Pearson Edexcel Physics

Unit 2: Waves and Electricity

2.3 Waves and Particle Nature of Light

Pearson Edexcel Physics

Unit 2: Waves and Electricity

2.3 Waves and Particle Nature of Light

Candidates will be assessed on their ability to::
33. Understand the terms amplitude, frequency, period, speed and wavelength
34. Be able to use the wave equation [math]v = fλ[/math]
35. Be able to describe longitudinal waves in terms of pressure variation and the displacement of molecules
36. Be able to describe transverse waves
37. Be able to draw and interpret graphs representing transverse and longitudinal waves including standing/stationary waves
38. CORE PRACTICAL 4: Determine the speed of sound in air using a 2-beam oscilloscope, signal generator, speaker and microphone
39. Know and understand what is meant by wavefront, coherence, path difference, superposition, interference and phase
40. Be able to use the relationship between phase difference and path difference
41. Know what is meant by a standing/stationary wave and understand how such a wave is formed, know how to identify nodes and antinodes
42. Be able to use the equation for the speed of a transverse wave on a string [math]v = \sqrt{\frac{T}{\mu}}[/math]
43. CORE PRACTICAL 5: Investigate the effects of length, tension and mass per unit length on the frequency of a vibrating string or wire
44. Be able to use the equation for the intensity of radiation [math]I = \frac{P}{A}[/math]
45. Know and understand that at the interface between medium 1 and medium 2 [math]n_1 sinθ_1 = n_2 sinθ_2[/math] Where refractive index is [math]n = \frac{c}{v}[/math]
46. Be able to calculate critical angle using [math]sinC = \frac{1}{n}[/math]
47. Be able to predict whether total internal reflection will occur at an interface
48. Understand how to measure the refractive index of a solid material
49. Understand what is meant by plane polarization
50. Understand what is meant by diffraction and use Huygens’ construction to explain what happens to a wave when it meets a slit or an obstacle
51. Be able to use [math]nλ = dsinθ[/math] for a diffraction grating
52. CORE PRACTICAL 6: Determine the wavelength of light from a laser or other light source using a diffraction grating
53. Understand how diffraction experiments provide evidence for the wave nature of electrons
54. Be able to use the de Broglie equation

[math]λ = \frac{h}{p}[/math]
55. Understand that waves can be transmitted and reflected at an interface between media
56. Understand how a pulse-echo technique can provide information about the position of an object and how the amount of information obtained may be limited by the wavelength of the radiation or by the duration of pulses
57. Understand how the behavior of electromagnetic radiation can be described in terms of a wave model and a photon model, and how these models developed over time
58. Be able to use the equation [math]E = hf[/math], that relates the photon energy to the wave frequency
59. Understand that the absorption of a photon can result in the emission of a photoelectron
60. Understand the terms ‘threshold frequency’ and ‘work function’ and be able to use the equation

[math]hf = \phi + \frac{1}{2}mv_{\text{max}}^2[/math]
61. Be able to use the electron-volt (eV) to express small energies
62. Understand how the photoelectric effect provides evidence for the particle nature of electromagnetic radiation
63. Understand atomic line spectra in terms of transitions between discrete energy levels and understand how to calculate the frequency of radiation that could be emitted or absorbed in a transition between energy levels.

  • 33) Understand the terms amplitude, frequency, period, speed and wavelength

  • ⇒ Wave:
  • A wave is a means for transferring energy via oscillations. Whilst energy moves from one place to another, the waves cause no net movement of any matter.
  • ⇒ Mechanical wave:
  • A mechanical wave is one in which there needs to be some sort of material medium – a substance that oscillates to allow the transfer of the energy.
  • For example,
  • – A sound wave of a human voice transfers energy from one person’s vocal cords to another’s eardrum by the repeated vibrations of air molecules.
  • ⇒ Electromagnetic waves:
  • Electromagnetic waves can transfer energy through repeated oscillations of electric and magnetic fields, but these fields do not need matter to support them. Indeed, the interaction between electromagnetic waves and matter generally slows their transfer of energy.
  • For example, light travels more slowly in water than it does in a vacuum.
  • Figure 1 The vibration of a wave over a certain distance, as if frozen at an instant in time.
  • ⇒ Displacement:
  • The position of a particular point on a wave, at a particular instant in time, measured from the mean (equilibrium) position. (Symbol: various, often x, SI units: m.)
  • ⇒ Amplitude:
  • The magnitude of the maximum displacement reached by an oscillation in the wave. (Symbol: A; SI units: m.)
  • ⇒ Frequency:
  • The number of complete wave cycles per second. This may sometimes be measured as the number of complete waves passing a point per second. (Symbol: f; SI units: hertz, Hz.)
  • ⇒ Wavelength:
  • The distance between a point on a wave and the same point on the next cycle of the wave, for example, the distance between adjacent wave peaks. (Symbol: λ; SI units: m.).
  • Figure 2 The oscillation of a single particle or point in a wave plotted against time.
  • ⇒ Period:
  • The time taken for one complete oscillation at one point on the wave. This will also be the time taken for the wave to travel one wavelength. (Symbol: T; SI units: s.)
  • ⇒ Phase:
  • The stage a given point on a wave is through a complete cycle. Phase is measured in angle units, as a complete wave cycle is considered to be the same as travelling around a complete circle, that is 360° or 2 radians. (No standard symbol; SI units: rad.)
  • ⇒ Wave speed:
  • The rate of movement of the wave – the same as speed in general. (Symbol: v, or c for speed of electromagnetic waves; SI units: [math]ms^{-1}[/math])

