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A LEVEL PHYSICS > 3.3 WAVES > Flashcards

3.3 WAVES Flashcards

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1
Q

What is the optimal size of a gap for a wave to diffract through?

A

One that is similar sized/slightly larger than the wavelength of the travelling wave. Waves bend around the gap, and interfere with each other (light has to bend around sharper corners)

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2
Q

Explain why there is little diffraction when a wave travels through a gap greater than its wavelength

A

Waves will not be able to combine and produce an interference pattern. This is because the gap is too large for the waves to bend around and interfere with each other

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3
Q

Explain why there is no diffraction when a wave travels through a gap smaller than its wavelength

A

Waves simply reflected back

Waves do not have to refract much and therefore there is limited interference according to some site

this is wrong app

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4
Q

What determines the resolution of a microscope?

A

The size of the objective lens relative to the wavelength of light

If the objective lens is much larger than the wavelength of the light, the diffraction pattern will be small and the resolution of the microscope will be high. However, if the objective lens is about the same size as the wavelength of the light, the diffraction pattern will be larger and the resolution of the microscope will be lower

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5
Q

Explain why a light microscope can’t view objects smaller than the wavelength of visible light

A

The light from the microscope will diffract around the edges of the object, causing the image of the object to blur. As the size of the object decreases, the diffraction of the light increases, causing the image to become even more blurred. This makes it impossible for a light microscope to resolve objects that are much smaller than the wavelength of light because the diffraction of the light causes the image to blur.

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6
Q

What is the optimal size for the wavelength of an electron used to investigate the size of gaps between atoms?
(in electron microscope)

A

in the order of 10^-10m (ask mrgray)

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7
Q

What makes something exhibit more wave like properties?

A

Having a longer wavelength

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8
Q

Why do objects travelling at higher speeds diffract less?

A

Shorter wavelength, less diffraction

λ = h/mv

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9
Q

What is the frequency range of visible light?

A

650nm (deep red) to 450nm (deep violet)

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10
Q

If i need saving

A

https://www.youtube.com/watch?v=TqTWSSBcV_4

https://www.youtube.com/watch?v=b6prlN4Fz90

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11
Q

What is a progressive wave?

A

A (moving) wave that transfers energy without transferring matter

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12
Q

What are examples of how waves transfer energy

A
  • EM waves cause things to heat up
  • X rays are ionising
  • Source of waves loses energy as
    wave propagates
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13
Q

Why can’t longitudinal waves be polarised?

A

polarisation restricts the vibrations to one
plane by absorbing the vibrations at right angles
to this plane
* longitudinal waves cannot be polarised
because the vibrations have to take place for
energy to be transmitted

The vibrations of a longitudinal wave occur along a single line, it is therefore not possible to confine that to a plane, i.e. a 1D shape cannot be confined within a 2D shape.

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14
Q

How does polarisation provide evidence for the nature of transverse waves?

A

Polarisation requires the direction of oscillation to be perpendicular to propagation

The reason that only transverse waves can be polarised is that their vibrations can potentially occur in all directions perpendicular to the direction of travel. It is therefore possible to confine the vibrations to a single plane

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15
Q

What is the plane of polarisation of an EM wave defined as?

A

The plane which the electric field oscillates in

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16
Q

Phase difference in radians

A

2πd/λ 2πt/T

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17
Q

What are puretone sound waves?

A

Sound waves of a single frequency

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18
Q

What does LASER stand for? [6]

A

Light amplification by stimulated emission of radiation

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19
Q

Describe how a standing wave is formed on a taut string

A
  • String is fixed at either ends (to a
    weight/wall) [lossless system]
  • Progressive wave sent along the
    string
  • When wave reaches the end of the string, reflected wave begins to travel in the opposite direction
  • Wave must have similar frequency/amplitude
    When the two waves meet, they superpose.

Nodes - where 2 progressive waves are always in antiphase, destructive

Antinodes - points where waves are always in phase, constructive

In between is a combo of both

As there is no oscillation at either end of the system, energy is “locked” and can’t be transferred through the standing wave

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20
Q

What are the equations for the nth harmonic frequency?

A

fn = v/λn = nv/2L

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21
Q

What is the first harmonic also referred to as

A

The fundamental mode

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22
Q

What is the equation to find the first harmonic frequency of a string?

A

f1 = (1/2L) * sqrt(T/μ)

where:

f1 is the frequency of the first harmonic mode of vibration (in hertz, Hz)
L is the length of the string (in meters, m)
T is the tension in the string (in newtons, N)
μ is the linear density of the string (in kilograms per meter, kg/m)

assuming the string is under ideal conditions, with negligible damping and a uniform linear mass density. It also assumes that the string is fixed at both ends and is vibrating in a single antinode

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23
Q

What is the principle of superposition?

