Sight and the Eye

To deal with light, we need to appreciate how we percieve it.

The eye is asophisticated optical bench. Light enters the eye through the cornea and is focussed by the pupil, which in combination act as a lens. The light is focussed on the retina, which detects the image formed and sends the information to the brain as electrical signals through the optic nerve.

The eye can adjust so as to perviece objects of different intensity and colours in different locations.

To cope with different light intensities, the pupil can contract and expand to allow less or more light into the eye, as is required for the retina to produce and image. It does this by contracting or relaxing the iris over the lens.

Task:

Observe this effect in each other, using torches to contract pupils in a darkened room.

Accomodation

Accomodation is the process of the eye changing shape so as to focus on objects which are near or far. The ciliary muscles relax or contract and in doing so change the shape of the lens and cornea. It becomes thicker so as to focus near objects and thinner to focus those further away:

The eye is considered to be able to focus betwen any two points, the near point and the far point. Outside this range it gets into difficulties. The near point varies from person to person, but is generally regarded to be about 25 cm (limited by how thick you can make your lens). The far point is supposedly infinity!

Depth Perception

Our ability to determine perspective comes primarily from experience. Using two eyes, we get two images, and the brain is able to judge distances and relative sizes from differences between the two images based on previous knowledge.

Studies have shown that people living for generations in dense forest (such as Amazonian tribes in South America) have very poor ability to percieve distances based on all experiences in the forest being relatively close to the observer.

Black and White and Colour

Images produced by the brain are based on the electrical impulses registered by the light sensitive retina. The retina is sensitive to light using its rod and cone sensors. Rods are sensitive to black and white (ie: they register light or no light), where as cones are sensitive to colour (ie: they register the quality of light at different wavelengths).

Different parts of the eye are sensitive to light in different ways because of the distribution of the rods and the cones in the retina.

This graph shows why we see colour (photopic vision) most clearly in our direct line of vision, and explains why our wide angle vision is more attuned to black and white (schotopic vision). Indeed in the dark, it is easier to make out objects in the "corner of your eye".

Photopic Vision

Demo:

LCD displays make up different colours by combining different proportions of red, green and blue light.

Our photopic vision works in a similar way. We have 3 different types of light sensitive cones. Cones sensitive to red light, cones sensitive to green light and cones sensitive to blue light. For any particular colour reaching the retina, the brain receives information from the different cones telling it how much of each type of light there is present, and ths the brain can interpret this to determine the colour it is 'seeing'.

Graph to show the spectral response of cones in the retina.

Colour mixing to determine colour

By combining 2 of each of the 3 primary colours (red,green and blue) it is possible to make the 3 secondary colours (magenta, cyan, and yellow). By combining all 3 of the primary colours in equal proportions you can make white, and by changing the proportions of each of the primary colours you can make all to the colours of the visible spectrum.

Playing with colours applet.

Colour filters on lights allow us to do this.

Colourblindness

This usually occurs from a genetic failure of the green and red cones. Most colourblind people cannot distinguish between these colours, and hence have trouble with yellow too.

The colourblindness test

Tasks:

1. Explain the process of accomodation in the human eye.

2. Distinguish between photopic and schotopic vision.

3. Sketch and explain the spectral responses of cones in the human eye.

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Doppler Effect

What is it?

Task 1:

Roll marbles down a track at a constant frequency, and make quantitative and qualitative observations of the outcome of changing the motion of the source of marbles, and the observer of marbles.

Questions:

a) In what way does the "frequency" of the marbles simulate the frequency of a sound wave?

b) How does the "frequency" of the marbles compare to the frequency of real sound waves?

c) Can the "frequency" of the marbles be altered? Can the frequency of real sound waves be altered?

d) How does the marble "frequency" change (higher or lower) as the receiver moves to and from the source?

e) Why did not all students observe the same "frequency" changes?

f) How would the results change if the source were moving and the receiver were stationary?

How can this phenomenon be applied to waves?

Hint: think in terms of each marble being a wavefront.

Task 2:

What observations can you make about sounds you hear when you listen to a moving source of sound both approaching you and receding from you?

The change in pitch of the sounds that we hear when a source of sound approaches us and recedes from us can be explained by the Doppler effect.

1. Consider a source emitting light with a wavelength of l metres and travelling at c metres per second.

2. The time for one complete wave to be emitted is l/ c seconds.

3. If the source is moving at a speed v away from the observer, then it will have moved a distance of , during the time it takes for one complete wave to be emitted.

4. This means that the wavelength as seen by the observer will have increased by this amount

 gives the change in wavelength,    gives the fractional change in wavelength           

The speed of the observer can also have an affect on the sound that you hear.