  • 34) Be able to use the wave equation v = fλ

  • The speed of a wave from distance divided by time, as with the speed of anything.
  • However, as each wave has a certain wavelength, and the frequency tells us how many of these wavelengths pass per second, you can also find the speed by multiplying frequency and wavelength together. This is often referred to as the wave equation:
  • [math]\begin{gather}
    \text{Wave speed (m·s}^{-1}\text{)} = \text{Frequency (Hz)} \times \text{Wavelength (m)} \\
    v = f \lambda
    \end{gather}[/math]

  • 35) Be able to describe longitudinal waves in terms of pressure variation and the displacement of molecules:

  • ⇒ Longitudinal:
  • A longitudinal wave in a fluid, such as air, is generated by squashing particles together and then stretching them apart from each other, repeatedly – thus vibrating them ‘longitudinally’.
  • ⇒ A compression:
  • The areas of higher pressure cause the particles to push apart from each other, but this makes the particles move and squash their neighbors. This higher pressure – a compression – then pushes them away to squash their neighbors.
  • Similarly, the areas where there are too few particles (compared with the uniform spread of the particles when the wave is not present) cause particles to move into the vacant space – this is known as the rarefaction.
  • Figure 3 A longitudinal wave generated in a long spring by repeatedly squashing and stretching one end. Areas of higher pressure are called compressions, and areas of lower pressure are called rarefactions.

  • 36) Be able to describe transverse waves

  • A transverse wave is one where the movements of the particles, or fields in an electromagnetic wave, are up and down, or left and right, whilst the energy travels forwards. This is illustrated in figure 5, where one student is vibrating the rope up and down, but the energy travels along the rope towards the other student.
  • As the first student moves the rope up and down, the particles pull their neighbors up and down. These then pass the vibration on to their neighbors, through their intermolecular forces, and the wave moves along the rope.
  • Figure 4 A transverse wave on a skipping rope.

  • 37) Be able to draw and interpret graphs representing transverse and longitudinal waves including standing/stationary waves

  • ⇒ Graphing waves:
  • In Figure 1 a graph of displacement versus distance. This is relatively easy with a transverse wave, as the graph appears like a picture of the wave, so it is easy to understand.
  • The wavelength can be found by measuring along the graph’s x-axis from one point on a wave cycle to the same point on the next wave cycle. Amplitude is measured from the x-axis vertically to a maximum displacement point.
  • Longitudinal waves are less easy to visualize from their graph, and making measurements from them can be difficult. Fig 4 illustrates how a longitudinal wave can be represented on a graph of displacement versus distance along the wave.
  • When a longitudinal wave is drawn like this, you can make measurements more easily. The wavelength and amplitude can be read from the graph just like for a transverse wave.
  • Figure 5 A longitudinal wave can be easier to measure if it is drawn on a displacement-distance graph.

  • 38) CORE PRACTICAL 4: Determine the speed of sound in air using a 2-beam oscilloscope, signal generator, speaker and microphone

  • Wave speed that is distance divided by time; we can use a twin beam oscilloscope to find the extra time a sound takes to travel a short extra distance.
  • One beam trace shows a sound picked up by a microphone held 50 cm from the loudspeaker. The other trace shows the same sound, picked up by a second microphone held further from the loudspeaker.
  • The difference in positions of the peaks on the two oscilloscope traces shows the time taken, t, for the sound to travel the extra distance, d. If we measure this carefully, then the speed will be given by:
  • [math]v = \frac{d}{t}[/math]
  • It can be difficult to make accurate measurements from the screen of an oscilloscope, so we need to synchronize the traces to minimize the effect of random error in taking such a measurement.
  • Firstly, with both microphones at the same distance from the loudspeaker, the two traces appear in identical phase positions (in phase).
  • If we slide the second microphone slowly away from the loudspeaker, we will move the traces out of phase with each other until eventually they come back to exact synchronization.
  • At this point, the distance between the two microphones is exactly one wavelength, A. We set the frequency on the signal generator, and so the wave equation can be used to find the speed:
  • [math]v = fλ[/math]
  • Figure 6 Investigating the speed of sound in air.

  • 39) Know and understand what is meant by wavefront, coherence, path difference, superposition, interference and phase

  • ⇒ Wavefront:
  • Diagrams of waves are often drawn as lines, as in figure 7, where all the points on a line represent points on the wave that are at exactly the same phase position, perhaps a wave crest. These lines are called wavefronts.
  • Figure 7 All points on each wavefront (black lines) are in the same phase position: 90°, or π/2. The green line showing direction of travel is called a ray. Rays must be perpendicular to wavefronts.
  • ⇒ Phase:
  • During a wave’s oscillation cycle, every point will be in a certain location. Some of these cycle locations, including peak (or crest), trough, compression, or rarefaction, have names of their own. Though there are too many to list, the remaining stages in a cycle are crucial. Each of their names. We use a number, known as the phase, to characterize each point through a cycle.
  • Since a full cycle is thought to be identical to turning around in a full circle, the phase position will be measured as an angle, with a full cycle being equal to 360°.
  • Figure 8 Wave phase represents the position through a complete cycle, measured in angle units.
  • ⇒ Wave Superposition:
  • When waves meet, each wave will be trying to cause a wave displacement according to its phase at that location.
  • The overall displacement will be the vector sum of the displacements caused by the individual waves. This is called wave superposition.
  • Figure 9 Wave Superposition
  • Coherence in waves refers to a constant phase relationship between two or more waves, meaning they maintain a consistent phase difference over time.
  • This constant phase relationship is crucial for producing observable interference patterns, as it ensures that the waves constructively and destructively interfere in predictable ways.
  • ⇒ Superposition of continuous Waves:
  • A single point along the path of the waves, we consider waves superposing over a large space, the outcome is a continuous wave that is the sum of the displacements over time in each location.
  • If the two waves are in phase, their effect will be to produce a larger-amplitude resultant wave. This is known as constructive interference.
  • If identical waves meet and are exactly out of phase – if their phase difference is 180° or radians – then the resultant is a zero-amplitude wave.
  • This idea of complete destructive interference can be confusing: imagine shining two beams of light to the same place, and at that point you see blackness.