A

When two waves meet, the total displacement at that point = the sum of the individual displacements at that point

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24
Q

What makes waves coherent

A

If they have a constant phase difference

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25
Q

Distance between adjacent node and anti/node

A

Node + node = 1/2λ
node + antinode = 1/4λ

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26
Q

What do pitch and loudness correspond to in a wave

A

Pitch - frequency
Loudness - amplitude

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27
Q

What is refraction

A

The changing of direction of a wave as it passes the boundary between 2 media

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28
Q

What is total internal reflection?

A

When all incident light reflects after hitting a boundary - Angle of incidence is greater than the critical angle (ray remains in original medium)

  • Only possible when ray travels from high to low refractive index
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29
Q

What is the refractive index (n) of a material

A

The ratio of the velocity of light in a vacuum to its velocity in a specified medium

(The capacity of the material to change the speed of light)

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30
Q

All versions of Snell’s law

A

n= sini/sinr
c1/c2 =sini/sinr
n1sinθ1 = n2sinθ2

where 1 is 1st medium, and 2 is 2nd medium

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31
Q

Difference between refraction of light vs other waves

A

Light is changing direction after passing between 2 TRANSPARENT media

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32
Q

What happens when light travels from a material with a high refractive index to a low one

A

It bends away from the normal
(lower refractive index = higher speed relative to c)

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33
Q

Relate the refractive index of a material (s) to the ratio of the wavelengths of light coming in and going out

A

Ns = c/cs =λ/λs

As frequency is constant

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34
Q

Do water waves get faster or slower as they travel from deep to shallow water

A

Slower, as same drag force from the “seabed” acts against a smaller mass of waver, decreasing rate of flow

Wavefronts move closer to the normal, decreasing λ, while f is constant, so v decreases

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35
Q

What is white light?

A

Light containing a continuous range of all wavelengths

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36
Q

Conditions for light interference

A
  • Waves must be coherent
  • Slit separation must be small
  • Light waves should have similar
    amplitude and intensity
37
Q

Conditions for greater diffraction

A

Smaller gap in relation to the wavelength of incident wave

38
Q

Young’s double slit formula

A

w = λD/s

Where w is fringe spacing (distance between 2 adjacent maxima or minima)
D is distance from slits to screen
s is distance between centres of each slit

39
Q

When does single slit diffraction occur

A

When a wave is incident on a slit approximately the same size of the wavelength

40
Q

Equation for single slit diffraction

A

W =λ 2D/a

where a is the size of the slit
W is the width of the central fringe

Consider conditions for first minimum, where rays from end of the slit meet with path difference of λ, as this means pairs of rays can be paired of in a way such that each pair destructively interferes. Angle from 0th order to minimum = angle between slit and shortest path between furthest apart rays. As D»a, angle is very small so sinθ = tanθ

41
Q

Single slit diffraction patterns

A

Central fringe is 2x as wide as each outer fringe (measured from minimum to minimum)
Each outer fringe is the same width
Outer fringes are much less intense than central fringe

As you move farther from the center, the path length differences between the waves from the two slits and the screen become larger. This causes the constructive and destructive interference to occur at different points, resulting in the bright fringes decreasing in intensity as you move away from the center

42
Q

Describe the structure of a fibre optic cable

A
  • Contains a core, highly transparent
    to prevent absorption of light which
    would reduce amplitude. +
    Propagates/guides light.
  • Cladding protects core from damage
    + prevents cross talk between
    touching fibres
    Prevents signal degradation through light escaping core, causing info to be lost

n Core > Cladding

43
Q

What is modal (multipath) dispersion and how is it prevented

A

Light travelling along the axis of the core travels a shorter distance per unit length than light continuously undergoing TIR. Larger range of path lengths - pulse would become longer than supposed to be, and could merge with next pulse.

Prevent by using narrow fibre + low difference in n between core and cladding

44
Q

What is chromatic/material dispersion and how can it be prevented

A

When white light is shone along an optical fibre, different wavelengths of light travel at different speeds. e.g. violet light travels slower than red light in glass. Causes pulse broadening as violet fails to catch up w red

Prevent by using monochromatic light

45
Q

How are optical fibres used in endoscopy?

A

Contains two bundles. Endoscope inserted into the body, then illuminated with an incoherent bundle. Lens at other end used to form an image of insides, and light is reflected back through a coherent bundle where the image can be observed.

Coherent bundle - fibres at each end are in same relative position

46
Q

In the double slit experiment, what would happen if you replaced the volume between the slits and the screen with water?

A

Maximas would be closer together/ lower fringe spacing, as wavelength of light decreases as it enters water

47
Q

What happens to a diffraction grating pattern when the number of slits is increased (with same spacing)

A

Increased intensity + sharper/ more defined maxima at respective angles

No more maxima/ diff in distance as wavelength isn’t changing

Light incident from lots more slits - more waves arrive at a particular point on the screen - so unless dsinθ = nλ, enough waves will be arriving out of phase to result in overall destructive interference.