This equation can be stated in terms of frequency instead...

Relative movement of the source and observer can complicate things.

Moving Source

If the source is moving towards the observer, then the frequency will increase, if it is moving away it will decrease.

If we consider a wave of speed, v, coming from a source moving with speed u_s then its frequency changes to

depending on whether the source is approaching or receeding.

Moving Observer

If the observer is moving towards the source, then the frequency will increase, if it is moving away it will decrease.

If we consider a wave of speed, v, moving towards an observer moving with speed u_o then its frequency changes to

depending on whether the observer is approaching or receeding.

Tasks:

Check out the police car applet to consolidate this.

Read text about the Doppler Effect applied to cosmology. Now get an appreciation of this by doing the questions linked to the spectral stellar data .

A simple applet also demonstrates what we are doing here.

Exercises

Questions on the Doppler Effect

Calculations of A-level Physics: pp139 q 1-5

Hutchings pp 537

Topic 11 teaching questions 8-17

Hutchings pp536-538, Kirk and Hodgson pp98-101.

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Standing Waves

Standing (stationary) waves are the result of the superposition of two waves (with the same speed, frequency and amplitude) travelling in opposite directions.

This situation is most easily attained by reflection.

Applet demonstration

There are two main points to note here.

1. At places where there is no motion is seen we have a node.

2. At places where maximum motion is seen we have a antinode.

Important notes:

• Standing wave patterns can only be established at certain frequencies of vibration.

• The point of reflection must always be a node.

Question:

For sound waves of frequency 2500Hz, it is found that two nodes are separated by 20.0cm, with three antinodes between them. What is the wavelength of these sound waves? What is the speed of sound in air?

Exercises:

Calculations for A-level Physics: pp127 questions 1 and 2.

Standing waves on strings

These are very much like the waves we have looked at so far, which when the frequency of vibration is just right can be set up by reflection at the fixed ends of a string.

We can have different patterns of standing waves set up like this, the simplest is called the fundamental.

We then get progressively more complicated waveforms as the frequency of vibration of the wave increases, look for the pattern.

How does the length of the string relate to the wavelength of the wave each time? How are the frequencies of successive harmonics related?

Standing waves in pipes

For a standing wave in a pipe with a closed end, the closed end must be a node, since it can't move. This means that harmonics are set up for frequencies of waves for which the end of the pipe is a node. So as to hear the maximum effect the open end of a pipe must be an antinode for a harmonic to be established.

Note: No even numbered harmonics exist for closed pipes.

The opposite condition is true for open ended pipes, ie, both ends must be antinodes.

Hutchings pp215-220, Kirk and Hodgson pp94-98.

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Experiments with Standing Waves

In each case answer the question posed using the apparatus show in the diagram.

1. Verify the frequencies of the tuning forks is correct (assume speed of sound is 330m/s).

PSOW sheet

2. Verify the frequency of sound that you have produced (assume speed of sound is 330m/s).

3. How does tension in the string affect standing waves?

PSOW sheet

4. Measure the frequency of the microwaves used.

5. Measure the speed of sound.

Hutchings pp215-220, Kirk and Hodgson pp94-98.

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Single Slit Diffraction

Recalling Huygens Principe from topic 4: "Every point on a wavefront actsas a source of secondary circular wavelets"

If we apply this theory to a wavefront passing through a gap, then not only do we have an explanation for diffraction, but also for the interference pattern that we see.

But how does Huygens' principle explain this?

We must consider light waves coming from either side of the slit.

When these wave meet and are in phase (path difference of a whole number of wavelengths) we would expect to see a maximum intensity. BUT we see a minimum...

To explain this we must consider all the other waves going through the gap. Going straight to the halfway point in the gap, a wave from here and the wave from the top edge must be out of phase by half a wavelength when the meet at P₁ causing deconstructive interference, as will a wave from just below the midpoint and one just below the upper edge. This arguement can be continued until you see that the wave from above the midpoint cancel out those from below it.

Therefore to determine the angular position of the first minima in a single slit diffraction pattern (given the small angle approximation of q = sinq ) ...

Question:

What would this equation become for the second, thrid and fourth minima in the series?

Question:

Kirk and Hodgson pp103.

Exercise:

Verify this result experimentally using the school's helium neon laser (wavelength is written on its base), and the distance to the first few minima in a pattern to determine the slit width of a gap you have made.

What other measurement do you need?

Verify your result using a microscope with graduation which you calibrate.

Investigate the effect of changing some of the other possible variables in the experiment using the applet

Hutchings pp226-228, Kirk and Hodgson pp101-103.