  • 40) Be able to use the relationship between phase difference and path difference

  • The relationship between phase difference (Δφ) and path difference (Δx) is given by:
  • [math]\Delta \phi = \left( \frac{2\pi}{\lambda} \right) \Delta x[/math]
  • Variables:
  • – Δφ: Phase difference (in radians)
  • – λ: Wavelength
  • – Δx: Path difference
  • Application:
  • This relationship is crucial in understanding wave phenomena like interference and diffraction. By calculating the phase difference, you can predict the resulting interference pattern (constructive or destructive) when waves overlap.
  • Phase difference describes the angular difference between two waves, often expressed in radians or degrees, while path difference refers to the difference in the physical distance traveled by two waves. These concepts are related: a larger path difference generally leads to a larger phase difference,
  • – Constructive interference: Δφ = 0, 2π, 4π, … (path difference is an integer multiple of λ)
  • – Destructive interference: Δφ = π, 3π, 5π, … (path difference is an odd multiple of λ/2)
  • This relationship helps analyze wave behavior in various contexts, including optics and acoustics.
  • Figure 10 Path difference and Phase difference

  • 41) Know what is meant by a standing/stationary wave and understand how such a wave is formed, know how to identify nodes and antinodes

  • ⇒ A stationary wave:
  • A stationary wave pattern, also referred to as a standing wave, can be created when waves moving in opposing directions constantly superimpose.
  • The waves must have a consistent phase connection, be the same speed and frequency, and have comparable amplitudes. Coherent waves have a consistent phase connection and the same frequency.
  • The reason stationary waves get their moniker is because their profile merely oscillates rather than moving.
  • Additionally, waves do not fit our formal definition of waves, which are more accurately described as progressive waves since wave energy does not travel along a standing wave.
  • In this situation, such as when a guitar string is played, the first wave will meet its own reflection. These two coherent waves set up a standing wave on the string.
  • Figure 11 A wave in which describe node and antinode
  • There are points where the resultant displacement is always zero. These points never move, and are called nodes. The points of maximum amplitude are called antinodes.
  • Note that all points between one node and the next are in the same phase at all times, although their amplitude of vibration varies up to the antinode and back to zero at the next node.

  • 42) Be able to use the equation for the speed of a transverse wave on a string  [math]v = \sqrt{\frac{T}{\mu}}[/math]

  • ⇒ String wave speeds:
  • Waves on a stretched spring travel at a speed that is affected by the tension in the string, T (in newtons) and the mass per unit length of the string, (in [math]kg.m^{-1}[/math]). The equation for the speed of a wave in a string is:
  • [math]v = \sqrt{\frac{T}{\mu}}[/math]
  • If this equation is combined with the wave equation, we get an equation that tells us how the frequency of string vibrations is affected by other factors:
  • [math]\begin{gather}
    v = \sqrt{\frac{T}{\mu}} \quad \text{or} \quad v = f \lambda \\
    f \lambda = \sqrt{\frac{T}{\mu}} \\
    f = \frac{1}{\lambda} \sqrt{\frac{T}{\mu}}
    \end{gather}[/math]
  • In the fundamental mode of vibration, this means the fundamental frequency. [math]f_o[/math], depends on the length of the string, its tension and its mass per unit length from:
  • [math]f_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}[/math]

  • 43) CORE PRACTICAL 5: Investigate the effects of length, tension and mass per unit length on the frequency of a vibrating string or wire

  • ⇒ Core Practical:
  • We can verify the equation
  • [math]f_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}[/math]
  • In order to confirm it, we need to do three separate investigations to verify each part of the relationship, whilst keeping the other variables as control variables. So, we need to verify:
  • [math]\begin{gather}
    f_0 \propto \frac{1}{L} \\
    f_0 \propto \sqrt{T}
    \end{gather}[/math]
  • And
  • [math]f_0 \propto \sqrt{\frac{1}{\mu}}[/math]
  • Figure 12 Sonometer experiments to verify the factors affecting the fundamental frequency of a stretched string.
  • A microphone attached to an oscilloscope may be used to track the noises made by a sonometer string and determine how frequently it vibrates.
  • Using a datalogging computer in place of an oscilloscope may make this simpler because the screen may be frozen for close inspection.
  • While the same string (constant u) and the same hanging mass (constant T) maintain control over the other variables, the sonometer’s string supports, also known as bridges, are movable, allowing us to determine the frequency with different lengths, L.
  • To confirm that [math]f_o[/math] on the y-axis is proportional to 1/L on the x-axis, we can create a graph.
  • A set length of the same string (constant L and constant μ) may be used to determine the fundamental frequency for different masses suspended over the pulley, or different T. To confirm that [math]f_o[/math] on the y-axis is proportional to VT on the x-axis, we can create a graph.
  • The final experiment requires a set of different strings (varying diameter metal wires could be used). Maintaining the same length and the same hanging mass (constant L and T) keeps the other variables controlled.
  • Measure the mass of each wire using a digital balance, and its full length, in order to calculate the mass per unit length for each wire or string used. We can plot a graph 1 to verify [math]f_o[/math] on the y-axis is proportional to [math]\sqrt{\frac{1}{\mu}}[/math] the x-axis.