48
Q

What increases the chances of overlapping maxima with a non monochromatic source?

A

A grating with a small number of slits

49
Q

Why can’t the first and second order visible spectra overlap?

A

The wavelength of red light is greater than 2x the wavelength of violet light

Because the second order maximum value is greater than the third order minimum, those would overlap. The first and second order do not overlap because the maximum of the 1st will never exceed the value for the 2nd order.

50
Q

Why does red light bend more than blue light in a diffraction grating, but bend less in a prism

A

Angle of refraction in a grating proportional to wavelength (dsinθ = mλ). Red light has a greater wavelength

n = c/c2 = λ/λs
Red has a greater wavelength so a lower refractive index, so speed is changed less in a prism, and therefore experiences less bending

51
Q

What could cause polarised light to pass through an analyser at an unexpected angle (badly worded)

A

If a transparent material is placed between the filters, it could rotate the plane of polarisation

52
Q

Why are sloping sides used in the ripple tank practical?

A

To prevent waves from reflecting off the sides

53
Q

What is a wavefront?

A

“A line of constant phase” e.g. from joining adjacent peaks

54
Q

Importance of satellite dish design

A

Larger dish - stronger signal, as more waves are reflected onto aerial, but a bigger dish reflects waves onto a smaller focus, as waves diffract less, so dish needs to be more carefully aligned due to smaller focus

55
Q

Why do EM waves lose energy as they propogate?

A

Waves “spread out” and lose intensity

56
Q

Difference between emission and absorption spectra

A

Line emission - sample is heated/given energy and photon excitation causes specific wavelengths of light to be given off (evidence for discrete energy levels)
Absorption -Continuous spectrum passed through sample which absorbs wavelengths of certain frequencies. Light is emitted but not in same direction

Line emission - coloured lines on dark screen
Absorption - Dark lines on coloured screen

57
Q

What happens when you use a larger wavelength in the single slit experiment?

A

Wider central fringe, + maxima further apart

(as subsequent fringes are also wider)

58
Q

Effect of using a range of wavelengths in single slit experiment

A

Central maximum unchanged in width
Broader maxima/range of angles for each maximum/order
Gradually getting broader/more spread out for greater order maxima

59
Q

Effect of using a range of wavelengths in single slit experiment

A

Central maximum unchanged in width
Broader maxima/range of angles for each maximum/order
Gradually getting broader/more spread out for greater order maxima

60
Q

Polarization by reflection

A

When unpolarized light reflects off an insulator, the light can become polarized in the plane parallel to the reflecting surface. (common example is glare from water and glass)

61
Q

Diff light sources

A

Discharge tubes/ vapour lamps
- Produce dominant colour

Filament/Sun
Produces continuous spectrum (white)

Laser - monochromatic

61
Q

Pros of diffraction grating

A

U get sharper fringes - each beam reinforced multiple times more

Multiple slits - fringes are more intense, but at the same angle if wavelength is unchanged

62
Q

Applications of diffraction gratings

A

Analysing composition of stars
Chemical analysis
Measure red shift / rotation of stars
Measure the wavelength / frequency of light from a star
Observe the spectra of materials
Analyse the absorption / emission spectra in stars

Diffraction gratings also play a role 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

63
Q

Assumptions in diffraction eqs

A

D»s - So rays arriving at a point are virtually parallel - can use similar angles + small angle approximations

Also for double slits assuming that slits are of equal width

64
Q

First minimum for single slit diff

A

path lengths differ by 3λ/2 for rays from the top and bottom of the slit. One ray travels a distance λ different from the ray from the bottom and arrives in phase, interfering constructively. Two rays, each from slightly above those two, will also add constructively. Most rays from the slit will have another to interfere with constructively, and a maximum in intensity will occur at this angle. However, all rays do not interfere constructively for this situation, and so the maximum is not as intense as the central maximum

Central max all interfere constructively so more intense

65
Q

Single slit diffraction pattern for white light

A

Central fringe white
Each wavelength forms own pattern
Bright fringes now a spectrum
violet at lowest angular separation for a given order, red at the greatest
Wider maxima
Smaller fringe spacing

(Violet/blue closest to central, red furthest)

66
Q

Effect of slit width + wavelength

A

Angle of diffraction proportional to wavelength
Width of bright maxima (fringe) proportional to wavelength

Narrower slits reduce intensity + increase fringe spacing

I/sinθ graph for red vs blue shows red wider if that makes senz

67
Q

Modulation of two slit interference pattern

A

If light emerging from each slit in two slit experiment may have alr formed its own diffraction pattern. So if light emerging has a minimum at a certain angle, due to OG diffraction, a minimum will also be formed, even if the two slit model predicts a maximum.

