## Centre of Mass (or Gravity) Imagine you could squash a pool ball up to the size of a pinhead. An arrow drawn vertically downwards from that point shows the centre of gravity, acting at the centre of mass.

Where might the centre of gravity of a pool cue be? Obviously at a point nearer the handle end where the wood’s diameter is greater.

Finding C of M is easy. Cut out a funny shape from a cereal packet. Make a tiny hole in one corner with a pin, making sure the piece of card can move freely and hang a piece of cotton with a weight on the end on it. Draw a line down the vertical string. Do exactly the same somewhere else on the card. Where the lines cross is the centre of mass. The force of this mass acting downwards acts along the line of the centre of gravity.

Some objects have a centre of mass outside of the material of the object, like  a wooden lab stool. All the mass is in the seat, so the centre of mass is in thin air somewhere underneath the seat.

Where’s the centre of mass of a Bunsen burner? Probably in the metal bit at the base. It’s in stable equlilbrium so if it’s tipped it’ll fall back on to its base again.

Remember the OSMIUM ROCKER from the density post. Think about what would happen. You might need your density table handout. Look back at it and suggest what might happen if you sat on the rocker Can you balance a potato on a knife edge? Sure you can. Two forks lower the centre of gravity to below the knife edge so that if it’s tilted, it’ll tend to return to the equilibrium position on the knife edge. Try it at home…ASK FIRST.

You might think about this as the ‘tightrope walker principle’. The only reason tightrope walkers don’t fall off is that the massive pole they carry lowers their centre of mass so it’s below the rope. in fact, they CAN’T fall off. In 1859, a man called Blondin first walked across Niagara Falls on a tightrope. He did it lots of times, once carrying a man on his back! Notice the pole!

London buses can tilt an awfully long way before toppling over also. See a bus doing the tilt test here. Why doesn’t it topple over? Because the centre of mass is low, almost between the wheels, so as long as the centre of gravity acts inside the pivoting wheel, it won’t topple over. Now, something for you to do. Tails on birds can lower their centre of mass so they can perch on wires. Cut out a parrot like this one from a piece of card about 25-30 cm high. Weight the tail with a heavy clip – it helps if you curl the tail down underneath the bird a bit. See if your parrot can stand on a wire on his own… Colour it in and bring them all to school and we’ll display them.

## Ladders and Tramlines: Electrons as waves and particles

What’s an eV?

eV stands for electron volt – that is, the amount of energy needed to move one electron through a potential difference of one volt. If that doesn’t mean much to you, don’t worry about it; the point is that it’s a small unit of energy, convenient for the scale of energy levels in an atom. To give you an idea of just how small an eV is…a 100 watt light bulb is putting out 100J of energy per second. That’s about 625,000,000,000,000,000,000eV per second!You may also have run across KeV (kilo-eV; that’s one thousand eV) and Mev (mega-eV, which is one million eV) in the med phys course.

Put another way, if we shine light on potassium, for example, photoelectric emission only happens if the energy of the incident light-bullets (or PHOTONS) is greater than 2.1eV – the work function energy of K Anything less, the electrons in the metal just get agitated.

Converting from eV to J:

2.1 x 1.6 x 10-19J = 3.36 x 10-19J is the minimum energy a light-bullet has to have to hook an electron out of the potassium surface. If the photons have more energy than this, the extra is picked up as the KE of emitted electrons.

Given h = 6.63 x 10-34 Js and E = hf = hc/λ, this corresponds to a wavelength of 592nm – a yellow-green colour.

Electrons in atoms

We know that electrons don’t just wander about – instead they exist in well-defined energy levels, much like the rungs of a ladder define how much gravitational potential energy it takes to climb them. But, the rungs aren’t evenly spaced. The ones near the nucleus are bigger jumps than the ones further away. For hydrogen the biggest jump an electron needs to get out is 13.6eV  – the first ionisation energy of hydrogen. If given 13.5eV, the electron just sits there, vibrating mutinously. To get it out it’ll have to get EXACTLY 13.6eV of energy.

These rungs of energy are a little bit like tramlines, the electrons almost run on rails. Why do electrons exist in these ‘tramlines in the sky’?

Because the wavelength of the electron has to fit inside the orbital like a curved guitar string, setting up a standing wave around the orbital. Any old orbital won’t do – only those allowing a whole number of wavelengths to fit is allowed. If you look carefully at the picture, you can see the circular standing waves on the loop of wire which is being driven by a small oscillator box at the bottom This is exactly how the electrons behave when orbiting a nucleus.

