Waves Flashcards

(39 cards)

1
Q

What is Superposition

A

When two or more waves overlap, the resultant displacement at a point is equal to the sum of the individual displacements at that point. This is the principle of superposition.

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2
Q

What are the rules of refraction

A

The angle of incidence = angle of reflection, i = r
The reflected ray is in the same plane as the incident ray.
All angles are measured from the normal, a line running at 90o to the surface.

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3
Q

What is refraction

A

Refraction is the change of direction a light ray undergoes when it enters a medium with a different optical density.
Light travels at different speeds in materials with different optical densities resulting in a change in the speed of the light which causes change in direction.

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4
Q

What is the critical angle

A

This angle is called the critical angle, C - it is the angle of incidence at which the corresponding angle of refraction is 90o (along the boundary).

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5
Q

What are the necessary conditions for total internal reflection to occur?

A

The wave must be travelling from a more dense medium to a less dense medium
Angle of incidence must be greater than the critical angle

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6
Q

What are optic fibres

A

Optical fibres are basically long thin bits of glass that carry electromagnetic waves over long distances (usually visible light).

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6
Q

come up with a short experimental method of identifying the refractive index of a solid.

A

Trace around the solid on a blank sheet of paper.
Use a ray box to draw a range of values for i and measure their corresponding values of r.
Plot sini vs sinr and calculate the gradient.
The gradient will be the refractive index of the solid.

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7
Q

How do optic fibres work?

A

Light is directed into the narrow fibre.
The light is travelling from air into a glass medium.
This means light is refracted and bends towards the normal.
In a normal glass fibre the light would refract out of the glass at this point.
With the cladding in place light is now travelling from a more dense medium to a less dense medium.
Provided the i > C, TIR will occur
One of the inherent problems with fibre optics is that light can take more than one path within the fibre.

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8
Q

What are the problems with optic fibres and how do you overcome them?

A

Using very high purity glass.

Using an infrared light source.

The glass is surrounded by another layer of glass that has a much lower refractive index. This CLADDING is optically less dense than the CORE glass.

The cladding is surrounded by a buffer coating that protects the fibre from damage and moisture.

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9
Q

What is multi path dispersion

A

Different rays of light take DIFFERENT PATHS
So they will arrive at DIFFERENT TIMES
Which will SMEAR/DISTORT the signal

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10
Q

How can multi path dispersion be corrected?

A

So how do we reduce multipath dispersion?

Use a monochromatic light source (more about these later in the course).

Use a very thin piece of optic fibre – this forces the light to travel almost entirely along the axis (the difference in the path of any reflected ray is negligible).

Choosing a cladding with a refractive index as close to the core as possible.

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11
Q

Optical Bench

A

Very simply, move the lens back and forth until the image is in focus
Measure u and v
Move the position of the object closer to the screen and repeat steps 1 and 2. do this for a range of object distances.
Plot 1/u verses 1/v
Find the intercepts of both axis. When 1/u = 0, 1/v = 1/f!

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12
Q

What is polarisation

A

Polarisation occurs when oscillations of electric and magnetic fields are in one plane only in the direction of the travel of light

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13
Q

Why is the EM wave polarised

A

The EM wave is oscillating in all directions.
Only the oscillations parallel to the first polaroid filter are transmitted.
the wave is now plane polarised.
When these waves arrive at the second the waves are perpendicular to the direction of propagation.
So no light will be transmitted

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14
Q

How do sunglasses polarise

A

Point 1 – light from the Sun is unpolarised.
Point 2 – light entering the glasses directly are only partially polarised so glare is reduced a little.

Point 3 – light that reflects off a surface (like water) is partially polarised in the horizontal plane.
Point 4 – partially polarised light is polarised further vertically

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15
Q

What is phase difference

A

The phase difference isthe difference in the phase angle of the two waves. It tells us how much a particle (or wave) is in front or behind another particle (or wave).

16
Q

What is path difference

A

Path difference isthe difference in the path travelled by two waves from their source.

17
Q

What is constructive interference

A

Constructive interference occurs whenever the path difference between coherent sources is nλ, where n is a whole number.

18
Q

What is coherent

A

COHERENT: Light that is composed of waves that are in phase or have a constant phase difference. They have the same amplitude and frequency.

19
Q

What is destructive interference

A

Destructive interference occurs whenever the path difference between coherent sources is nλ/2, where n is a whole number.

20
Q

Difference between standing and progressive waves.

A

Standing Wave:
Energy is stored
Amplitude Varies
Oscillations are in phase
Progressive Wave
Energy is transferred
Amplitude stays constant
Phase varies continuously

21
Q

What factors affect the frequency at which the fundamental frequency will occur?

