Wave Properties

Waves have some properties that aren't always intuitive.

  • Waves only transmit energy, not matter.
  • Multiple waves can be in the same position at the same time.
  • Waves move at a constant speed determined by their medium.
  • Waves expand in all directions.
  • Interference

    Unlike particles, waves pass through each other without any effect. When two waves are in the same position we say they are in superposition.

    The observed amplitude for waves in superposition is the sum of each wave's amplitude.

    Ears are able to detect separate sound frequencies, even though they enter the ear in a superposition. This spectrogram lets you visualize the frequencies that make up various sounds.

    Waves add constructively when both wave amplitudes are positive or both are negative. This produces a higher observed amplitude. Constructive interference is how two speakers playing the same song are louder than just one.

    Waves add destructively when one wave amplitude is positive and the other is negative. This produces a lower observed amplitude. Destructive interference is put to use in noise cancelling headphones.

    The Wedge is a famous surf spot that has some dangerously fun waves because of constructive interference from a wave reflected off a jetty.

    Question: A wave is created on each side of a rope with an amplitude of 0.3m. The waves interfere as they pass through each other.

    What is the amplitude of the rope when a crest lines up with a crest?
    What is the amplitude of the rope when a crest lines up with a trough?
    solution

    Two crests add constructively to make a larger amplitude.

    $$ 0.3\, \mathrm{m} + 0.3\, \mathrm{m} = 0.6\, \mathrm{m}$$

    A crest and trough add destructively to cancel out.

    $$ 0.3\, \mathrm{m} - 0.3\, \mathrm{m} = 0\, \mathrm{m}$$
    Example: Draw the superposition of these waves. They have the same wavelength, so they only add constructively.
    solution
    Example: Draw the superposition of these waves. They have the same wavelength but they are out of phase with each other. They can only add destructively.
    solution
    Example: Draw the superposition of these waves.
    solution
    Example: Draw the superposition of these waves.
    solution
    Example: Draw the superposition of these waves.
    solution
    Example: Draw the superposition of these waves.
    solution
    Example: Draw the superposition of these waves. When waves have similar frequencies they produce a pattern called beats.
    solution

    See the Pen Beats by Landgreen by Landgreen ( @lilgreenland) on CodePen.

    Adding many sine waves together can produce some interesting waveforms. The two below are starting to look like sawtooth and triangle, but they need more contributing sine waves.

    Interference is even more complex and interesting in higher dimensions. Take a look at these 2-D ripple tank simulations:

    Dipole source: Two wave sources interfere.
    Intersecting Planes: Two wave planes interfere at 90°.
    Beats: Two wave sources with slightly different frequencies.

    Polarization

    Polarization is a property of transverse waves that describes the angle of oscillation. The polarization can be any angle perpendicular to the direction the wave is moving.

    Most light sources are unpolarized. They oscillate a bit in every direction. Light can become polarized after passing through a polarization filter that only removes the light that oscillates at a certain angle.

    A polarization filter that blocks vertical light will let horizontally polarized light through. Polarization filters are often used in sunglasses or 3-D movies.

    Unpolarized light can also become partially polarized after reflecting off some shiny surfaces. The reflected light becomes polarized parallel to the surface. For example, a flat road reflects horizontally polarized light.

    Question: What polarization does the sun have?
    solution
    Most light sources, including the sun, don't have one polarization. They have a bunch of random polarizations.

    Although, the refracted blue sun light from Rayleigh scattering is polarized towards the sun.
    Question: What classifications of waves can and can't be polarized?
    solution
    Transverse wave, like light, can be polarized.

    Longitudinal waves, like sound, can't be polarized.
    (except for sound waves in solids, which can be transverse)
    Question: What angle would a polarization filter need to be to block the glare from a highway road?
    solution
    A horizontal polarization filter will block the glare from a road.
    Question: Imagine looking at sunlight through a horizontal polarization filter and a vertical polarization filter. What color would you see?
    solution
    It would just be black, since all the light would be blocked.

    Diffraction

    Waves tend to spread out in every direction possible within their medium. Diffraction is when a wave spreads out after some of the wave hits a barrier. This explains how sound can be heard through an open door even when there is no line of sight.

    Diffraction can be explained by thinking of every point on a wavefront as the source of spherical waves.

    You can see examples of diffraction in these 2-D ripple tank simulations:

    Single Slit: Waves spread out in a circle as they pass through a gap.
    Double Slit: Notice the pattern of constructive and destructive interference.
    Half Plane: Waves curve around a wall.
    Obstacle: Waves of the right wavelength can almost ignore a small barrier.

    Example: Draw the pattern of the waves that would pass through the gap as the waves propagate to the right.
    solution
    Example: Draw the pattern of the waves that would pass through the gap as the waves propagate to the right.
    solution
    This is the famous double slit experiment.
    Notice the alternating constructive and destructive interference pattern.

