point where the ray entered to the point where it left: that line is the path of the ray inside the glass.
Draw the normal at the entry point with a protractor, then measure $i$ and $r$ from it. Repeat for several angles of incidence. Crosses far apart give the direction of each ray more accurately than crosses close together.
ray diagram — not drawn yet
Top view of a rectangular block outline drawn on paper. A ray from a ray box on the left strikes the top long face at an angle, bends towards the dashed normal inside the block, then leaves the bottom face bending away from a second dashed normal. The emergent ray is drawn parallel to the incident ray but shifted sideways, with a small double-headed arrow marking the sideways shift. Two crosses are marked on the incident ray and two on the emergent ray.
The ray bends towards the normal going in and away from it coming out. Because the two faces are parallel, the emergent ray is parallel to the incident ray, only shifted sideways.
What you see with the rectangular block: the ray bends towards the normal as it enters, away from the normal as it leaves, and comes out parallel to the incident ray but displaced sideways.
A semicircular block is used to study the flat face on its own. Aim the ray through the curved side at the centre of the flat face. A ray aimed at the centre travels along a radius, and a radius is the normal to a curved surface, so the ray enters with no change in direction. All the bending you see happens at the flat face.
A triangular prism shows two bends at faces that are not parallel. The ray bends towards the normal as it enters and away from the normal as it leaves, and both bends turn it towards the thicker base of the prism, so the emergent ray is not parallel to the incident ray.

Part 4 of 8

Refractive index: a ratio of two speeds

The refractive index, $n$ , compares the speed of a wave in two regions. For light going into a material from air (or a vacuum):
$n = \frac{speed of light in air (or vacuum)}{speed of light in the material}$
The definition is a ratio of speeds, and the order matters: the faster speed (in air) goes on top. The same idea works for any wave, for example water waves passing from deep water to shallow water.
Because it is a speed divided by a speed, $n$ has no unit. For a material, it is greater than $1$ , because light is slower in the material than in air. For air itself it is almost exactly $1$ .
Example: glass with $n = 1.5$ . The speed of light in it is $\frac{3.0 \times 10^{8}}{1.5} = 2.0 \times 10^{8}$ m/s.
A larger $n$ means a slower speed in the material, and so a larger bend towards the normal for the same angle of incidence.

Part 5 of 8

Using n = sin i / sin r

The angles connect to the refractive index through this equation. It is not given on Paper 4, so you must recall it.
$n = \frac{\sin i}{\sin r}$
Here $i$ is the angle in air and $r$ is the angle in the material. Learn it that way round and you will not go wrong when the light travels the other way.
Set your calculator to degrees before you start. A check: $\sin 30 °$ must give $0.5$ .
To find $r$ , rearrange first, then use $\sin^{- 1}$ . Divide $\sin i$ by $n$ because $n$ is what divides it in the equation, then undo the sine.
$\sin r = \frac{\sin i}{n} r = \sin^{- 1} (\frac{\sin i}{n})$
When the light goes from glass into air, the angle in air is still on top. So the angle in air comes from $\sin (angle in air) = n \times \sin (angle in glass)$ .
Sense check after every answer: the angle in the material must be smaller than the angle in air. If it is not, you have the equation upside down.

