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## Reflection and refraction

Lesson for 16-19

This topic gives an overview of reflection and refraction, with the emphasis on an interpretation in terms of waves. Applications, particularly those involving total internal reflection, are considered.

## Episode 316: Preparation for reflection and refraction topic

Teaching Guidance for 16-19

- Level Advanced

The demonstration of refraction using a ripple tank is tricky, and it is a good idea to practise this.

## Main aims of this topic

Students will:

- know how to justify the law of reflection by a wave diagram
- know how to justify Snell’s law in terms of wave velocities
- be able to perform calculations involving the refractive index
- be able to perform calculations involving critical angle
- know the benefits of fibre optic communication

## Prior knowledge

Students should know:

- the law of reflection
- the law of refraction, or at least have a qualitative knowledge of refraction

## Where this leads

This work explains the behaviour of light in terms of waves; this is particularly important as diffraction and interference follow (as the result of superposition of waves).

### Up next

### Reflection and refraction of waves

## Episode 317: Reflection and refraction of waves

Lesson for 16-19

- Activity time 145 minutes
- Level Advanced

This episode starts from the phenomenon of refraction and moves on to Snell’s law.

Lesson Summary

- Demonstration and discussion: Reflection and refraction with ripple tank (15 minutes)
- Discussion and student activity:
Marching soldiers

model of refraction (10 minutes) - Student experiment: Ray tracing through rectangular block (30 minutes)
- Discussion: Refractive index and Snell’s law (20 minutes)
- Worked example: Using refractive index (10 minutes)
- Student questions: Calculations involving refractive index (30 minutes)
- Discussion: Summary (10 minutes)

## Discussion and demonstrations: Reflection and refraction with ripple tank

Show reflection of ripples at a straight barrier. Start with straight ripples striking a straight barrier, at an angle. Continue with a single straight ripple, then a curved ripple.

To show refraction with a ripple tank, you need to show how ripples change speed when travelling from deeper into shallower water (or vice versa). Submerge a sheet of glass in the water to provide an area of shallower water; the shallower, the better. Start with ripples arriving head on

to the boundary between deep and shallow water. You should be able to see that the separation of the ripples has decreased; this is because they are travelling more slowly.

Now alter the position of the glass so that the ripples enter the shallower area at an angle. It can help to concentrate on one ripple at a time; simply depress the vibrating bar and release it. You should see that the ripples change direction.

Now show diagrams to summarise these observations.

## Reflection

When the first part of the ripple touches the barrier a semicircular wave starts to travel away from the point of contact at the same speed as the incoming wave. This happens for every point on the ripple. The tangent to the new circles is the new wavefront. In the time taken for the end of the ripple farther away to reach the barrier, the reflected wave has travelled outwards the same distance so the equal angles can be seen.

## Refraction

Snell’s Law may have been covered previously but probably obtained experimentally by ray tracing with no reason given for the form of the law. Closer examination of the wave behaviour shows clearly the relationship between the two velocities and the sine of the angles.

Fermat’s principle of least time can be useful here – a ray of light, travelling between two points, takes the path of shortest time. An analogy of seeing someone in a river works well; to rescue them, you would run along the bank (greater speed) until you judged it sensible to swim (lower speed). The path taken is that followed by light.

Remind the students of the reversibility of light too. It often helps with problems.

## Discussion and student activity: Marching soldiers’ model of refraction

Here is a practical way of explaining refraction. Students form a line, arms linked. They march forwards in step, so that they meet a boundary between, say, tarmac and grass obliquely. On the grass, they march more slowly. The outcome is that the line bends and changes direction.

A comparable effect is the skidding of a car if its wheels on one side leave the road and start slipping on a grassy verge. The car slews round.

Another analogy uses soldiers: As the soldiers walk onto the sand they slow down and so each line of soldiers bends. They end up moving in a different direction – this is just what happens when a beam of light hits a block of glass at an angle – it refracts.

Episode 317-1: Refraction: soldiers walking from tarmac onto sand (Word, 60 KB)

## Student experiment: Ray tracing through rectangular block

Students can gain practice in ray tracing through using a ray box and rectangular glass block. They can investigate Snell’s Law or make measurements of refractive index.