    Reflection and Refraction

    Reflection and refraction occur when the wave speed changes as it enters a new medium. Part of the wave refracts into the new medium, and part of the wave reflects back into the old medium.

    These effects are produced because media have different propagation speeds.

    Incident ray Normal Reflected ray Refracted ray θ i θ r θ R

    Refraction = After entering a new medium, a wave will change its direction of propagation.

    When slowing down, the wave will bend towards the normal.
    When speeding up, the wave will bend away from the normal.

    Reflection = When hitting a new medium, part of a wave stays in the original medium. The angle between the incident ray and the normal line equals the angle between the reflected ray and the normal. $$\theta_i = \theta_r$$

    Most substances are bumpy at the molecular level. Because of the bumpiness, they scatter reflected light in many directions. Scattered light is called diffuse light. Materials like dirt, skin, and wallpaper reflect diffuse light.

    Smooth materials produce a clear mirror image, called a specular reflection. Specular reflections are found in materials like metal, glass, and water. Mirrors are typically made by coating one side of a plate of glass with metal.

    Example: How will the light ray bend as it enters the denser media? Complete the path of light in the diagram above.
    solution

    It will refract up, towards the normal.


    Simulation: Try out this light bending simulation. Can you figure out how to produce total internal reflection? This is when the light only reflects and doesn't refract.
    solution
    Total internal reflection occurs when the light starts in the slower media and hit the faster media at a small angle.
    Simulation: Use the simulation to find out how the angle of incidence relates to the angle of reflection? These are the angles between the wave and the line normal to the media barrier.
    solution
    They are equal.
    Simulation: Dispersion is a phenomenon where different frequencies refract different amounts. Use the simulation to see how refraction works differently for high vs. low frequency?
    solution
    The wavelength/color/frequency of the light changes the refraction. A blue light has a shorter wavelength and it will refract more.

    When you shine a light on a substance the color coming from the substance is reflected light.

    When a wave is reflected the interference pattern can produce standing waves.

    Here are some examples of reflection and refraction in a 2-D ripple tank:

    Snell's Law

    When a wave enters a new medium several ratios have equal proportions.

    θ 1 2 θ

    $$\frac{sin \theta_1}{sin \theta_2} = \frac{v_1}{v_2} = \frac{ \lambda_1}{\lambda_2} =\frac{n_2}{n_1} $$

    \( \theta_1 \) = angle of incidence
    \( \theta_2 \) = angle of refraction
    \(v\) = propagation speed [m/s]
    \(\lambda\) = wavelength [m, meters]
    \(n\) = index of refraction [no units]

    It helps to draw a quick diagram for each problem.
    Label which medium is going to be 1 and which is 2.
    Focus on just two ratios, and set them equal to each other.

    Example: 7.0 mm light travels from air to water. Use the index of refraction to find the wavelength of the light in water.
    solution $$\text{1 = air} \quad \quad \text{ 2 = water}$$ $$\frac{n_2}{n_1} = \frac{\lambda_1}{\lambda_2}$$ $$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1}$$ $$\lambda_2 = \frac{\lambda_1 n_1}{n_2} $$ $$\lambda_2 = \frac{(0.0070) (1)}{1.33} $$ $$\lambda_2 = 0.0053\, \mathrm{m}$$ $$\lambda_2 = 5.3\, \mathrm{mm}$$
    Example: Why can't we compare the ratio of the frequencies in each medium?
    solution
    Wavelength, speed, and angle change in each medium, yet frequency doesn't change.
    Example: Light enters air from glass at 36.0º to the normal. This glass has an index of 1.52. What angle does the light leave relative to the normal?
    solution $$\text{1 = glass} \quad \quad \text{ 2 = air}$$ $$\frac{\mathrm{sin} \theta_1}{\mathrm{sin} \theta_2} = \frac{n_2}{n_1} $$ $$n_1 \mathrm{sin} (\theta_1) = n_2 \mathrm{sin} (\theta_2) $$ $$1.52 \mathrm{sin} (36.0) = 1 \mathrm{sin} (\theta_2) $$ $$ \mathrm{sin}^{-1} (0.893) = \mathrm{sin}^{-1}(\mathrm{sin} (\theta_2)) $$ $$63.3^{\circ} = \theta_2 $$
    Example: Diamonds have a very high index of refraction. Its high index is why they seem sparkly. How fast does light move through a diamond?
    hint
    Setup a situation where the light goes from air to diamonds.
    Build an equation with the index of refractions and the velocities.

    solution $$\text{1 = diamond} \quad \quad \text{ 2 = air}$$ $$\frac{v_1}{v_2} = \frac{n_2}{n_1} $$ $$v_1 = \frac{n_2 v_2}{n_1}$$ $$v_1 = \frac{(1)(3\times 10^8)}{2.42}$$ $$v_1 = 1.24\times 10^8 \mathrm{\tfrac{m}{s}}$$

    As the light wave enters a new medium some of the light is reflected, and some is refracted. If the angle of refraction is 90º or higher, the light will only reflect. This is called total internal reflection. The angle of incidence that produces a 90º angle of refraction is called the critical angle.