  • 44) Be able to use the equation for the intensity of radiation [math]I = \frac{P}{A}[/math]

  • ⇒ Intensity of radiation:
  • When a lamp emits light, we could measure its intensity. This is the amount of energy it carries, per unit area, and per unit time. Power is the rate of transfer of energy, so this becomes:
  • [math]\begin{gather}
    \text{Intensity (W·m}^{-2}\text{)} = \frac{\text{Power (W)}}{\text{Area (m}^2\text{)}} \\
    I = \frac{P}{A}
    \end{gather}[/math]
  • Example:
  • On a particular day, sunlight lands on Earth with an intensity of 960 W m-2. What is the power received by the solar cell on a calculator placed in this sunlight, if the cell is a rectangle 1 cm by 4 cm?
  • Solution:
  • [math]\begin{gather}
    I = \frac{P}{A} \\
    P = I \times A \\
    P = 960 \times 0.01 \times 0.04 \\
    P = 0.38 \, \text{W}
    \end{gather}[/math]

  • 45) wKnow and understand that at the interface between medium 1 and medium 2 [math]n_1 sinθ_1 = n_2 sinθ_2 [/math] Where refractive index is [math]n = \frac{c}{v}[/math]

  • ⇒ Refraction:
  • When waves pass from one medium into another, there is a change in speed. The frequency remains constant, so the change in speed causes a change in wavelength.
  • If the waves are approaching the interface between the two media at an angle, then the change in speed causes a change in direction as well. This is called refraction.
  • Figure 13 Wave speed depends on the medium, and refraction is caused by a change in medium changing the wave speed.
  • Any ray crossing an interface along the normal line does not change direction at all, as the wavefronts are parallel to the edge.
  • Therefore, their wavelength is equally changed along the length of the wavefront. If the wavefront is at an angle to the interface, then the part that hits first will change speed first.
  • Then the wavefront becomes bent because different parts of it are travelling at different speeds. The changes in direction caused by refraction are the basis for the functioning of lenses, and can lead to optical illusions.
  • ⇒ Refractive index:
  • A measure of the amount of refraction caused by different materials is called the refractive index, and its symbol is n.
  • The refractive index, n, is equal to the ratio of the speed of light in a vacuum to the speed of light in the material:
  • [math]n = \frac{c}{v}[/math]
  • Whilst it is difficult to measure the underlying change in the speed, at least for light waves, the effect on direction can be measured easily. The relationship between direction and refractive index is given by Snell’s law:
  • [math]n_1 sinθ_1 = n_2 sinθ_2[/math]
  • The values of [math]n_1[/math] and [math]n_2[/math] are the refractive indices in each medium.
  • The values of [math]θ_1[/math] and [math]θ_2[/math] are the angles that the ray of light makes to the normal to the interface between the two media at the point the ray meets that interface, as shown in figure 14
  • Figure 14 Refraction
  • Example:
  • What is the angle of refraction when a light ray passes from water ([math]n_1[/math] = 1.33) into glass ([math]n_2[/math] = 1.50), hitting the interface between the two at an angle of 48° to the normal?
  • Solution:
  • Refractive index of first medium = [math]n_1[/math] = 1.33
  • Refractive index of second medium = [math]n_2[/math] = 1.50
  • Angle of incident = [math] θ_1 = 45^0[/math]
  • [math]\begin{gather}
    n_1 \sin \theta_1 = n_2 \sin \theta_2 \\
    \sin \theta_2 = \frac{n_1 \sin \theta_1}{n_2} \\
    \sin \theta_2 = \frac{1.33 \sin 45^\circ}{1.50} \\
    \sin \theta_2 = 0.659 \\
    \theta_2 = \sin^{-1}(0.659) \\
    \theta_2 = 41.2^\circ
    \end{gather}[/math]

  • 46) Be able to calculate critical angle using

  • [math]sinC = \frac{1}{n}[/math]
  • Figure 15 Moving from refraction to total internal reflection.
  • ⇒ Critical angle:
  • The angle in the less dense medium is greater than the incident angle inside the denser medium. In the second diagram, the incident angle has been increased.
  • From Snell’s law, we find that at this critical angle, the ray would emerge in the less dense medium at an angle of 90° – it would emerge exactly along the interface.
  • If the angle within the denser medium is increased further, the emergent angle should increase also, but this would then be greater than 90°.
  • This means the ray emerges within the denser medium. This is not refraction, as that requires a change in medium. This means that Snell’s law cannot apply.
  • ⇒ Critical angle calculations:
  • From Snell’s law, we could find the critical angle for a material:
  • [math]n_1 sinθ_1 = n_2 sinθ_2[/math]
  • If we take medium 1 to be the optically denser material, then [math]θ_2[/math] must be 90° when the light is at the critical angle, [math]θ_1[/math] in medium 1.
  • [math]\begin{gather}
    n_1 \sin \theta_1 = n_2 \sin \theta_2 \\
    n_1 \sin \theta_c = n_2 \sin 90^\circ \\
    \sin 90^\circ = 1 \\
    n_1 \sin \theta_c = n_2 \\
    \sin \theta_c = \frac{n_2}{n_1}
    \end{gather}[/math]
  • If the situation involves a light ray emerging into air, then the equation becomes:
  • [math]sinθ_c = \frac{1}{n_1}[/math]
  • If we know the critical angle, then that will give us the refractive index for the material:
  • [math]n_1 = \frac{1}{sinθ_c}[/math]

  • 47) Be able to predict whether total internal reflection will occur at an interface

