Waves W8 Flashcards
(24 cards)
Define and explain refraction
Refraction: The change in direction of a wave when it passes through a boundary between mediums of different density.
Refraction is caused by a change in speed of different parts of the wavefront as they hit the boundary of the medium (a transparent material). When a wave refracts, its speed and wavelength change, but its frequency remains the same. When the light ray is incident on the boundary at 90°: The wave passes straight through without direction, because the whole wavefront enters the boundary at the same time.
What are the conditions for refraction?
The angles of incidence and refraction are measured from the normal line. This is drawn at 90° to the boundary between the two media.
When light passes from a less optically dense medium to a more optically dense medium, (e.g. air → glass): The refracted light has a lower speed and a shorter wavelength than the incident light, therefore resulting in a smaller angle of refraction bending towards the normal and vice versa into a more dense medium.
- Explain the refractive index and what is the refractive index formula?
- What is the Snell’s law formula?
- The refractive index, n of a material tells us how optically dense it is. The refractive index of air is n = 1. Media that are more optically dense than air will have a refractive index of n > 1.
The refractive index of a material is calculated using the equation: n = c/c(small s)
Where: n = refractive index of material; c = the speed of light in a vacuum (m s–1); cs = the speed of light in a substance (m s–1) - The refractive index of a material is calculated using the equation: n1 sin theta1 = n2 sin theta2
Where:
n1 = the refractive index of material 1 (the material that the ray goes through first)
n2 = the refractive index of material 2 (the material that the ray goes through second)
θ1 = the angle of incidence of the ray in material 1
θ2 = the angle of refraction of the ray in material 2
- Explain the Critical Angle
- What is the critical angle formula?
- The larger the refractive index of a material, the smaller the critical angle. As the angle of incidence is increased, the angle of refraction also increases, until the angle of incidence reaches the critical angle.
When the angle of incidence = critical angle then: Angle of refraction = 90°; The refracted ray is refracted along the boundary between the two materials.
When the angle of incidence < critical angle then: the ray is refracted and exits the material.
When the angle of incidence > critical angle then: the ray undergoes total internal reflection. - Sin theta c = n2 / n1
Where: n1 = refractive index of material 1; n2 = refractive index of material 2; θc = critical angle of material 1
The formula finds the critical angle of the denominator of the fraction
The critical angle can also be calculated using the angles of incidence and refraction:
Sin theta c = n2 / n1 = sin theta 1 / sin theta 2
Explain TIR
Total internal reflection is a special case of refraction that occurs when: The angle of incidence within the denser medium is greater than the critical angle (I > θc), and the incident refractive index n1 is greater than the refractive index of the material at the boundary n2 (n1 > n2). Total internal reflection follows the law of reflection; angle of incidence = angle of reflection. TIR is more likely in optically dense materials.
- Define diffraction
- Describe the conditions for diffraction
- List some examples of diffraction
- Diffraction: the spreading out of waves after they pass through a narrow gap or around an obstruction.
After passing through a gap: the waves spread out so they have curvature, and the wave has a smaller amplitude because the barrier on either side of the gap absorbs wave energy. The only property of a wave that changes when it diffracts is its amplitude, while the wavelength of the wave remains the same. - No diffraction occurs when gaps are very small or very big in comparison to the wavelength of the waves. When the wavelength of the wave and the width of the gap are similar in size, then diffraction occurs. When the wavelength is bigger than the gap, more diffraction occurs, and the wave spreads out more after passing through.
- Radio waves moving in between or around buildings
Water waves moving through a gap into a harbour
Describe the single slit monochromatic diffraction pattern
The diffraction pattern of light passing through a single slit is a series of light and dark fringes on a screen.
The bright fringes are areas of maximum intensity, produced by the constructive interference of each part of the wavefront as it passes through the slit
The dark fringes are areas of zero or minimum intensity, produced by the destructive interference of each part of the wavefront as it passes through the slit
The central maximum is:
Much wider and brighter than the other bright fringes
Much wider than that of the double-slit diffraction pattern
On either side of the wide central maxima are the much narrower and less bright maxima. These get dimmer as the order increases.