    Air Water n 1 n 2 Critical angle Refracted ray θ 1 θ 1 θ 2 θ 2 θ c Total internal reflection Incident ray

    You can see total internal reflection if you swim under some water and look upwards.

    Example: A 600 nm wave of red light starts in water and enters air. What is the critical angle for total internal reflection?
    hint
    Look up the index of refraction for air and water.
    Set the angle of refraction to 90º.
    Solve for the angle of incidence.

    solution $$\text{1 = water} \quad \quad \text{ 2 = air}$$ $$\frac{\mathrm{sin} \theta_1}{\mathrm{sin}\theta_2} = \frac{n_2}{n_1} $$ $$n_1\mathrm{sin} \theta_1 = n_2\mathrm{sin} \theta_2$$ $$(1.33)\mathrm{sin} \theta_1 = (1)\mathrm{sin}(90)$$ $$\mathrm{sin} \theta_1 = \frac{1}{1.33}$$ $$\mathrm{sin}^{-1}\mathrm{sin} \theta_1 = \mathrm{sin}^{-1}(0.751)$$ $$\theta_1 = 48.6 \degree$$
    Simulation: Use the simulation and Snell's law to calculate the index of refraction for mystery substances A and B. Use the simulated protractor to measure the angles or use the speed tool.
    solution A $$\text{1 = air} \quad \quad \text{ 2 = mystery A}$$ $$\theta_1 = 39.0^{\circ} \quad \quad \quad n_1 = 1.0$$ $$\theta_2 = 15.0^{\circ} \quad \quad \quad n_2 = \ ?$$ $$\frac{\mathrm{sin} \theta_1}{\mathrm{sin} \theta_2} =\frac{n_2}{n_1} $$ $$n_2 = \frac{n_1 \mathrm{sin} \theta_1}{\mathrm{sin} \theta_2} $$ $$n_2 = \frac{(1.0) \mathrm{sin} (39.0)}{\mathrm{sin} (15.0)} $$ $$ n_2 = 2.43 $$
    solution B $$\text{1 = air} \quad \quad \text{ 2 = mystery B}$$ $$\theta_1 = 39.0^{\circ} \quad \quad \quad n_1 = 1.0$$ $$\theta_2 = 26.5^{\circ} \quad \quad \quad n_2 = \ ?$$ $$\frac{\mathrm{sin} \theta_1}{\mathrm{sin} \theta_2} =\frac{n_2}{n_1} $$ $$n_2 = \frac{n_1 \mathrm{sin} \theta_1}{\mathrm{sin} \theta_2} $$ $$n_2 = \frac{(1.0) \mathrm{sin} (39.0)}{\mathrm{sin} (26.5)} $$ $$ n_2 = 1.41 $$

    The Doppler Effect

    The Doppler effect is heard when a car moves past a stationary observer as a "vvvvvvvvvvvvvVVVVVVVVRRRRROOOOOOMMMMMMMMmmmmmmmmmm". The observed sound drops in frequency very quickly as the car zooms past.

    When an observer is moving towards a wave source, the observer will see the wave's frequency as higher and the wavelength as shorter.

    When an observer is moving away from a wave source, the observer will see the wave's frequency as lower and the wavelength as longer.

    A sonic boom occurs when the source of sound is moving at the speed of sound. Can you produce a sonic boom in the simulation below?

    See the Pen Doppler effect by Landgreen ( @lilgreenland) on CodePen.

    Light can also be Doppler shifted, but it is harder to notice because light moves so fast. A light source near the speed of light will noticeably shift in color.

    The convention is to say that light is blueshifted as it moves towards you and redshifted as it moves away.

    Astronomers use this color shifting to calculate the relative motion of far away objects. In 1929 Edwin Hubble published the observation that almost all galaxies are redshifted. This was the initial evidence that the universe is expanding.

    Redshifted light from stars expanding away from us is actually the main reason why the sky is dark at night.

    The Doppler effect is used in laser cooling to produce temperatures near absolute zero.

    Radar guns use the Doppler effect to measure relative speed. They bounce radio waves off moving objects to measure the frequency shift of the reflected waves. The frequency shift indicates the relative speed.

    Question: Imagine you are driving towards a cop with a radar gun. How will your motion affect how they see your reflected radio waves?
    solution
    The cop will observe your light at a higher frequency and a shorter wavelength.
    Back