A

The length of the string
The tension of the string
The thickness of the string – the mass per unit length

22
Q

What types of waves can be diffracted? When will the greatest amount of diffraction occur?

A

All types of waves can be diffracted, but the amount depends on the ratio of the wavelength to the size of the opening or obstacle.
Diffraction is greatest when the wavelength is approximately the same as the width of the gap.

23
Q

Why is the diffraction greatest when wavelength is approximately same as the width of the gap

A

Why this happens can be explained by Huygen’s principle: every point on a wave front may be considered as a source of secondary waves.

24
What will happen if the secondary wave fronts produce overlap?
If they are in phase there will be constructive interference If they are out of phase there will be destructive interference
25
The intensity of the fringes against distance from the centre can be plotted on a graph. What are the two most important features of this graph. Challenge: how will the pattern change if a larger slit is used or a longer wavelength?
the central maximum is twice as wide as the other fringes. the central maximum is much brighter than the other fringes. if a narrower slit or larger wavelength of light are used, the pattern would spread out more.
26
Why don't we all diffract
In other words a particle with a mass larger than its de Broglie wavelength will not exhibit any wave-like properties
27
What is a photon
A photon is a particle of electromagnetic radiation that has zero mass and carries a “discrete packet” quantum of energy
28
What were Hertx observations
If the frequency of the incident light was too low (red light, for example), then nothing happened even if the intensity of the light was very high or it was shone onto the surface for a long time. If the frequency of the light was higher (green light, for example), then zinc lost its charge even if the intensity was very low or it was shone for only a short time.
29
What did Einstein explain about Hertz findings
He suggested that photons, these “discrete packets of quantum energy” carry a certain fixed amount of energy (or quanta) that depends on the frequency of light. Their interaction – in other words, their behaviour as a particle – with the metal causes electrons to be released from the surface of the metal This minimum frequency, needed to cause electron ejection is referred to as the threshold frequency. (fo) Einstein realised that light – being composed of these discrete packets of energy (our photons) are bombarding the surface of the metal. He figured out that if one of these photons collides with a free electron it transfers all of its energy to it giving it enough energy to be released.
30
What is the work function
– the minimum photon energy required to emit an electron is dependent on the metal and is called the work function, Φ
31
What is an electron volt
An electronvolt is equal to the amount of energy transferred to a single electron if it is accelerated through a potential difference of 1 V: 1 eV = 1.6 × 10-19 J
32
What did Einstein realise
Einstein realised that: the photon energy is dependent on the frequency of the radiation according to the equation: E= hfo The photon energy must be at least equal to the work function for electrons to be released. hfo ≥ φ When radiation of sufficiently high frequency, f, is incident on a metal surface of work function, φ. The photon, of energy E, is absorbed by the electron. This electron leaves with kinetic energy: KEmax = hfo – φ
33
How would you plot this on a graph KEmax = hfo – φ
fo is the x-intercept -φ is the y-intercept h will be the gradient
34
How does the photoelectric effect provide evidence of the particle nature of light.
The photoelectric effect provided evidence of the particle nature of light as observations could not be explained by wave theory. - Only one photo is absorbed by one electron (all of the photon energy is transferred to one electron– this is evidence of the particle model - The wave model says that energy can be supplied to electrons continuously allowing energy levels to build up. There is no photoemissions below the threshold frequency The rate of photoemissions is proportional to the intensity of the incident radiation (i.e. higher intensity = more photon which increases the number of photons that can be absorbed thus increasing the number of photoelectrons that can be released. The maximum kinetic energy of photoemission is independent of the intensity but is proportional to frequency of the incident radiation.
35
Why does the wave model not support the particle nature of light? Evidence One - frequency of waves
The wave model says that frequencies of waves gradually build up This means the energy absorbed by electrons will gradually increase with each coming wave meaning any frequency of wave should cause photoelectrons to be emitted This can not explain threshold frequency This can not explain one electron absorbing only one photon
36
Evidence 2 - Takes time for..
The wave model suggests that it takes time for energy to be absorbed This means it would take time for the energy supplied to electrons to reach the work function This can not explain the immediate release of photoelectrons
37
Evidence 3 - Increasing intensity...
The wave model suggests that increasing the intensity would increase the speed of the photoelectrons This can not explain the fact that increasing the intensity increases the number of photons released per second.
38
Evidence 4...Kinetic energy...
The wave model suggests that photoelectrons would all have the same kinetic energy This can not explain the photoelectrons are released with a range of kinetic energies that is linked to their frequency