Here’s the ‘energy ladder’ for hydrogen. Remember, no intermediate steps are allowed. Each step corresponds to the emission of a photon having the same energy as the ‘jump’ between the rungs. Those falling to n=1 are in the UV, (Lyman series) those falling to n=2 are the Balmer series, shorter, visible wavelengths. n=1 is sometimes called the GROUND STATE and in books is sometimes given a – sign denoting that energy is required to be given to the electron (think of it stuck down a well and has to climb out).

## Polarisation, Brewster’s Law and Malus’ Law

If a wave is plane polarised, all the oscillation directions of the electric or magnetic field vector  (it’s important in an exam to actually say what’s oscillating)  except one are absorbed. This can only happen with transverse oscillations since these are oscillations at right angles to the direction of travel and there are lots of ways for (example) an EM wave can oscillate perpendicular to the travel direction; in a longitudinal wave, there’s only one.

This flash animation shows a polarised EM wave, the electric field is restricted to one direction.

This shows what happens when a polariser is placed in the path of an unpolarised wave. The direction of oscillation is restricted to one. Another crossed polariser reduces the intensity to zero.

Using 3cm microwaves, we see something like this. The microwaves produced from the klystron are already plane polarised – vertical electric oscillations only. We can direct them towards a metal grid, arranged vertically, as shown. We might imagine that the polarised waves would pass through the grid, much like a wave on a string would pass through the gaps. Quite the reverse happens- the energy is absorbed by the free electrons in the metal of the grid, setting up a stationary wave in the vertical bars, thus absorbing the energy. When the bars are horizontal, the microwave energy passes through unimpeded. Applied stress rotates the plane of polarised light. This image shows what happens when a plastic hook is loaded and viewed through crossed polaroids. This phenomenon, known as photoelasticity was discovered by David Brewster (see later). In summary, the effect is due to the fact that the magnitude of the refractive indices at each point in the material is directly related to stresses experienced by the material at that point.

Light reflected from surfaces like a flat road, smooth water or glass is generally horizontally polarized. This horizontally polarized light is blocked by the vertically oriented polarizers in Polaroid sunglass lenses, substantially reducing glare.

At a particular angle, however:

Reminding ourselves that weak reflections accompany refraction, when light is refracted at an interface and the reflected and refracted components are at 90 degrees, the surface acts like a polarizer – reflected component is perfectly polarized one way, refracted is partially polarized the other. This happens at Brewster’s angle. For a glass/air interface, and using Snell’s Law, Brewster’s Angle is about 56 degrees. Malus’ Law  says that when a perfect polarizer is placed in a polarized beam of light, the intensity, I, of the light that passes through is given by where I0 is the initial intensity, and θi is the angle between the light’s initial polarization direction and the axis of the polarizer.

A beam of unpolarized light can be thought of as containing a uniform mixture of linear polarizations at all possible angles. Since the average value o is 1/2, the transmission coefficient becomes Put another way, if a polarizing filter is placed at some angle theta to a vertically oriented plane of polarized light, its amplitude E will be reduced by a factor cos theta  and since intensity is amplitude squared, then, as above, This is known as Malus’ Law.

It can be determined experimentally quite easily. Taping a piece of Polaroid (oriented so as to pass only the vertical component) over a beam of white light from an overhead projector will polarize the beam. Orient a second Polaroid at the same vertical angle and rotate it, using a calibrated LDR circuit to measure the intensity (seen here as a current), taking measurements at suitable angles, all the way around to 360 degrees.

The graph will look something like this, peaking at 0, 180 and 360 (cos squared=1) and zero at 90 and 270, cos squared =0. ## The Inverse Square Law

This works for all waves but it’s convenient to think about EM waves. Imagine a point source of EM wave energy, such as a candle, a light bulb or even the Sun, radiating energy in all directions. We know that  a sphere has surface area = 4πr². If we move twice as far away, the energy is then smeared out over an area FOUR times the size, reducing the intensity by a factor of FOUR. Thus, intensity is inversely proportional to the square of the radius.

Here’s a problem. Imagine a 40W light bulb, which we can approximate to a point source. How far away would we have to be for the intensity (in watts per square metre) to drop to a) 5mW/m² b) 0.05mW/m²?

a) 40W/0.005W = area of the sphere = 8000m² = 4πr², hence r=25.23m

b) 40W/0.00005W = area of sphere = 800,000m², thus r = 252.3m

If I could measure the intensity at a distance of 100km, what would it be?

r = 100,000m,  I = 40/4π.(100,000)² = 0.32nW/m². Is this measurable?