- Describe the single slit intensity pattern for monochromatic light
- Describe the intensity pattern for white light
- The features of the single slit intensity pattern are:
The central maximum is the central bright fringe with the greatest intensity
The dark fringes are regions with zero intensity
As you move away from the central maxima either side, the intensity of each bright fringe reduces. - A source of white light diffracted through a single slit will produce the following intensity pattern:
The central maxima is equal in intensity to that of monochromatic light
The non-central maxima are wider and less intense
The fringe spacing between the maxima get smaller
The amount of red wavelengths in the pattern increases with increasing maxima, n increases from n = 1, 2, 3…
The amount of blue wavelengths decrease with increasing maxima
Describe the single slit diffraction pattern for white light
The central maximum is bright white because constructive interference from all the colours happens here. It is wider and brighter than the other bright fringes, and wider than the central maximum of the double-slit diffraction pattern. All the other maxima are composed of a spectrum: where the shortest wavelength (violet) will appear closest to the central maximum as it is diffracted less, while the longest wavelength (red) will appear furthest from the central maximum as it is diffracted more.
- Explain the effect of wavelength on fringe width
- Explain the effect of slit width on fringe width
- An increased wavelength will increase in the angle of diffraction, which increases the width of the bright minima. Red light (with the longest wavelength) will produce a diffraction pattern with wide fringes, while blue light will produce narrow fringes.
- If the slit was made narrower: The angle of diffraction is greater, so the waves spread out more beyond the slit
The intensity graph will show that: the intensity of the maxima decreases, so the width of the central maxima increases, and the spacing between fringes is wider.
- What is diffraction grating?
- What is the diffraction grating equation?
- A diffraction grating is a piece of optical equipment that also creates a diffraction pattern when it diffracts monochromatic light into bright and dark fringes, and white light into its different wavelength components. A diffraction grating consists of a large number of very thin, equally spaced parallel slits carved into a glass plate. Diffraction gratings are useful because they create a sharper pattern than a double slit. This means their bright fringes are narrower and brighter while their dark regions are wider and darker.
- The regions where constructive interference occurs are also the regions of maximum intensity. Their location can be calculated using the diffraction grating equation:
d sin θ = n λ
Where: n is the order of the maxima, the number of the maxima away from the central (n = 0); d is the distance between the slits on the grating (m); θ is the angle of diffraction of the light of order n from the normal as it leaves the diffraction grating (°); λ is the wavelength of the light from the source (m).
- How do you calculate slit spacing?
- What is angular separation?
- Diffraction grating sizes are determined by the number of lines per millimetre (lines / mm) or lines per m. This is represented by the symbol N. d is the distance between the slits on the grating. d can be calculated from N using the equation: d = 1 / N
- The angular separation of each maxima is calculated by rearranging the grating equation to make θ the subject. The angle θ is taken from the centre meaning the higher orders of n are at greater angles. The angular separation between two angles is found by subtracting the smaller angle from the larger one. The angular separation between the first and second maxima at n1 and n2 is θ2 – θ1.
- What is the order of maxima?
- How do you calculate the angle between two order lines?
- The maximum angle of diffraction with which maxima can be seen is when the beam is at right angles to the diffraction grating. This means θ = 90 degrees and sin θ = 1. The highest order of maxima visible is therefore calculated by the equation: n = d / λ. Since n is an integer number of maxima, if the value obtained is a decimal it must be rounded down to determine the highest-order visible, e.g If n is calculated as 2.7 then n = 2 is the highest-order visible.
- Calculate the two angles between the order lines and the normal, then add these together (do not directly measure the angle between the lines).
What are the applications of diffraction gratings?
They are used in spectrometers to analyse light from stars, to determine their composition, and carry out chemical analysis to observe the spectra of materials.
They are also used in X-ray crystallography: X-rays are directed at a thin crystal sheet which acts as a diffraction grating to form a diffraction pattern. This is because the wavelength of X-rays is similar in size to the gaps between the atoms. This diffraction pattern can be used to measure the atomic spacing in certain materials.
- Define and explain the principle of superposition
- Define interference and explain the types of interference
- Define coherence and give some examples
- Superposition: two or more waves arrive at the same point and overlap, and their amplitudes combine.
The principle of superposition states that: When two or more waves overlap at a point, the displacement at that point is equal to the sum of the displacements of the individual waves. - Interference is the effect of superposition, which can be seen clearly when waves overlap in phase or antiphase.
When two waves are in antiphase, their combined effect means they cancel each other out; this is destructive interference
When two waves are in phase, their combined effect makes the resultant wave amplitude larger; this is constructive interference. - Coherence occurs when waves are the same frequency and have a constant phase difference.