  • ⇒ Total internal reflection (TIR)
  • The wave energy is reflected inside the denser medium, and the angles follow the law of reflection. This is total internal reflection (TIR).
  • ⇒ Application of total internal reflection:
  • Fig 16 shows the movement of light through 45° prisms. The left-hand example is commonly used in periscopes, and the right-hand diagram is the fundamental basis for reflective signs.
  • Figure 16 Total internal reflection helps us direct light usefully.
  • A more complex use for TIR is in fiber optics. A thin glass fiber can guide light along its length by the repeated TIR at the internal edges. This may just be used for decorative lighting;
  • Fiber optics can be used to guide sunlight to the interior of large buildings. Alternatively, optical fibers can be used to carry information as light pulses (as in fiber broadband) or as actual images (in uses such as a medical endoscope – see fig 17)
  • Figure 17 A medical endoscope is used to view the body’s internal organs without cutting the patient open. Light is sent in along one optic fiber, and the reflection is carried away along the other for viewing by medical staff.

  • 48) Understand how to measure the refractive index of a solid material

  • ⇒ Refractive index:
  • A measure of the amount of refraction caused by different materials is called the refractive index, and its symbol is n.
  • The refractive index, n, is equal to the ratio of the speed of light in a vacuum to the speed of light in the material:
  • [math]n = \frac{c}{v}[/math]
  • Examples:

  • 49) Understand what is meant by plane polarization

  • ⇒ Plane polarization:
  • Transverse waves have oscillations at right angles to the direction of motion. In many cases, the plane of these oscillations might be in one fixed orientation.
  • Figure 18 shows the electric (red) and magnetic (blue) fields in an electromagnetic wave. In this example of a light wave, the electric fields only oscillate in the vertical plane.
  • The wave is said to be plane polarized or, more precisely, vertically plane polarized. For electromagnetic waves, the plane of the electric field’s oscillations is the one that defines its plane of Polarization
  • Figure 18 A polarized electromagnetic wave.
  • ⇒ Polarizing filters:
  • Unpolarized radiation can be passed through a filter that will transmit only those waves that are polarized in a particular plane.
  • Waves on a string could be polarized simply by passing the string through a card with a slit in it, which will then only allow oscillations to pass through if they are in line with the slit.
  • For light waves, the polarizer is a piece of plastic soaked with chemicals with long chain molecules, called a Polaroid sheet.
  • The Polaroid filter will only allow light waves to pass if their electric field oscillations are orientated in one direction.
  • Figure 19 Polaroid filters transmit light waves if their plane of polarization matches with the orientation of the filter.
  • Figure 19 illustrates the effects of Polaroid filters on light from a bulb. It starts unpolarized, with vibrations in all directions.
  • The first filter only permits vertical vibrations, so they are selected, and the light is transmitted vertically plane polarized.
  • The second filter is oriented in the same vertical direction, and the light will pass through this without change. The third Polaroid filter is oriented horizontally.
  • This blocks vertically polarized light, and so no waves are transmitted beyond it. The second and third filters are referred to as ‘crossed Polaroids’, as their orientations are at right angles and, together, they will always block all light.

  • 50) Understand what is meant by diffraction and use Huygens’ construction to explain what happens to a wave when it meets a slit or an obstacle

  • ⇒ Huygens’ Principle:
  • The Dutch scientist Christiaan Huygens came up with a principle for predicting the future movement of waves if we know the current position of a wavefront.
  • The basic idea is to consider that any and every point on the wavefront is a new source of circular waves travelling forwards from that point.
  • When the movement of these numerous circular waves is plotted, and then their superposition considered, the resultant wave will be the new position of the original wavefront.
  • Figure 20  Huygens’ construction for explaining the movement of waves.
  • Huygens’ geometrical system exactly explains all the basic phenomena that we see with light waves.
  • We can make drawings like figure 20 to correctly predict the movement of wavefronts in reflection, refraction, diffraction (figure 21), interference and straight-line propagation of light.
  • Figure 21  Huygen’s construction correctly predicts diffraction through a gap.

  • 51) Be able to use nλ = dsinθ for a diffraction grating

  • ⇒ Diffraction grating:
  • A diffraction grating is a device that will cause multiple diffraction patterns, which then overlap. This creates an interference pattern with a mathematically well-defined spacing between bright and dark spots.
  • It is a collection of a very large number of slits through which the waves can pass. These slits are parallel and have a fixed distance between each slit.
  • Figure 22 Diffraction grating
  • The pattern produced by each color passing through a diffraction grating follows the equation:
  • [math]nλ = dsinθ[/math]
  • where θ is the angle between the original direction of the waves and the direction of a bright spot, A is the wavelength of the light used, dis the spacing between the slits on the grating, and n is called the ‘order’. The order is the bright spot number from the central maximum (which is n = 0).
  • The grating spacing is often quoted as a number of lines per metre (or per millimetre), which means that to find d you will need to do an initial calculation in which:
  • [math]d = \frac{1}{\text{Number per metre}}[/math]
  • OR
  • [math]d = \frac{1 \times 10^{-3}}{\text{Number per millimetre}}[/math]

  • 52)CORE PRACTICAL 6: Determine the wavelength of light from a laser or other light source using a diffraction grating

  • ⇒ Experiment Overview
  • This experiment uses a diffraction grating to determine the wavelength of light from a laser or other light source.
  • ⇒ Formula
  • The wavelength (λ) can be calculated using the diffraction grating equation:
  • [math]dsin(θ) = nλ[/math]
  • Where:
  • – d: Distance between grating lines (grating spacing)
  • – θ: Angle of diffraction
  • – n: Order of diffraction (integer)
  • Figure 23 Diffraction grating
  • ⇒ Procedure
  • – Shine the laser through the diffraction grating.
  • – Measure the angle of diffraction (θ) for a specific order (n).
  • – Calculate the wavelength (λ) using the formula.
  • ⇒ Calculation
  • Rearrange the formula to solve for λ:
  • [math]λ = d * sin(θ) / n[/math]
  • By measuring θ and knowing d and n, you can determine the wavelength of the light.