Part 6 of 8

Internal reflection, the critical angle and total internal reflection

Go back to the semicircular block, with the ray passing through the curved side to the centre of the flat face. Now the light is inside the glass, travelling towards a boundary with air, where it is faster.
Small angle of incidence: most of the light refracts out into the air, bending away from the normal. A weak ray is also reflected back into the glass. Reflection from the inside of a boundary like this is called internal reflection.
Increase the angle of incidence: the refracted ray bends further from the normal and gets closer to the flat face. The internally reflected ray gets brighter.
At one particular angle, the refracted ray travels along the flat face: its angle of refraction is $90 °$ . That angle of incidence is the critical angle, $c$ .
The definition to write: the critical angle is the angle of incidence, for light going from a material into one where it travels faster (glass to air), at which the angle of refraction is $90 °$ .
Above the critical angle, no light leaves the glass. All of it reflects back inside, obeying the law of reflection (angle of incidence = angle of reflection). This is total internal reflection (TIR).
ray diagram — not drawn yet
Three semicircular blocks side by side, flat face at the top, each with a ray entering the curved side and aimed at the centre of the flat face, with a dashed normal at that point. Left, labelled 'i less than c': a refracted ray leaves into the air bending away from the normal, plus a faint reflected ray inside the glass. Middle, labelled 'i = c': the refracted ray runs along the flat face, angle of refraction marked 90°, plus a reflected ray. Right, labelled 'i greater than c': no ray in the air, one bright reflected ray inside the glass with equal angles marked either side of the normal.
Below c: refraction plus weak internal reflection. At c: the refracted ray grazes the surface. Above c: total internal reflection.
Total internal reflection needs both conditions: the light is travelling towards a material where it is faster (glass to air, water to air), and the angle of incidence is greater than the critical angle. It cannot happen as light goes from air into glass.
Everyday example: a prism periscope uses two right-angled glass prisms with $45 °$ corners. Light enters a short face along the normal and meets the long face at $45 °$ . That is greater than the critical angle of typical glass (about $42 °$ ), so the light is totally internally reflected and turned through $90 °$ . The prism acts as a mirror.
Everyday example: under water in a swimming pool, look up at the surface at a large angle and it looks like a silvery mirror. Light from below meets the water–air boundary at more than the critical angle and is totally internally reflected.
The optical fibre is an example with its own objective, so it has its own section below.

Part 7 of 8

Using n = 1 / sin c

This equation comes straight from $n = \frac{\sin i}{\sin r}$ , so you do not have to learn it as a separate fact. At the critical angle the angle in air is $90 °$ and the angle in the glass is $c$ .
Put those into the equation, with the air angle on top as always. Since $\sin 90 ° = 1$ :
$n = \frac{\sin 90 °}{\sin c} = \frac{1}{\sin c}$
You must still recall it on Paper 4. Rearranged for the critical angle:
$\sin c = \frac{1}{n} c = \sin^{- 1} (\frac{1}{n})$
Examples: glass with $n = 1.5$ gives $c = \sin^{- 1} (0.667) = 41.8 °$ . Water with $n = 1.33$ gives $c = 48.8 °$ . A larger refractive index gives a smaller critical angle.
The equation also shows why TIR happens. For glass with $n = 1.5$ and an angle of $45 °$ inside the glass, the angle in air would need $\sin = 1.5 \times \sin 45 ° = 1.06$ . No angle has a sine greater than $1$ , so there is no refracted ray. All the light reflects.
Use $n = \frac{1}{\sin c}$ for a material meeting air (or a vacuum). That is the boundary the equation describes.

Part 8 of 8

Optical fibres in telecommunications

An optical fibre is a very thin, flexible strand of glass (or plastic). Light goes in at one end and comes out at the other, even if the fibre curves.
How it works: the light meets the side of the fibre at an angle of incidence greater than the critical angle. It is totally internally reflected, crosses the fibre, meets the other side at more than the critical angle again, and reflects again. It travels the whole length by repeated total internal reflection, and no light escapes through the sides.
If the fibre bends too sharply, the angle of incidence at the side can fall below the critical angle, and then light leaks out. Gentle curves keep the angles large enough.
ray diagram — not drawn yet
A gently curved optical fibre drawn in long section, with an inner core and a thin outer cladding layer labelled. A single ray enters the left end and zigzags along the core, reflecting at the core–cladding boundary several times, with the angles of incidence at each reflection marked as larger than c. The ray leaves the right end.
Each reflection is total internal reflection, because every angle of incidence is greater than the critical angle.
A real fibre has a glass core surrounded by a layer of glass called cladding, which has a lower refractive index than the core. Total internal reflection happens at the core–cladding boundary, so the light stays in the core even if the outside of the fibre is scratched or touching something.
In telecommunications, fibres carry telephone calls, internet data and television signals as very fast pulses of light or infrared. One fibre can carry a very large amount of information at the same time.
Advantages of optical fibres over copper cables: they carry much more data; the signal weakens less, so it travels further before it needs boosting; and they are not affected by electrical interference.
Fibres are also used in medicine. An endoscope carries light down a bundle of fibres into the body and carries an image back, so a doctor can see inside without a large cut.

