Light

Question 1

What happens to our wave equation if we apply boundary conditions in two hard walls separated by length L?

Answer 1

boundary conditions demand 0 amplitude

Question 2

in the unbound case, what happens to our wave equation if the wave moves in time?

Answer 2

if it moves in time it must also have a period

Question 3

What is special about a time varying wave with a frequency=1/T?

Answer 3

in one period wave moves by one wavelength, so that the phase velocity |as d=v×t| v=λ/T| ω=vk (=λ2π/Tλ)|this is the dispersion relation of the wave

Question 4

What do we need to add to complete our definition of the wave?

Answer 4

we need to specify its initial state usually at t=x=0|we do this by introducing a phase factor

Question 5

What is the principle of superposition? |What is constructive interference? |what is destructive interference?

Answer 5

waves that overlap add up to their vector sum ||occurs when the two waves are in phase creating a wave with twice the amplitude ||occurs when the waves are completely out of phase creating a wave with zero amplitude.

Question 6

What are coherent waves?

Answer 6

source that emits waves that all have the same phase relationship is said to be a coherent source.

Question 7

How was young’s double slit experiment proof for the wave nature of light?

Answer 7

1. if light was a particle, it would pass through the first slit and hit the second wall so none would hit the screen.|2. instead he observed an interference pattern with maxima and minima. A pattern which we’d expect from a wave…||- the first slit turns the light into a coherent source (as slit smaller than wavelength)|-the coherent light then proceeds to the next two slits which both set up different coherent light sources |-the two spherical waves will cause an interference pattern because at any given angle the two wavefronts have travelled a different distance from the slits and are therefore out of phase. |- If we assume the screen is far away then we can assume the angle between each slit and a point on the screen is the same.||∆L=L1-L2|sinθ=∆L/d|∆L=dsinθ

Question 8

How can we make calculations from the path difference and positions of maxima and minima in youngs double slit experiment.

Answer 8

maxima at path difference= integer multiples of wavelength |we must have d≥λ to get a maxima ||minima at path difference is a half a wavelength more than an integer multiple of the wavelength||we can use trig to work out the angle to a minima/maxima |-we can then use knowledge of which order maxima it is, distance between slits and angle to work out the wavelength of light from the interference pattern.

Question 9

geometric optics indicates light as a particle but youngs experiment indicates it as a wave, how can both be true?

Answer 9

if wavelength of light much smaller than any of the relevant physical distances, then we can ignore most of the wave properties and think of light as acting as a particle. wavelength 500nm so this applies to all geometric optics||but in experiment, slit small compared to the wavelength of light so we must treat light as a wave.

Question 10

What is the set up of michelson-morely experiment?

Answer 10

1. beam of collumnated light shone at a partially reflective mirror |2. part of the light is reflected at one mirror and the rest is transmitted to another |3. the light bounces off each mirror and then comes back to the partially reflective mirror |4. part of the reflected beam is diverted to a screen. The two beams are out of phase (due to reflection) so an interference pattern will occur.|-a series of interference fringes are produced due to mirrors not being perfectly parallel so phase mismatch depends on part of the mirror light bounced off.

Question 11

How do we use the set up in the michelson-morely experiment to disprove the existence of the ether?

Answer 11

In the michelson morley experiment… |they theoretically knew that when light in static conditions interfere they will form the maxima and minima at the calculated values i.e. the fringe width or the distance between two maximas was known.||so if light travels through ether, then due to the ether drag (as earth was moving through it) the speed of light must change in both the perpendicular and longitudinal direction. |(this can be understood by taking the analogy of a boat moving in a river, when boat travels in longitudinal direction to the river its direction is governed by its velocity in still water plus/ minus the velocity of stream(In this case the speed of earth travelling through ether) based upon its direction.|whereas when it travels perpendicular to the stream of water, its velocity is simply found by the vector addition of the two velocities.) ||now since the velocity of light is now changed in two directions the wavelength must change (c=fλ); frequency remains same as it is the property of source not the medium through which the light is travelling) |now since the wavelength changes thus we can compute the interference pattern of the two sources by simple calculation and hence the new fringe width or distance between two maximas. therefore the distance between two maxima should have changed. ||from that change it could have been confirmed that light travels trough ether but the result came out that there was no change or shift in the two maximas and the result was same as in static conditions. |Therefore no ether

Question 12

What are the implications of the michelson-morely experiment?

Answer 12

1. there is no ether (since we are not moving through it) so there is no medium in which light is propagating.|light does not need a medium to propogate|2. The speed of light is the same regardless of the relative motion of the observer||these implications led Einstein to go onto discover the world of special relativity

Question 13

what are the consequences of the theory of special relativity

Answer 13

TIME DILATION |a clock that is moving with respect to an observer appears to tick more slowly than a clock at rest |LENGTH CONTRACTION|if an object is moving with respect to an observer, it appears to get smaller |LOSS OF SIMULTANEITY |two events that occur at the same time for a stationary observer do not necessarily occur simultaneously for a moving observer.