Episode 317-2: Measuring refractive index (Word, 28 KB)

## Discussion: Refractive index and Snell’s law

Explain the meaning of refractive index, and state Snell’s law. Points to mention concerning refraction:

Angles are measured from the normal (because this works if the interface is curved).

It is useful to remember which way the bending occurs (towards the normal when slowing down).

Rays only bend at points where their speed is changing (usually at interfaces between different media).

The refractive index of a material (i.e. the *absolute* refractive index) means relative to a vacuum, but this value is a good enough approximation where the other material is air.

## Worked examples: Using refractive index

A useful equation derived from Snell’s Law using absolute refractive indices is:

n_{1}sin( θ _{1}) = n_{2}sin( θ _{2})

A ray of light travelling from water to glass is incident at 50 ° to the normal. Calculate the angle of refraction in the glass.

(
n_{water} = 1.33

n_{glass} = 1.5
)

n_{1}sin( θ _{1}) = n_{2}sin( θ _{2})

θ _{glass} = 42.8 °

## Student questions: Calculations involving refractive index

Episode 317-3: Questions on refractive index (Word, 27 KB)

## Discussion: A summary

Show diagrams of wavefronts going through a prism and both types of lens.

Remind students that this behaviour applies to other waves e.g. sound (a balloon filled with CO_{2} can act as a converging lens).

Summarise the ideas you have been looking at: reflection and refraction can both be explained with the idea of waves.

Episode 317-4: More about Snell's law (Word, 110 KB)

Episode 317-5: The invisible man (Word, 26 KB)

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### Up next

### Total internal reflection

## Episode 318: Total internal reflection

Lesson for 16-19

- Activity time 95 minutes
- Level Advanced

Total internal reflection (TIR) is a consequence of refraction.

Lesson Summary

- Student experiment: Ray tracing using semicircular block (30 minutes)
- Discussion: Uses of TIR (15 minutes)
- Demonstration: Transmitting a radio programme using fibre optic cable (10 minutes)
- Worked example: Critical angle (10 minutes)
- Student questions: On TIR (30 minutes)

## Student experiment: Ray tracing using semicircular block

Students can shine rays of light into the curved face of a semicircular glass or perspex block.

Ask students to mark the centre of the straight edge with a fine permanent pen before they start. There are a few qualitative questions and answers included.

Emphasise that total internal reflection can only happen when the light goes from high to low refractive index (from the denser

to the less dense

medium). It will show on their calculators as an error message if the refracted angle would be greater than 90 ° .The critical angle is always in the denser medium.

Episode 318-1: Ray tracing on the way out (Word, 44 KB)

## Discussion: Uses of TIR

Consider some of the uses of TIR. Prisms are used in cameras and binoculars. A 90° prism can be used to turn light through 90 ° (reflects at the hypotenuse) or 180° (in through the hypotenuse and reflects off both short sides). If critical angle for the prism material is about 42 ° then it is totally internally reflected because the angle of incidence will be 45 ° from geometry. The reflection is more efficient than with a silvered mirror.

Optical fibres are the most important use nowadays. A simple fibre or glass rod will lead the light along because the air outside is less dense than the glass. The shape ensures that the angle of incidence is greater than critical, around 42 ° . The addition of cladding improves the efficiency. In order to confine the light to paths almost straight down the centre and reduce time differences due to different paths or wavelengths, it is best to have the critical angle large, e.g. over 80 ° . This is achieved by the cladding. Usually for transmission of signals a monochromatic semiconductor laser is used.

Episode 318-2: Fibre optics (Word, 77 KB)

## Demonstration: Transmitting a radio programme using fibre optic cable

Show data transfer on an optical fibre. This can be very quick if it is set up before hand.

It is convincing to see the light in the pipe and hear the broadcast start as it is inserted into the transmitter or receiver.

Episode 318-3: Data transfer on an optical fibre (Word, 43 KB)

## Worked examples: Critical angle

Calculate the critical angle in water at an air boundary.

The critical angle in a medium is related to the refractive index of the medium by

n = 1sin(C)

sin(C) = 1n

or

sin(C) = 11.33

sin(C) = 0.752

C = 48.8 °

## Student questions: On TIR

Some practice questions on total internal reflection

Episode 318-4: Questions involving total internal reflection (Word, 27 KB)