Worked examples

Example 1. Describe an experiment to show how a ray of light is refracted as it passes through a rectangular glass block, and how you would measure the angle of incidence and the angle of refraction.

Show the working

Start with the set-up: place the block on white paper and draw round it, so the outline stays on the paper after the block is moved.
Name the light source: a ray box with a single slit gives one narrow ray. Direct it at one long face at an angle (not along the normal, or nothing bends).
Say how the path is recorded: two crosses on the incident ray and two on the emergent ray, well apart, because a long baseline gives the direction more accurately.
Remove the block, draw the rays through the crosses, and join the entry point to the exit point to show the ray inside the glass.
Now the measurement: draw the normal at the entry point at $90 °$ to the face, then use a protractor to measure $i$ and $r$ from the normal.
Finish with a repeat and the observation: repeat for different angles of incidence. Each time the ray bends towards the normal on entering and away on leaving, and the emergent ray is parallel to the incident ray.

Answer: Draw round the block on paper, shine a single ray from a ray box at one face at an angle, mark the incident and emergent rays with two crosses each, remove the block and join the entry and exit points. Draw the normal at the entry point and measure $i$ and $r$ from it with a protractor. Repeat for several angles of incidence. The ray bends towards the normal entering the glass and away from it leaving, and emerges parallel to the incident ray.

Example 2. A ray of light in air strikes a flat face of a glass block. The ray makes an angle of $50 °$ with the surface of the glass. Inside the glass, the refracted ray makes an angle of $25 °$ with the normal. (a) Calculate the refractive index of the glass. (b) A second ray strikes the same block with an angle of incidence of $60 °$ . Calculate its angle of refraction.

Show the working

The $50 °$ is measured from the surface, not the normal, so convert it first.
$i = 90 ° - 50 ° = 40 °$
The $25 °$ is already from the normal, so $r = 25 °$ . Now use the equation, with the angle in air on top.
$n = \frac{\sin 40 °}{\sin 25 °} = \frac{0.643}{0.423} = 1.52$
No unit: $n$ is a ratio. Check it: a value between $1$ and $2$ is sensible for glass, and $r$ is smaller than $i$ , as it must be going into glass.
For (b), rearrange for $\sin r$ before you substitute.
$\sin r = \frac{\sin 60 °}{1.52} = \frac{0.866}{1.52} = 0.5697$
Undo the sine with $\sin^{- 1}$ .
$r = \sin^{- 1} (0.5697) = 34.7 °$
Check: $34.7 °$ is smaller than $60 °$ , so the ray has bent towards the normal, as expected.

Answer: (a) $n = 1.52$ (no unit). (b) $r = 34.7 °$ .

Example 3. Glass has a refractive index of $1.5$ . (a) A ray inside a glass block meets the glass–air boundary with an angle of incidence of $30 °$ . Calculate the angle between the ray and the normal after it leaves the glass. (b) Explain why a ray inside the glass meeting the boundary at $45 °$ cannot leave the glass, and state what happens to it.

