Waves Tutorial 6 – Reflection, Refraction, and Optical Fibres

We have seen in Unit 1 how light behaves as a particle and how particles behave as waves.  In this tutorial, we are going to look at how light shows wave behaviour.  All other waves behave in the same way.

 

Some Revision

In this section we will consider light to be a wave carrying energy that travels in straight lines at a speed of 3.0 × 108 m/s in air.  When light hits an object it is:

 

Reflection

You should be familiar with reflection which you did in early secondary school.  All reflection depends on the smoothness of the surface.  If the surface is smooth, then parallel rays are reflected parallel (specular reflection).  If the surface is rough, then parallel rays are scattered.  However each individual ray obeys the law of reflection.

 

High quality mirrors are silvered on the front to prevent multiple images.  Here is a ray of light striking a mirror at an angle:

 

We observe the following:

 

The image in a mirror has these features:

 

Curved mirrors obey the same rules of reflection as flat (plane) mirrors do.

 

Reflections can give a lot of atmosphere to a photograph - what an excuse to share one of my favourites!

 

 

Refraction

Light rays are bent when they travel from a medium of one optical density into another, for example from air to glass.  In air light travels at 3.0 × 108 m s-1, while in glass its speed is 2.0 × 108 m s-1.  We say that glass in an optically denser medium than air.  We can see that the direction of the ray is abruptly changed, or deviated as it passes the boundary:

 

We should note the following:

 

Question 1

Can you complete these diagrams showing the refracted ray and the emergent ray?

Answer

 

 

Refractive Index

The refractive index given the physics code n and has no units.  We can also describe the absolute refractive index as the ratio of the speed of light in a vacuum to the speed of light in an optical medium.  Glass has a refractive index of 1.5, so the speed of light is:

 

3.0 × 108 m s-1 ÷ 1.5 = 2.0 × 108 m s-1.

 

Before we carry on, we need to be sure of what these terms mean:

From this we can write:

 

and          

 We can rearrange each equation and combine the two to write:

 

  So we can write:

 

Note that:

Question 2

What is the speed of light in glass, refractive index 1.5, when the speed of light in air is 3.0 x 108 m s-1?

Answer

 

 

Snell’s Law

It was the Dutch physicist Willebrord Snel van Royen (1580 – 1626), who in 1621 rediscovered the quantitative relationship between the angle of incidence and the angle of refraction.  It was initially worked out by by Ibn Sahl (940 – 1000), an Arabic scholar, in 984.  If we double the angle of incidence, we do NOT double the angle of refraction.

 

 

 

The ratio n2/n1 is the relative refractive index, which we can also define as the ratio of the sines of the angles of incidence and refraction:

 

We can write this formula in a more useful form:

n1 sin q1 = n2 sin q2

 

The ratio n2/n1 can also be written as as 1n2. This means the refractive index of light going in from material 1 to 2.

 

 

Question 3

What does 2n1 mean?

Answer

Question 4

A ray of light strikes an air-glass boundary at an incident angle of 30o.  If the refractive index of glass is 1.5, what is the angle of refraction?

Answer

 

Remember that each wavelength of light has a slightly different refractive index.  Generally we use yellow light.

 

The table shows some absolute refractive indices for some common materials:

 

Medium

Refractive Index

Vacuum

1.000

Air

1.003

Ice

1.30

Water

1.33

Glass

1.50

Diamond

2.42

 

The absolute refractive index will always be always be greater than 1.0 as light cannot travel any faster then it does in a vacuum.  It is possible for particles to travel faster than light in a material (but NOT in a vacuum).  The result of this will be little flashes, the light equivalent of a sonic boom, called Cherenkov radiation.  This is why radioactive materials appear to make water glow.

 

 

Photo: United States Nuclear Regulatory Commission, Wikimedia Commons

 

Now let us have a look at what happens when a ray passes from a dense medium to a less dense medium.  Diagram A shows the ray going from air to glass, while Diagram B shows the ray leaving the glass and going into air.

 

 

In Diagram A we have established that:

 

In Diagram B we see:

  The relative refractive index going from medium 2 to medium 1 is 2n1, so we can write:

 Therefore:

 1n2 = 1 ÷ 2n1 

 

For example, the refractive index from air to water is 1.33, while the refractive index from water to air = 1/1.33 = 0.75

 

Question 5

What is the relative refractive index when a light ray passes from glass (RI = 1.5) to air?

Answer

Question 6

What is the angle of refraction and what is the angle through which the ray is deviated when light passes at an incident angle of 48o into water of refractive index 1.33 from air at a refractive index of 1.00? 

Answer

 

 

Critical Angle

We have seen how a ray of light passing from glass to air bends away from the normal.  If we increase the angle of incidence, the angle of refraction increases more (Diagrams A and B):

Angle of refraction getting bigger:

 

At a particular value of angle of incidence, the angle of refraction is 90o, as shown in Diagram C.  This particular angle of incidence is called the critical angle.