Examples of interference from coherent light sources are: Monochromatic laser light; Sound waves from two nearby speakers emitting sound of the same frequency.
- Define path difference
- What is the difference between path and phase difference?
- Path difference is defined as: The difference in distance travelled by two waves from their sources to the point where they meet. (TIP: The path difference is how much longer, or shorter, one path to a point is than the other. In other words, the difference in the distances. Make sure not to confuse this with the distance between the two paths where the slits start.)
- Path difference compares the amount of progress made by waves along a path, while phase difference compares the distance between the phases (peaks and troughs) of coherent waves that are normally travelling parallel to each other at a point.
- What are the properties of lasers, and what is the diffraction pattern produced by a laser?
- What are the safety issues with lasers?
- Lasers form light that is coherent and monochromatic.
It consists of areas of constructive interference – the bright fringes, and areas of destructive interference – the dark fringes. - Lasers produce a very high-energy beam of light, which can cause permanent eye damage or even blindness. Therefore, you must take the following precautions: never look directly at a laser or its reflection; don’t shine the laser towards a person; wear laser safety goggles.
- What are the conditions for constructive and destructive interference?
- Explain wavefront diagrams
- The condition for constructive interference is: path difference = n lambda
The condition for destructive interference is:
path difference = (n + ½) lambda
Where: lambda is the wavelength of the wave in metres, and n is any integer. - A curved line represents each wavefront (peak or trough), where you can count the number of wavelengths to determine the type of interference that occurs. This is done by counting the number of wavelengths from the point to the centre of each individual wavefront.
- Explain interference using sound waves
- Explain interference using microwaves
- Sound waves are longitudinal waves made up of compressions and rarefactions. Constructive interference occurs when the compressions and rarefactions from each wave line up and the sound appears louder. Destructive interference occurs when a compression from one wave lines up with a rarefaction from the other and vice versa. The two waves cancel each other out, so zero sound is heard. This is the technology used in noise-cancelling headphones.
- Two-source interference for microwaves (and other electromagnetic waves) can be detected with a moveable microwave detector. The detector picks up a maximum amplitude or intensity in regions of constructive interference. The detector picks up a minimum or zero amplitude, so no signal in regions of destructive interference.
- What is the formula of intensity variation with amplitude?
- How do you calculate the number of quiet spots heard between two speakers in phase in a specified distance?
- The intensity (power per unit area) of a wave is proportional to the energy transferred by the wave. The intensity of a wave is proportional to the square of the amplitude
I ∝ A^2
Where: I = intensity of the wave in W m–2; A = amplitude of the wave in metres (m) - Calculate the wavelength (with the speed of sound in air and frequency given). Use the destructive interference formula to calculate the smallest path difference (n) – the location of the first quiet spot. Then add the path difference multiple times up to and including the maximum distance.
W1-5
How to calculate the superposition and find the resultant wave of waves that are:
1. In phase?
2. Out of phase?
- For waves that are in phase, you add together the amplitude to get the total amplitude. If one amplitude is above the rest point and the other is below the rest point, then subtract the one below the rest point from the one above. If the value is positive, then the amplitude is above the rest point. If negative, then the amplitude is below the rest point.
- For waves that are out of phase, where they intersect will be their maximum and minimum amplitudes. Multiply the amplitude of where they intersect by 2 to find the maximum amplitude.
W1-5
1. What is the name of the effect that causes a wave to return, and what do you need to remember when calculating the speed for these waves?
2. Describe what the observer would see if one polarising filter is placed in front of another and rotated around 360 degrees.
- A reflection is when a sound wave changes direction when it hits the boundary between two different materials. When calculating the speed of these waves, you need to remember to double the distance, or half the time, as the wave is travelling away from the observer and back again.
- There is a variation in intensity between maximum and minimum amount of light allowed through. There are two maxima and two minima in a 360 degrees rotation.
W1-5
1. How you would describe the motion of a particle during a complete transverse wave cycle just before it moves vertically upwards
2. What do you need to remember when calculating the speed of sound based on a reflection?
- The oscillations are perpendicular to the direction of wave travel. (1) It oscillates from equilibrium to maximum positive displacement, and back to equilibrium. (1) (Do not state that the particle moves up and down).
- As the wave is travelling to the wall and back, don’t forget to half the time or double the distance.
Define all of the terms in the formula for the double slit experiment.
The fringe separation formula is:
w = λD / s
λ is the wavelength of light.
D is the distance from the slits to the screen.
s is the slit spacing.