Show the working

The light is going from glass into air, but the angle in air still goes on top of the equation. Call the angle in air $θ$ .
$1.5 = \frac{\sin θ}{\sin 30 °}$
Multiply both sides by $\sin 30 °$ , because that is what divides $\sin θ$ .
$\sin θ = 1.5 \times \sin 30 ° = 1.5 \times 0.5 = 0.75$
Undo the sine.
$θ = \sin^{- 1} (0.75) = 48.6 °$
Check: $48.6 °$ is larger than $30 °$ . Going into a faster material, the ray bends away from the normal, so the angle in air must be the larger one.
For (b), try the same method with $45 °$ .
$\sin θ = 1.5 \times \sin 45 ° = 1.5 \times 0.707 = 1.06$
No angle has a sine greater than $1$ , so there is no possible refracted ray. The light cannot leave the glass.
State what happens: all the light is reflected back into the glass, which is total internal reflection. The angle of reflection equals the angle of incidence, $45 °$ .

Answer: (a) $48.6 °$ to the normal. (b) The calculation needs $\sin θ = 1.06$ , which is impossible, so there is no refracted ray; the light is totally internally reflected at $45 °$ to the normal.

Example 4. A transparent plastic has a refractive index of $1.60$ . The speed of light in air is $3.0 \times 10^{8}$ m/s. (a) Calculate the speed of light in the plastic. (b) Calculate the critical angle for the plastic. (c) A ray inside a block of the plastic meets the plastic–air boundary at an angle of incidence of $30 °$ . State whether it is totally internally reflected, and explain your answer.

Show the working

For (a), start from the definition: $n$ is the speed in air divided by the speed in the plastic. Rearrange for the speed in the plastic.
$v = \frac{3.0 \times 10^{8}}{1.60} = 1.875 \times 10^{8} m/s$
The data are given to two significant figures, so give $1.9 \times 10^{8}$ m/s. Check it: the speed must be less than in air, and it is.
For (b), use $\sin c = \frac{1}{n}$ , then undo the sine.
$\sin c = \frac{1}{1.60} = 0.625 c = \sin^{- 1} (0.625) = 38.7 °$
For (c), compare the angle of incidence with the critical angle. $30 °$ is less than $38.7 °$ .
So the second condition for TIR fails. The ray refracts out into the air, bending away from the normal, and only a weak ray is internally reflected.

Answer: (a) $1.9 \times 10^{8}$ m/s. (b) $c = 38.7 °$ . (c) No: $30 °$ is less than the critical angle of $38.7 °$ , so most of the light refracts out into the air, with only a weak internal reflection.

Where people go wrong

Mistake: Measuring the angle of incidence or refraction from the surface of the block.

Instead: Both angles are measured from the normal. If you are given the angle to the surface, subtract it from $90 °$ .

The equations $n = \sin i / \sin r$ and $n = 1 / \sin c$ only work with angles from the normal, so an angle from the surface gives a wrong refractive index.

Mistake: Putting the angle in the glass on top when light goes from glass into air.

Instead: The angle in air always goes on top: $\sin (angle in air) = n \times \sin (angle in glass)$ .

$n$ is greater than $1$ , so the angle in air is always the larger angle. Swapping them gives a refractive index less than $1$ , which is impossible for glass.

Mistake: Saying a ray that hits a boundary along the normal is refracted with a change of direction.

Instead: Along the normal the ray goes straight through. Its speed changes, but its direction does not.

The bend comes from one side of the wavefront slowing before the other. Along the normal the whole wavefront reaches the boundary at the same moment.

Mistake: Saying total internal reflection can happen as light goes from air into glass, or whenever the angle is large.

Instead: TIR needs light going towards a material where it is faster (glass to air) and an angle of incidence greater than the critical angle.

Going from air into glass, the ray bends towards the normal, so a refracted ray always exists. There is no critical angle in that direction.

Mistake: Defining the critical angle as 'the angle at which light is totally internally reflected'.

Instead: It is the angle of incidence at which the angle of refraction is $90 °$ . TIR happens at angles greater than it, so 'the angle of incidence above which total internal reflection happens' is also correct.

At exactly the critical angle a refracted ray still travels along the boundary. The definition is about the $90 °$ refracted ray.