  • 53) Understand how diffraction experiments provide evidence for the wave nature of electrons

  • ⇒ Electron diffraction:
  • As a diffraction pattern is a wave phenomenon, to observe this from a beam of electrons means that they must be behaving as waves. This is true whether we are passing waves through a gap
  • In 1927, Davisson and Germer tried to detect diffraction of electron ‘waves’ when they reflected from a crystal of nickel. Figure 24 illustrates the experimental setup. They measured the intensity of the beam at different angles for various accelerating voltages, and plotted a graph of their results.
  • Figure 24 Davisson and Germer reflected a beam of electrons from a nickel crystal and measured the intensity of the reflection at different angles.
  • Not only did Davisson and Germer prove experimentally that electrons can behave as waves, but their results also allow calculation of the distance between atoms in the nickel crystal.
  • This has given rise to advances in the study of atomic structures using electron beam crystallography.
  • Figure 25 Davisson and Germer’s electron beam reflection from a surface of atoms, acting as a reflection grating, showed variable intensity at different angles, exactly as is observed with waves.

  • 54) Be able to use the de Broglie equation

  • [math]λ = \frac{h}{p}[/math]
  • ⇒ De Brogile Equation:
  • In 1924, a French prince called Louis de Broglie suggested electrons could behave as waves and proposed an equation to calculate their wavelength.
  • Their wavelength is inversely proportional to the momentum they have when considered as particles.
  • [math]\begin{gather}
    \text{Electron wavelength (m)} = \frac{\text{Planck’s constant (J·s)}}{\text{Momentum (kg·m·s}^{-1}\text{)}} \\
    \lambda = \frac{h}{p}
    \end{gather}[/math]
  • Example
  • Calculate the wavelength of electrons travelling at 10% of the speed of light.
  • [math]\begin{gather}
    \lambda = \frac{h}{p} \\
    p = mv \\
    p = (9.11 \times 10^{-31}) \times (0.1 \times 3 \times 10^8) \\
    p = 2.73 \times 10^{-23} \, \text{kg·m·s}^{-1} \\
    \lambda = \frac{6.63 \times 10^{-34}}{2.73 \times 10^{-23}} \\
    \lambda = 2.43 \times 10^{-11} \, \text{m}
    \end{gather}[/math]

  • 55) Understand that waves can be transmitted and reflected at an interface between media

  • ?

  • 56) Understand how a pulse-echo technique can provide information about the position of an object and how the amount of information obtained may be limited by the wavelength of the radiation or by the duration of pulses

  • ⇒ Pulse-Echo Measurement:
  • The natural habitat for some bats is woodland, and they fly through the trees at 10-15 [math]kmh^{-1}[/math], depending on species. In order to catch insects, the bats need to be able to sense their location precisely.
  • Their well-known echolocation system, using very high frequency (50-100 kHz – ultrasound) sound pulses, gives a detailed perception of the world at distances of less than 5 metres.
  • At greater distances, the echo is too quiet for the bat to use.
  • Figure 26 Eco-Pulse equipment
  • The bat will make a ‘chirp’ through its nose. This sound pulse will typically last 3 milliseconds. When the sound hits nearby objects, it will be reflected back to the bat’s sensitive ears and its brain can accurately measure the time between making the sound and hearing the echo.
  • The bat’s brain has also evolved to calculate the distance to the reflecting object using the equation:
  • [math]\text{Distance} = \text{Speed} \times \text{Time}[/math]
  • Experiments with dolphins have shown that they use a similar echolocation system, but for them it is so good that dolphins can build up an image of the shapes of nearby objects.
  • They can ‘see’ with sound. This is probably also true for bats, to ensure that they eat an insect and not an insect-sized leaf, but dolphins can respond to experimental scientists better than bats.
  • Figure 27 Dolphin echolocation is good enough to make out the shapes of objects in murky waters.
  • For example, Venus has a very hot, high pressure, atmosphere with sulfuric acid clouds that block our view of the surface. Mapping the surface using a remote radar technique has avoided the need to land on the planet.

  • 57) Understand how the behavior of electromagnetic radiation can be described in terms of a wave model and a photon model, and how these models developed over time

  • 58) Be able to use the equation E = hf, that relates the photon energy to the wave frequency