 

Above the critical angle we get total internal reflection.  There is no transmission of light at all. See diagram D.

 

 

Above the critical angle we get total internal reflection.  There is no transmission of light at all.  We know that:

 

n1 sin q1 = n2  sin q2

At the critical angle:

n2 sin qc = n1  sin 90

Since sin 90 = 1, we write:

n2 sin qc = n1

 

Therefore

We can write this as:

Since n2/n1 is the relative refractive index 1n2, we can write:

Or we can write:

If you struggle with this, the best way to tackle it is to use:

n2 sin qc = n1  sin 90

Note:

 

Question 7

What is the critical angle of water, of which the refractive index is 1.33? 

Answer

Question 8 

Rays from a point source of light at the bottom of a swimming pool 1.8 m deep strike the water surface and only emerge through a circle of radius r, as shown in the diagram.  If the refractive index between water and air is 1.33, calculate:

(a) the critical angle, qc;

(b) the value of r.

 

Give your answers to an appropriate number of significant figures.

Answer

 

 

Total Internal Reflection in Optical Fibres

An example of the use of total internal reflection is in optical fibres.  At their simplest, optical fibres are long thin strands of glass that carry light from one end a long distance to the other.  The light can be guided round corners using total internal reflection.

 

 

There are two problems:

 

The problems can be overcome by:

 

Optical fibres are amazingly flexible and strong.  They mounted in a polythene tube for further strength.  Optical fibres offer many advantages over wires.

 

Fly by Light

If we use a copper wire to transmit data from several sensors, we have to be very careful to screen it from unwanted signals.  Ordinary wires are very prone to this, especially in long runs.  Screening is easy enough; we simply surround the wire with an insulating layer and then copper braiding to shield the wire.  We have a coaxial cable.

 

 

The problem with coaxial cable is that:

More recently there have been projects that use fly-by-light technology.  This uses optical fibres, which are much less heavy and take up a lot less space; several optical fibres can take the space of one coaxial cable.  An optical fibre looks like this:

 

The fibres are 50 mm across (including the sheath).  The core has a diameter of 5 mm (5 × 10-6 m).

 

Optical Fibres

Optical fibres work by total internal reflection.  The light ray makes a certain angle of incidence when it hits the boundary of an optically dense material (like glass) and an optically less dense material (like air).  If this angle is greater than the critical angle, the ray is totally internally reflected.  The critical angle, qc, is determined by the formula:

 

 

Where 1n2 is the physics code for the refractive index going from material 1 to material 2.  We can rewrite the equation in a more user-friendly way:

 

Or we could go back to our old friend:

n1 sin q1 = n2 sin q2

 

The rays of light should travel like this:

 

 

But instead light rays can travel several paths:

 

 

This means that the light rays can arrive at different times, resulting in dispersion or smearing.  The signal that was sharp when it left the transmitter is smeared.

 

 

The picture shows us how the signal can be unacceptably distorted and even produce spurious signals that were not there.  The problem can be resolved by cladding the core with a material of slightly lower refractive index.  For example the core might have a refractive index of 1.6, while the cladding has a refractive index of 1.4.  This is called a step index fibre.

 

 

Dispersion can be reduced further by use of a graded index or multimode fibre.  Some light is passes down the middle, which has a higher refractive index, therefore slower rate of travel.  With clever manipulation of the refractive indices, the ray travelling down the middle can be made to arrive at the same time as the ray that goes from side to side.  They can meet with a time difference of about 1 ns km-1.  In aircraft where the distances are generally less than 50 metres, this is not too bad.

 

Monomode fibres are designed such that the rays pass only down the middle.  If the light were perfectly monochromatic, i.e. of one wavelength only, the rays would all arrive at the same time.  However even the best lasers produce a slight spread, and since refractive index varies with wavelength, there can be slight differences in arrival times, leading to smearing.

 

Worked Example

The picture shows light entering into a straight length of step-index optical fibre.

The critical angle between the core, refractive index 1.52, and the cladding is 59o.

(a) Calculate the refractive index of the cladding.

(b) What happens to a light ray that strikes the core/cladding boundary at an angle of less than 59o?

(c) What happens to a light ray that strikes the core/cladding boundary at an angle of greater than 59o?

Answer

(a)  Use our old friend:

n1 sin q1 = n2 sin q2

 

n1 sin 90 = 1.52 sin 59

 

n1 = 1.52 × 0.857 = 1.30

(b) The light ray will be refracted into the outer cladding.

(c) The light ray will be totally internally reflected.

 

Question 9

 A stepped index optical fibre has a central core of refractive index 1.50, and a cladding of refractive index 1.45.

 

 

A single fine beam of monochromatic light enters the core at an incident angle of 30o

(a) Calculate the angle of refraction of the light ray as it passes into the fibre.

(b) Calculate the critical angle between the core and the cladding.

(c) Explain whether or not the light will continue to go down the core.

Answer

 

Although we have looked at light, all these phenomena can be observed with other forms of waves, e.g.