Question 14

What is relativistic momentum?

Answer 14

for a particle moving with velocity v, the classical momentum is defined as p=mv |but after taking relativity into account…||for very small v, γ∼1 so is the same as the classical definition. |if the particle moves close to the speed of light, its momentum will be quite different.

Question 15

What is relativistic energy?

Answer 15

the classical definition of kinetic energy of a particle is given by K=1/2mv² |relativity requires this to change to |K=γmc²-mc²||total energy…|E=mc² + K = mc² +γmc² - mc² | =γmc²||with some manipulation, we can write this as |E²=(pc)² + (mc²)²

Question 16

What are the implications of the equation |E²=(pc)² + (mc²)² ?

Answer 16

particle with mass=0 can still have a non zero energy, E=pc if the particle moves at the speed of light. |if v→c, then γ→∞ and the combination of this allows p to remain finite even thought m→0.

Question 17

What is a blackbody?

Answer 17

A hypothetical perfect radiator of light that absorbs and re-emits all radiation incident upon it. |Such objects give off black body radiation.|It’s light output depends only on temperature and this output acts as to maintain its temperature.

Question 18

What do we find if we measure energy radiated as a function of wavelength and then plot a graph of intensity against wavelength for blackbodies at different temperatures?

Answer 18

1) total power radiated per unit area (integral of intensity over all wavelengths) increases dramatically with T,|it can be described by STEPHANS LAW|P/A=σεT⁴||2) The peak of the wavelength distribution is found at lower wavelengths as T increases. |This is described by Wein’s law…|λmaxT=2.898×10-³ (mK)

Question 19

What is the box model of blackbody radiation?

Answer 19

a useful model of blackbody radiation is a box with a hole inside.|any radiation that enters the hole is unlikely to come out as the hole is small, so approximated to absorbs all light and the box is in equilibrium with the surroundings with which is absorbs from so must also be a perfect emitter.

Question 20

How did Rayleigh and Jean use the box model to try to calculate the intensity of the emitted radiation as a function of temperature

Answer 20

using the assumption that each standing wave mode inside the box has an average energy proportional to KbT, where Kb=1.381×10-²⁷ J/K (boltzman constant). |By counting the number of modes they derived the equation on the right..||at long wavelengths this is reasonable but at shorter wavelengths this doesnt work|it diverges (goes to infinity) as λ→0||This represents a complete breakdown of the classical description and was called the ultraviolet catastrophe.

Question 21

What assumptions does Plancks theory of blackbody radiation have?

Answer 21

1) the radiation in the box came from a set of tiny oscillators that had a discrete energy with spacing E=hv||2)energy was emitted or absorbed when the light interacted with oscillators and caused them to make transitions between their states.|The energy wasn’t emitted or absorbed continuously but rather in discrete chunks.||3) instead of having the energy of each mode proportional to KbT, Planck made it proportional to the energy difference E between the oscillator levels. |Furthermore, rather than having equal populations of all modes, Planck weighted them accordingly to the boltzman distribution law, |making the probability of being in a mode equal to e^(-E/KbT). This means mode with higher energy are less likely to be occupied.

Question 22

What was Planck’s equation for blackbody radiators?

Answer 22

1) Produces the same curve when we plancks constant, h=6.626×10-³⁴ Js||2) Reproduces Stephans law if we integrate over all wavlengths, ||3)Reproduces Wein’s law if we differentiate with respect to λ and set it equal to zero.

Question 23

Describe the set up of the experiment, the photoelectric effect

Answer 23

1. place two metal plates in an evacuated quartz tube and shine light of a fixed frequency on one plate |2. measure the current between the two plates

Question 24

What is the physics behind the set up of the experiment?

Answer 24

1. light is transferring energy to electrons inside the metal plates and causing them to be ejected from the plates |2. since the particles carry a charge, they create a current when they travel between the plates.|3. We vary…|-intensity of light |-frequency of light |-voltage between plates.|4. Since the electrons will be ejected with a variety of energies, the current gradually decreases as more negative voltage is applied until there is a critical voltage V0 beyond which no current is observed.|(emitter is the +ve charged plate so attracted back towards terminal) |5. There is voltage at which an electron with max K.E wont escape and thus no current detected,|E(k.e.max)=eV0

Question 25

What were the findings of the photoelectric experiment?

Answer 25

1) If we increase the intensity…|- we get more current which makes sense as we are putting more energy into the system |- V0, the critical voltage stays the same so there is no change to the max k.e. that an electron carries ||2) Should be a tiny delay between the absorption of light and the emission of a photoelectron, since it takes time to collect enough energy from light (especially true at lower intensities)…|- but photoelectrons appear almost instantaneously |regardless of the intensity ||3) When we change the frequency of the light, we observe a linear relationship between Vo and frequency. |below a certain frequency, no current is observed regardless of the intensity of light.|- different metals are observed to have different cut off frequencies but the slope is always the same||Kmax=eV0 =h(f-f0) | =hf-φ|φ is the workfunction of the metal

Question 26

What were Einstein’s explanations for the inconsistencies of the photoelectric effect?