## Waves within Boundaries : Strings

A guitar string is fixed at both ends. If we pluck a guitar string in the middle, near the twelfth fret, the string oscillates with SHM. The tension and mass per unit length defines how fast the string moves in both directions down the string to either end, where it reflects with a phase change of π radians, superposing with the wave coming the other way. This sets up a resonant condition in the string, each part of the string oscillating in phase with variable amplitude down its length, maximum amplitude being in the middle. The string length defines the fundamental frequency ( v = fλ) and is half a wavelength . Plucking the first harmonic on the twelfth fret sets up a standing wave an octave higher, with a NODE or position of no disturbance in the middle and antinodes on either side.

The fundamental, and first to fourth harmonics are shown.

## Huygens and Single Slit Diffraction

Huygens’ Constructions are geometric diagrams which are a fast way of getting to grips with several progressive wave properties like reflection, refraction, diffraction and interference, particularly with EM waves, but modelled in a ripple tank. Here’s an overview.

A plane wavefront consists of a large number of secondary circular wavelets whose speed is constant, if the medium is isotropic (uniform density, like glass or water). These superpose algebraically to form a new wavefront elsewhere in time and space.

Let’s look at diffraction at a single slit. When we see diffraction effects in a ripple tank, we see something like this…the wave smears out horizontally. The most circular diffraction occurs when the gap size is of the order of 1 wavelength.

If we try the same trick with monochromatic laser light, however, we don’t just see the wave energy ‘smearing out’, as we can see here. Instead, we see a series of light and dark bands.  This applet shows the simplest single slit diffraction very well, but you need the Wolfram CDF player to run it. You can change the colour (wavelength in air) of the light and also change the width of the slit. If we use red light, the pattern is broader than with blue light. Of course, we can’t physically do this, since we usually only have a laser with one colour available. Wider slits allow more light in but produce a narrower set of light and dark bands. Think about what it would look like if we were able to use light of any color, and be able to vary the slit width. Notice the central bright maximum is twice as wide as the others.

Now, how does it work? Imagine dividing the slit into two and the slit to consist of lots of secondary wavelets, as in the diagram. The diffracted rays can be thought of as being parallel because the screen is a very, very long way away. Now think about the first dark fringe. The light from a wavelet near the top end of the slit has a partner exactly halfway down the slit with a path difference of exactly half a wavelength.

The next wavelet down is similarly partnered, and so on all the way down the slit. This means that  every pair of wavelets is superposing destructively, or cancelling out, causing the first dark fringe. in some syllabuses written as:  Similarly, the light bands are formed because each wavelet is partnered with another with a path difference of  a whole wavelength.so, this time, or: We notice that the bright bands are much less bright, which this little construction explains. Imagine this time that the slit is divided into three. At a particular angle, two-thirds of the slit has a wavelet partner half a wavelength out of phase, leaving just one third of unpartnered contributions to form the first or subsequent maxima, so it is much less bright than the central bright maximum. Interference happens when wave displacements (amplitude = max displacement) add together algebraically in space and time. This is a fancy way of saying that if two waves/pulses/ripples meet, their amplitudes add up at a particular place and a particular moment. When two crests meet, we get – what a surprise, a BIG crest – interference is CONSTRUCTIVE. When a crest meets a trough, we get zero disturbance- DESTRUCTIVE interference. Easiest to see with circular water waves because the wavelengths are of the order of centimetres.

Diffraction is the spreading of a wave round an edge. There’s only a change in the wave direction at the edge of the gap(s). If we imagine (à la Huygens) that a plane wave consists of a  infinite number of closely packed circular ripples, at the edge of the gap, there’s nowhere else for the wave energy to go except in a circle. NB: the most circular pattern is observed when the wavelength is the same size as the gap. This applies whatever wave type we think about. It explains sound diffraction around doorways,  also radio wave diffraction in the Welsh hills, so even the Welsh get to watch the telly….Shorter wavelengths won’t diffract as well around large obstacles, so if the gap size is large compared to the wavelength, we don’t see much diffraction. The thing to now get hold of is this. We can’t have one without the other. This means that diffraction can be observed in a double-slit interference pattern. Essentially, this is because each slit emits a diffraction pattern, and the diffraction patterns interfere with each other.  The shape of the diffraction pattern is determined by the width of the slits, while the shape of the interference pattern is determined by d, the distance between the slits. If W is much larger than d, the pattern will be dominated by interference effects; if W and d are about the same size the two effects will contribute equally to the fringe pattern. Generally what you see is a fringe pattern that has missing interference fringes; these fall at places where dark fringes occur in the diffraction pattern. So, put another way, we see the broad diffraction envelope and underneath it, the equally spaced interference fringes. When an interference fringe sits underneath a diffraction minimum, we can’t see it. These so-called missing orders are a favourite in exams. This treatment is detailed, but worth some study, as is this Java applet from Walter Fendt which just illustrates single slit diffraction. You can vary both the slit width (narrow slit = more diffraction) and the wavelength (longer wavelength [redder] = broader pattern.