Check yourself

A ray of light strikes a glass surface, making an angle of $58 °$ with the surface. Where do you draw the normal, and what is the angle of incidence?
Answer
Draw the normal at $90 °$ to the surface, at the point where the ray meets it. The angle of incidence is $90 ° - 58 ° = 32 °$ .
A ray of light passes at an angle from water into air. Which way does it bend, and why?
Answer
Away from the normal, because light travels faster in air than in water.
In the rectangular block experiment, how does the direction of the emergent ray compare with the direction of the incident ray?
Answer
It is parallel to the incident ray, but shifted sideways.
State the meaning of critical angle.
Answer
The angle of incidence, for light going from a material such as glass into air, at which the angle of refraction is $90 °$ .
State the two conditions needed for total internal reflection.
Answer
The light must be travelling towards a material in which it is faster (for example glass to air), and the angle of incidence must be greater than the critical angle.
A ray in air has an angle of incidence of $35 °$ and an angle of refraction of $22 °$ in a liquid. Calculate the refractive index of the liquid.
Answer
$n = \sin 35 ° / \sin 22 ° = 0.574 / 0.375 = 1.53$ (no unit).
A glass has a refractive index of $1.47$ . The speed of light in air is $3.0 \times 10^{8}$ m/s. Calculate the speed of light in the glass and its critical angle.
Answer
Speed $= 3.0 \times 10^{8} / 1.47 = 2.0 \times 10^{8}$ m/s. $\sin c = 1 / 1.47 = 0.6803$ , so $c = 42.9 °$ .
In telecommunications, what does an optical fibre carry, and in what form?
Answer
Information such as telephone calls, internet data and television signals, carried as very fast pulses of light or infrared that travel along the fibre by total internal reflection.

Exam questions

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

Question 1. A student directs a ray of light from a ray box into a semicircular glass block. The ray enters through the curved surface and is aimed at the centre of the flat face. The student can change the angle of incidence at the flat face.

(a) Explain why the ray does not change direction as it enters the block through the curved surface. 1 mark No calculator

Mark scheme

B1 the ray travels along a radius, which is the normal to the curved surface / the angle of incidence there is zero, owtte

(b) Describe what happens to the ray at the flat face when the angle of incidence is smaller than the critical angle. 2 marks No calculator

Mark scheme

B1 the ray refracts into the air, bending away from the normal
B1 a weak (partial) ray is reflected back into the glass

(c) State the meaning of critical angle. 2 marks No calculator

Mark scheme

B1 the angle of incidence (in the glass / the material in which light is slower)
B1 at which the angle of refraction is 90° / the refracted ray travels along the boundary; accept: above which total internal reflection occurs, owtte

(d) The critical angle for this glass is $41 °$ . Calculate the refractive index of the glass. 2 marks

Mark scheme

C1 n = 1 / sin c, or n = 1 / sin 41°
A1 1.52 (accept 1.5); no unit

(e) The student increases the angle of incidence at the flat face to $50 °$ . Predict what happens to the ray at the flat face, and give a reason. 2 marks No calculator

Mark scheme

B1 total internal reflection: all the light is reflected back into the glass and no ray enters the air
B1 because 50° is greater than the critical angle (41°)

Question 2. A ray of light in air strikes the flat surface of a transparent plastic block. The angle of incidence is $55 °$ and the angle of refraction is $33 °$ . The speed of light in air is $3.0 \times 10^{8}$ m/s.

(a) Define refractive index. 1 mark No calculator

Mark scheme

B1 speed of light in air (or vacuum) divided by speed of light in the material, owtte (e.g. ratio of the speed in air to the speed in the material); the ratio the wrong way round: no mark

(b) Calculate the refractive index of the plastic. 2 marks

Mark scheme

C1 n = sin i / sin r, or n = sin 55° / sin 33°
A1 1.50 (accept 1.5); no unit

(c) Calculate the speed of light in the plastic. 2 marks

Mark scheme

C1 speed = 3.0 × 10^8 / n; ecf from (b)
A1 2.0 × 10^8 m/s; unit required

Question 3. Optical fibres carry telephone calls and internet data between cities. Each fibre has a glass core surrounded by cladding.