Answer 26

1) light itself is a discrete particle even though it has wave like properties- we refer to these quanta of light as photons.|The energy of each photon is given by E=hf|The intesntiy of light must be proportional to the number of photons/m²s |increasing the intensity of light increases the number of photons, more photons should produce more photoelectrons, but at the same energy.||2) because photons are particles, they transfer their energy through collisions. Therefore, the energy should be transferred quickly, and the electrons should emerge almost instantaneously regardless of intensity.||3) since each metal should hold onto its electrons with a characteristic energy, it makes sense that |Kmax=Elight-φ|but by quantising light we find that…|Kmax=h(f-f0) (=eV0 )| =hf-φ|therefore we see that if the energy of the photons is lower than φ, then no photoelectrons will be created regardless of intensity/total energy.||-gives value for h which was a universal constant.

Question 27

What does the Youngs double slit experiment tell us about the nature of light?

Answer 27

It tells us that light must behave as a wave.|This and the phenomena of the photoelectric effect are distinct (one is not limiting the other)||In Young’s double slit experiment, if we limit the the intensity of light incident on the double slit to the point where only one photon enters the device at a time, we can see it hit the screen at a single point.|Remarkably, we still see an interference pattern. ||-a single photon must be interfering with itself and passing through both slits simultaneously. |-if we try to observe which slit it goes through the interference pattern disappears! |-we are changing the outcome of the experiment just by making an observation in the middle of it. This has no analogue in classical physics.

Question 28

What is compton scattering?

Answer 28

Compton, is the scattering of a photon by a charged particle, usually an electron. It results in a decrease in energy (increase in wavelength) of the photon

Question 29

In compton scattering, come up with useful expressions by considering the conservation of energy and momentum…

Answer 29

If a photon is a particle, it must move at the speed of light and therefore must obey this E=pc.|We know E=hf |so p=hf/c = h/λ

Question 30

What can we conclude from the equation ∆λ=λc(1-cosθ)?

Answer 30

that the max shift in wavelength for a compton scattering event occurs at θ=π so the back scattering of the photon.|in this case…|∆λ=2λc||since λc is directly proportional to 1/m, this shift is easier to observe for an object of small mass e.g. an electron.

Question 31

Why would it be hard to detect the compton scattering effect with visible light?

Answer 31

1. visible light in the region of 300-700nm where as the compton scattering of an electron, |λc=h/mc≈2.43pm for the electron ||2. this would be incredibly difficult to detect in visible light, as is such a small fractional increases of the wavelength of the initial wave||3. we first detected Compton scattering with the use of x-rays and electrons as electrons can go down to 10 pm.

Question 32

Describe the process of pair production

Answer 32

pair production is the process in which energy is converted to mass.|It is the formation of an electron and a positron from a quantum of electromagnetic radiation.|- This can either be in the form of two photons or |- Occur if there is another particle in the area to absorb the photon’s momentum. If the particle is very massive, it will not absorb a significant amount of the photons energy, so that all of the photons energy is transferred to the electron and positron.

Question 33

What is the opposite of pair production?

Answer 33

pair annihilation, a positron and electron meet and annihilate to form two identical photons.

Question 34

What is the minimum photon energy requirement for the pair production of an electron and positron?

Answer 34

E=hv describes energy of the photons |E=mc² describes the energy of the electron |we consider the case γ→e+ + e-|∴hv=2mc²| v=2mc²/h| ∴λ=h/2mc|λ≈1.21pm|this is equivalent to gamma rays which are found in nature but in cosmic rays.

Question 35

What is Bragg Scattering?

Answer 35

When x-rays are scattered from a crystal lattice, peaks of scattered intensity are observed which correspond to the following conditions:|1) The angle of incidence = angle of scattering.|2) The path length difference is equal to an integer number of wavelengths.||The condition for maximum intensity contained in Bragg’s law above allow us to calculate details about the crystal structure, or if the crystal structure is known, to determine the wavelength of the x-rays incident upon the crystal.

Question 36

What is Bragg’s scattering experiment?

Answer 36

1. if we vary the incident and scattered angle, θ, we will find an intereference pattern similar to that of young’s double slit experiment|2. we can find the maxima again by looking at the path difference, 2∆l|3. there should be a maxima when the path difference is an integer multiple of wavelength.|2∆l=nλ=2dsinθ n=0,1,2…….||- doesn’t just occur in the horizontal plane, it can occur in any plane and for a full understanding of the interference pattern we must consider the 3cd nature of the lattice and symmetry condition.||- light must be comparible to the lattice spacing, visible light too large so we must use small wavelengths like that of x-rays to probe this regime.