  • ⇒ Wave Model:
  • The wave model describes electromagnetic radiation as oscillating electric and magnetic fields. Key features include:
  • – Wavelength (λ): Distance between successive peaks.
  • – Frequency (f): Number of oscillations per second.
  • – Speed (c): Product of wavelength and frequency ([math]c = λf[/math]).
  • ⇒ Photon Model:
  • For electromagnetic radiation, the energy of the photon can be calculated by multiplying the frequency by Planck’s constant, h.
  • This constant has an extremely small value because it represents the fundamental minimum possible step in energy. In SI units, [math]h = 6.63 × 10^{-34} J s[/math]. If we work on very small scales, photons cannot have energy values that differ by less than the Planck constant.
  • This means that there are some energy values that are impossible in our universe. Such a system of minimum sized steps is called quantization.
  • The photon model describes electromagnetic radiation as particles (photons) with energy:
  • [math]\begin{gather}
    \text{Photon energy (J)} = \text{Planck’s constant (J·s)} \times \text{Frequency (Hz)} \\
    E = hf
    \end{gather}[/math]
  • Where:
  • – E: Energy of the photon
  • – h: Planck’s constant
  • – f: Frequency of the radiation
  • ⇒ Historical Development:
  • The understanding of electromagnetic radiation evolved over time:
  • – Wave theory: Initially, light was thought to be a wave (Huygens, Young).
  • – Particle theory: Einstein’s photoelectric effect experiments supported the photon model.
  • – Wave-particle duality: Both models are now recognized as complementary descriptions of electromagnetic radiation.
  • ⇒ Electrons are particles:
  • Experiments that produce ions can demonstrate electrons behaving as particles because a fixed lump of mass and charge is removed from the atom in order to change the atom into an ion.
  • The charge to mass ratio is a unique identifying property of particles, and was first demonstrated for the electron by J.J. Thomson in 1897. Robert Millikan, in an experiment finally published in 1913, took this one step further to find the electron charge itself.
  • The fact that electrons hold a fixed amount of charge and a fixed mass indicates they are localized particles.
  • ⇒ Electrons are waves:
  • If electrons are made to travel at very high speeds, they will pass through gaps and produce a diffraction pattern.
  • They will also interact with a double-slit apparatus to produce the interference pattern seen when waves pass through two slits.
  • Diffraction and interference are not expected by classical particles, as they should simply travel straight through the slits. Observation of these experimental results proves that electrons can behave as waves.

  • 59) Understand that the absorption of a photon can result in the emission of a photoelectron

  • Photoelectron:
  • If ultraviolet light is shone onto a negatively charged zinc plate, the plate loses its charge. The explanation for this is that the light causes electrons to leave the metal, removing the negative charge.
  • The electrons that are released are called photoelectrons. However, variations on this experiment produce some surprising observations.

  • Figure 28 Threshold frequency and threshold wavelength of metal

  • There is a certain minimum amount of energy that an electron needs in order to escape the surface of the metal. This energy is called the work function, and has the symbol, ф.
  • The wave theory of light would allow some of the wave energy to be passed to the electrons to enable them to gain the work function and escape.
  • In the wave theory, the energy carried by a wave depends on its amplitude, which for light means its brightness.
  • So, the wave theory would predict that any color of light, if made sufficiently bright, should enable the release of photoelectrons.
  • Alternatively, if the wave were of smaller amplitude, it could shine for longer and slowly pass energy to the electrons, until they had gained as much as the work function and escape. However, this is not what is observed.
  • The same negatively charged zinc plate cannot be discharged by red light, no matter how bright or how long the illumination. No photoelectrons are emitted under red light.
  • There is a maximum wavelength for the light, above which no photoelectrons are ever emitted. This can be thought of equivalently as a minimum frequency for the light, known as the threshold frequency.
  • The explanation for these experimental observations can be summarized as:
  • – Light travels as photons, with a photon’s energy proportional to the frequency.
  • – When a photon encounters an electron, it transfers all its energy to the electron (the photon ceases to exist).
  • – If an electron gains sufficient energy – more than the work function – it can escape the surface of the metal as a photoelectron.
  • – Brighter illumination means more photons per second, which will mean a greater number of photoelectrons emitted per second.
  • – If an electron does not gain sufficient energy from an encounter with a photon to escape the metal surface, it will transfer the energy gained from the photon to the metal as a whole before it can interact with another photon. Thus, if the photon energy is too low, no photoelectrons are observed.
  • Figure 29 The emission of a photoelectron

  • 60) Understand the terms ‘threshold frequency’ and ‘work function’ and be able to use the equation

  • [math]hf = \phi + \frac{1}{2}mv_{\text{max}}^2[/math]

  • 61) Be able to use the electron-volt (eV) to express small energies

  • 62) Understand how the photoelectric effect provides evidence for the particle nature of electromagnetic radiation