(a) Explain how light travels along the fibre, even when the fibre is curved. 3 marks No calculator

Mark scheme

B1 light meets the boundary (side) of the core at an angle of incidence greater than the critical angle
B1 total internal reflection occurs; insufficient: 'the light reflects' with no mention of total internal reflection
B1 this happens repeatedly along the length of the fibre, so no light escapes through the sides

(b) State two advantages of optical fibres over copper cables for telecommunications. 2 marks No calculator

Mark scheme

B1 any one of: carry more data / information at the same time; less signal loss, so signals travel further before boosting; not affected by electrical interference; lighter / thinner; more secure (harder to tap); accept any other correct advantage, owtte
B1 a second, different advantage from the list

How this is examined

Calculating a refractive index or an angle with n = sin i / sin r: a two- or three-mark Calculate with the two angles given in words or on a ray diagram; an angle may be given from the surface instead of from the normal Writing $n = \sin i / \sin r$ with the angle in air on top earns the first mark; the correct value earns the final mark, with no unit for $n$ and degrees for an angle. An angle given from the surface must be converted ( $90 ° -$ angle) before it goes in.
Describing internal reflection, the critical angle and total internal reflection in a semicircular block: a Describe or Explain about a ray inside a block, worth two or three marks, which can be split into what happens below, at and above the critical angle Below $c$ : 'the ray refracts out, bending away from the normal, with a weak internally reflected ray'. At $c$ : 'the angle of refraction is $90 °$ ; the refracted ray travels along the surface'. Above $c$ : 'all the light is reflected back into the glass; no refracted ray'.
How light travels along an optical fibre, and why fibres are used for telecommunications: a three-mark Explain of how light stays inside a curved fibre, followed by a State asking for advantages over copper cables Three points: 'the angle of incidence at the side of the fibre is greater than the critical angle'; 'total internal reflection occurs'; 'repeatedly, all along the fibre'. For advantages, two separate ones such as 'carries more data', 'less signal loss over long distances' or 'not affected by electrical interference'.
Defining the critical angle, then using n = 1 / sin c: a one- or two-mark State of the meaning, followed by a two-mark Calculate of $n$ or $c$ The definition needs both parts: 'the angle of incidence' and 'at which the angle of refraction is $90 °$ '. The calculation needs $\sin c = 1 / n$ and the correct angle in degrees, or the correct $n$ with no unit.
Defining refractive index and using it to find the speed of light in a material: a one-mark Define, then a two-mark Calculate of a speed 'Speed of light in air (or vacuum) divided by speed of light in the material', in that order. For the speed: dividing $3.0 \times 10^{8}$ m/s by $n$ , with m/s on the answer.

Words to know

Refraction: The change in direction of light as it crosses a boundary between two materials, caused by a change in its speed.
Normal: A line at right angles ( $90 °$ ) to a boundary, at the point where a ray meets it.
Angle of incidence: The angle between the incident ray and the normal.
Angle of refraction: The angle between the refracted ray and the normal.
Refractive index, n: The speed of light in air (or vacuum) divided by the speed of light in the material. It has no unit; $n = \sin i / \sin r$ .
Critical angle, c: The angle of incidence, for light going from a material into one where it is faster, at which the angle of refraction is $90 °$ ; $n = 1 / \sin c$ .
Internal reflection: Reflection of light from the inside of a boundary, back into the material it is travelling in.
Total internal reflection: Reflection of all the light back into a material, which happens when light travels towards a material where it is faster at an angle of incidence greater than the critical angle.
Optical fibre: A thin, flexible glass or plastic strand that carries light along its length by repeated total internal reflection.
Cladding: The outer layer of an optical fibre, with a lower refractive index than the core.

All Physics lessons Next: 3.2.3 Thin lenses