Note the separation of the centres of the slits must always be less than the width of each slit

Can be said that single slit pattern modulates double slit pattern (see pg 396 of IB textbook)

68
Q

Angles for minima of single slit pattern, vs maxima for 2 slit pattern

A

Minimum - θ = nλ/a (a is slit separation)
Maximum θ = nλ/s (s is distance between centres of slits)

sma θ = sinθ

69
Q

Which derivations involve sma vs don’t

A

Single slit + double slit, however diffraction grating u can’t use sinθ = θ as angles are larger

70
Q

A submarine is using ultrasound of frequency 50 kHz to measure the height of objects on the seabed. The submarine encounters object X. (X is a height hx above the seabed). The phase difference between a signal from the seabed and the returned signal from object X is fraction numerator 2 pi over denominator 3 end fraction radians.

Assuming there is no change of phase when the signals reflect off either the seabed or the object, which of the following heights, A to D, is not a possible height hx of object X?

The speed of sound in water is 1500 m s–1.

A

The wavelength of the ultrasound can be found using the wave equation:
λ = v / f = 1500 / 50000 = 0.03 m.
Since a phase difference of π radians (or 180º) is equivalent to the waves being separated by λ / 2, 2π / 3 radians is equivalent to a separation of λ / 3,
λ / 3 = 0.01 m or 1 cm.
Therefore the path difference Δx between signals reflected off the seabed and signals reflected off the object must be equal to an integer number of wavelengths (which has no effect on phase) ± one-third of a wavelength:
Δx = nλ ± (λ / 3).
Substituting the value of λ gives
0.03n ± 0.01.
However, because the wave is reflecting and returning to the submarine, the path difference Δx is actually equal to twice the height h𝑥 of the object.
Therefore,
Δx = 2h𝑥 = 0.03n ± 0.01;
h𝑥 = 0.015n ± 0.005

71
Q

Critical angle for diamonds

A

The diamond sparkles more than the fake because more light is reflected back out of the front face (there is more internal reflection);
A smaller critical angle will allow more total internal reflection;
This is because there would be more angle of incidences that would be greater than the critical angle to produce TIR

72
Q

Uses of polarisers

A

3D Glasses, Sunglasses to reduce glare, Polaroid cameras

Vertically polarised so that plane polarised light is absorbed

Btw in reality light intensity decreases when passing through filter

73
Q

How polarising filters helps to take picture of underwater fish

A

Partially plane polarised in vertical plane, light reflected of water polarised in horizontal, blocks glare. Light from fish isn’t reflected but refracted and therefore isn’t plane polarised

74
Q

Types of waves that can be polarised

A

EM radiation,Water waves, S(shear) waves

75
Q

What would cause the displacement of diff particles in phase in a longitudinal wave to be different

A

If there’s no absorption/spreading out of the wave

76
Q

Why can’t coherent waves produce constant patterns

A

They interfere, however the pattern randomly fluctuates as the phase diff is changing

77
Q

Condition for standing waves

A

Two waves moving in opposing directions along the same path and in the same plane.
The waves must travel at the same velocity.
The waves must have the same frequency.
The waves should have a similar amplitude (idk y they still produce standing wave but cancellations are diff)

78
Q

n=ci/cs derivation

A

Consider horizontal line XY’. Wavefront at angle i from X to point above Y moving a distance ct from Y to Y’. Once inside new wavefront refracted towards normal (angle r) moving distance cst from XY’ to X’Y’. You can form right triangles using the fact that the direction of propagation is ppd to wavefronts, and eliminate XY’, then sub in distances

79
Q

Angle of incidence for a wavefront vs a ray

A

Complements of each other, prove to yourself sasageyo

80
Q

Relationship between wavelength and amount of diffraction

A

n=l/ls

Smaller l greater n greater diffraction (capacity to change speed)

81
Q

Why diamonds sparkle in white light

A

Highest refractive index - separates white light more than other substances, low critical angle - may be TIRd many times before emerging - spreading out colours even more

82
Q

What can prevent TIR in an optical fibre (bend)

A

If radius of bend is too small, COULD potentially not refract

83
Q

Pulse broadening and absorption

A

Absorption - part of signals energy absorbed by fibre, signal is attenuated, reducing amplitude of signal - loss of info

Pulse broadening caused by modal/chromatic dispersion

Reduce broadening by making core narrow,using monochromatic source, using repeaters so pulse is regenerated before significant broadening has occurred

84
Q

Measuring fringe width practically

A

Measure across a few dark fringes, centres of dark easier to find that centres of bright

85
Q

What determines the “single slit envelope”

A

Width of single slit

86
Q

Path difference as

A

Integer number of wavelengths + path difference (nλ + x)

also to convert from radians to fractions of T/λ divide by 2pi

87
Q
A

A LEVEL PHYSICS (17 decks)

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