  • ⇒ The Photoelectric effect equation:
  • Einstein developed an equation to explain the photoelectric effect, which is based on the conservation of energy.
  • The energy of a photon can be found from E = hf. During the photoelectric effect, the energy of a photon is transferred to an electron.
  • To escape a metal surface, an electron requires at least an amount of energy equal to the work function. It may also lose energy before escaping the metal due to other physical processes.
  • Only the remaining energy will be available for the electron emission kinetic energy to be emitted from the surface.
  • Therefore, the kinetic energy that the electron can have on emission from the metal surface is less than or equal to the difference between the photon energy and the work function.
  • This maximum possible kinetic energy mv²max determines the maximum initial velocity Vmax of the photoelectron. Einstein’s photoelectric effect equation then is:
  • [math]\frac{1}{2}mv_{\text{max}}^2 = hf – \phi[/math]
  • The conservation of energy, the photon energy transferred to the electron is used partly to escape the metal surface and partly for its kinetic energy on leaving the metal. So, a rearrangement of the equation is:
  • [math]hf = \frac{1}{2}mv_{\text{max}}^2 + \phi[/math]
  • ⇒ The Photoelectric cell Experiment:
  • The photoelectric effect equation to measure Planck’s constant and the work function for a metal. In a vacuum, we place the metal as the anode in a cell that has a gap to the cathode see figure 30.
  • When we shine light of a known frequency onto the anode, photoelectrons will be emitted and the current registered on the ammeter.
  • Figure 30 A photoelectric effect cell experiment circuit.
  • If we slowly increase the pd across the photoelectric cell, eventually the anode will become sufficiently positive that all photoelectrons will be stopped and attracted back to it, so the photoelectric current will be zero.
  • This stopping voltage, V, will give us the maximum kinetic energy of the photoelectrons, from the definition of voltage:
  • [math]\frac{1}{2}mv_{\text{max}}^2 = eV_s[/math]
  • Figure 31 Graphical analysis of results from a photoelectric cell experiment can determine Planck’s constant and the work function for the anode metal.
  • A range of light frequencies and find the stopping voltage for each, we can plot a graph of the photoelectron maximum kinetic energy, on the y-axis, against frequency, on the x-axis.
  • Comparison with the equation for a straight line shows us that the graph should produce a straight best-fit line and the gradient will be equal to Planck’s constant.
  • [math]\begin{gather}
    \frac{1}{2}mv_{\text{max}}^2 = hf – \phi \\
    y = mx + c
    \end{gather}[/math]
  • The y-intercept will represent the value of the work function, ф. Also, when the value of y is zero (the x-intercept) then the photon energy must equal the work function.
  • [math]\begin{gather}
    \frac{1}{2}mv_{\text{max}}^2 = hf – \phi \\
    0 = hf – \phi \\
    hf = \phi
    \end{gather}[/math]
  • This means that the value of the x-intercept will give the threshold frequency for the anode metal.
Metal Work Function / eV
Cadmium 4.07
Caesium 2.10
Iron 4.50
Nickel 5.01
Zinc 4.30
  • Example:
  • What is the work function of potassium if green light with a wavelength of 510 nm shining on it produces photoelectrons that have a maximum kinetic energy of 0.14 eV?
  • Solution:
  • [math]\begin{gather}
    \frac{1}{2}mv_{\text{max}}^2 = hf – \phi \\
    \phi = hf – \frac{1}{2}mv_{\text{max}}^2 \\
    \frac{1}{2}mv_{\text{max}}^2 = 0.14\, \text{eV} = 0.14 \times 1.6 \times 10^{-19} \\
    \frac{1}{2}mv_{\text{max}}^2 = 2.24 \times 10^{-20} \, \text{J} \\
    hf = \frac{hc}{\lambda} \\
    hf = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{510 \times 10^{-9}} \\
    hf = 3.9 \times 10^{-19} \, \text{J} \\
    \phi = 3.9 \times 10^{-19} – 2.24 \times 10^{-20} \\
    \phi = 3.68 \times 10^{-19} \, \text{J}
    \end{gather}[/math]

  • 63) Understand atomic line spectra in terms of transitions between discrete energy levels and understand how to calculate the frequency of radiation that could be emitted or absorbed in a transition between energy levels.

  • ⇒ Electron Energy Levels:
  • Electrons in semiconductors can have varying amounts of energy, and that the energy they have can put them into the valence band or conduction band.
  • These energy bands are wide in solids – there is a large range of values of energy that the electron could have and still be in that band.
  • In free atoms, such as those in a gas, the energy values that the electrons could have are limited to a small number of exact values, often called energy levels.
  • Figure 32 An energy level diagram for hydrogen. The energy values are negative, as we have to put energy in to lift the electron up from the ground state.
  • ⇒ Excitation and De-excitation:
  • Figure 32 illustrates the energy values that electrons of hydrogen could have. Under normal circumstances, an electron in an atom of hydrogen would be in its ground state. This is the lowest energy level, with a quantum number (or level) of n = 1.
  • In order to move up energy levels, the electron must take in some energy. This is called excitation. Electrons can become excited if the atom collides with another particle.
  • Alternatively, if the electron absorbs a photon that has exactly the correct amount of energy, the electron can jump to a higher energy level.
  • For example, the difference in energy between the ground state ([math]-2.18 x 10^{-18} J[/math]) and the n = 2 state ([math]-5.45 x 10^{-19} J[/math]) is [math]1.635 x  10^{-18} J[/math]
  • A photon with exactly this energy could be absorbed by an electron in the ground state of hydrogen, and this would lift the electron up to the energy level above the ground state, n = 2, and the photon would no longer exist. This is illustrated in figure 33.
  • Figure 33 Electron excitation in a hydrogen atom. For ease of reference, the energy levels are numbered with integers from n = 1 for the ground state, upwards.
  • An incident photon that does not have the energy exactly equivalent to a jump between the current position of the electron and one of the higher levels will not be absorbed – the photon and electron will not interact at all.
  • If gas atoms are illuminated by a range of frequencies (colors), those with the correct frequency values will be absorbed, so there will be some colors missing from the light after it passes through the gas.
  • Figure 34 Electron de-excitation in a hydrogen atom.
  • If an electron is already excited, after a random amount of time it will de-excite. This may involve dropping straight down to the ground state, or it may drop to an intermediate level if there is one. Figure 34 shows an excited hydrogen electron dropping from energy level n = 4 down to n = 3.
  • As the electron ends up with less energy than it had, the conservation of energy requires that this energy is emitted as a photon with exactly the energy difference between the levels.
  • The frequency of the emitted photon can be calculated from the equation for photon energy, [math]E = hf[/math]. So, a collection of gas atoms that are excited will emit light with a particular collection of frequencies, dependent on which element the gas atoms are.
  • ⇒ Line Spectrum:
  • Light made up of multiple wavelengths (colors) can be split up to show which colors are present. This could be done using a diffraction grating in which the amount of diffraction is dependent on the wavelength, and so the various colors will spread different amounts.
  • The resulting spectrum will often be a series of individual lines, if the original light contained only a few wavelengths. Such a line spectrum is the typical result of exciting the atoms of a gas, perhaps by heating the gas.
  • Figure 35 Hydrogen emission line spectrum.
  • The gas has electricity passed through it, and this will excite the hydrogen atoms. Each colored line is a wavelength of light given off as a result of an electron dropping between two energy levels.
  • The different energy gaps cause the difference in wavelengths emitted. When viewed all together, the colors merge, and the gas glows a purple color.
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