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Resolution

Resolution means the ability to distinguish between two seperate images. This can be problematic in some fields such as astronomy and microscopy, since the objects in question are either very far away or very small, and thus produce quite large central maxima in a single slit diffraction by an optical device.

Astronomy	Unresolved image	Resolved image
Microscopy

The Rayleigh Criterion

To be able to resolve two images viewed through an optical device, their central maxima must be separated by a distance greater then the distance to each's first minima.

The condition set for this criterion to be satisfied is.

Where:

q is the angular separation of the two maxima

l is the wavelength of the light

d is the aperture width of the instrument used.

Demonstrations:

How resolution changes with angular separation

How resolution changes with wavelength

How resolution changes with aperture diameter

So the smaller q, the greater the resolving power of the instrument.

Question:

Why do the 'dishes' of radio telescopes have to have much larger diameters than the objectives of optical telescopes?

Questions:

Kirk and Hodgson pp104.

Topic 11 teaching questions

Hutchings pp228-, Kirk and Hodgson pp103-104.

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Polarisation

Light is an electromagnetic wave, which is caused by the vibration of charged particles. It is made of an oscilating electric and magnetic field.

Normally objects emitting light do so in all planes, just like it is possible to shake a rope in different planes to change the plane of the transmitted transverse wave.

If the wave is polarised, then it means that only waves in one plane exist. A polaroid filter acts to cut out all waves except those in one plane.

Analogy with a picket fence doing this with waves on a rope.

Vertical rope waves can be transmitted through vertical pickets.	Vertical rope waves can be transmitted through vertical but not hotizontal pickets

Using a polaroid filter we can create plane polarised light

We can also see how using two polaroids in succession gives us the ability to cut out light.

Investigation:

Adapt the experiment suggested by Hutchings pp175 to investigate the intensity of received microwaves changes with angle of angle of reveiver. Plot a graph to show this. Can you deduce the relationship?

Malus Law of Transmitted Polarisation

To understand this we need to be familiar with the ideas of the polariser and analyser filters in an optical bench arrangement to investigate polarisation.

Since only one component of the em wave is allowed to pass through a polariser, then we can deduce the size of that component from simple trigonometry if we know the angle of the analyser with respect to the polariser.

We can see that transmitted component of the amplitude of plane polarised wave at angle, q, to the analyser is...

a = a₀ cos q

However to relate this to the intensity of the wave that is observed, we must consider the relationship between intensity and amplitude for a wave...

I a a²,

Thus Malus Law is quoted as...

I = I₀cos² q

Polarisation by Reflection

Many substances have the effect of polarising light. If the long crystals of quinine iodosulphate are all aligned in the same direction, then they constitute a polariod filter such as we have used. Basically they absorb all the light incident except in the diection of their alignment.

Shiny surfaces are natural polarisers, and light reflected from them is usually at least partially horizontally polarised. A good example of this is light relected from the surface of water. Reflected light is partially horizontally polarised, with a little remaining vertical light.

Glare is often a problem to photographers, anglers, and sailors, and the need to cut out this glare is achieved with the use of polaroid lenses. By introducing a horizontal polariod filter, the horizontally polarised light can be cut out, leaving only the small amount of vertical light that the water didn't cut out.

Demonstration applet- rotaing the polariser to reduce the glare.

Brewster's Law (Kirk and Hodgson pp107)

Investigations:

Kirk and Hodgson pp106. The data presented is good enough for analysis, but I want you to verify that the effect exists. Using an adapted polarimeter: laser directed through a polariser, a sample of sugar solution and a polariser onto a light intensity meter, verify that:

i) the angle of rotation of polarised light changes as the length of the sugar solution changes.

ii) the angle of rotation of polarised light changes as the concentration of sugar solution changes.

Tasks:

1. Explain and draw a diagram with appropriate table of results to show how you have verified these effects.

2. Analyse the data presented by Kirk and Hodgson pp106.

3. Research and explain in a presentation to the class one of the following:

• What we mean by an optically active substance, how it is optically active, and how we can use the method of polarimetry to determine concentrations of optically active substances.

• How we use polarimetry in stress analysis of different materials.

• The mode of operation of LCD displays, and how polarizers are critical to this effect.

(Each presentation should be no more than 5 minutes long, constitute of a powerpoint with handouts for the other students detailing the essential points, and sources appropriately referenced with appropriate hyperlinks to pertinent internet sites included)

Questions:

Topic 11 teaching questions on Polarisation

Hutchings pp174-177, Kirk and Hodgson pp104-108.

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anrophysics 2008-09