3.3 WAVES

Question 1

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

Answer 1

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)

Question 2

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

Answer 2

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

Question 3

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

Answer 3

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

Question 4

What determines the resolution of a microscope?

Answer 4

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

Question 5

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

Answer 5

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.

Question 6

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

Answer 6

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

Question 7

What makes something exhibit more wave like properties?

Answer 7

Having a longer wavelength

Question 8

Why do objects travelling at higher speeds diffract less?

Answer 8

Shorter wavelength, less diffraction

λ = h/mv

Question 9

What is the frequency range of visible light?

Answer 9

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

Question 10

If i need saving

Answer 10

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

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

Question 11

What is a progressive wave?

Answer 11

A (moving) wave that transfers energy without transferring matter

Question 12

What are examples of how waves transfer energy

Answer 12

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

Question 13

Why can’t longitudinal waves be polarised?

Answer 13

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.

Question 14

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

Answer 14

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

Question 15

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

Answer 15

The plane which the electric field oscillates in

Question 16

Phase difference in radians

Answer 16

2πd/λ 2πt/T

Question 17

What are puretone sound waves?

Answer 17

Sound waves of a single frequency

Question 18

What does LASER stand for? [6]

Answer 18

Light amplification by stimulated emission of radiation

Question 19

Describe how a standing wave is formed on a taut string

Answer 19

• 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

Question 20

What are the equations for the nth harmonic frequency?

Answer 20

fn = v/λn = nv/2L

Question 21

What is the first harmonic also referred to as

Answer 21

The fundamental mode

Question 22

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

Answer 22

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

Question 23

What is the principle of superposition?

Answer 23

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

Question 24

What makes waves coherent

Answer 24

If they have a constant phase difference

Question 25

Distance between adjacent node and anti/node

Answer 25

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

Question 26

What do pitch and loudness correspond to in a wave

Answer 26

Pitch - frequency Loudness - amplitude

Question 27

What is refraction

Answer 27

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

Question 28

What is total internal reflection?

Answer 28

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

Question 29

What is the refractive index (n) of a material

Answer 29

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)

Question 30

All versions of Snell's law

Answer 30

n= sini/sinr c1/c2 =sini/sinr n1sinθ1 = n2sinθ2 where 1 is 1st medium, and 2 is 2nd medium

Question 31

Difference between refraction of light vs other waves

Answer 31

Light is changing direction after passing between 2 TRANSPARENT media

Question 32

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

Answer 32

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

Question 33

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

Answer 33

Ns = c/cs =λ/λs As frequency is constant

Question 34

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

Answer 34

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

Question 35

What is white light?

Answer 35

Light containing a continuous range of all wavelengths

Question 36

Conditions for light interference

Answer 36

- Waves must be coherent - Slit separation must be small - Light waves should have similar amplitude and intensity

Question 37

Conditions for greater diffraction

Answer 37

Smaller gap in relation to the wavelength of incident wave

Question 38

Young's double slit formula

Answer 38

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

Question 39

When does single slit diffraction occur

Answer 39

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

Question 40

Equation for single slit diffraction

Answer 40

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

Question 41

Single slit diffraction patterns

Answer 41

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

Question 42

Describe the structure of a fibre optic cable

Answer 42

- 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

Question 43

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

Answer 43

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

Question 44

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

Answer 44

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

Question 45

How are optical fibres used in endoscopy?

Answer 45

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

Question 46

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

Answer 46

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

Question 47

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

Answer 47

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.

Question 48

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

Answer 48

A grating with a small number of slits

Question 49

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

Answer 49

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.

Question 50

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

Answer 50

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

Question 51

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

Answer 51

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

Question 52

Why are sloping sides used in the ripple tank practical?

Answer 52

To prevent waves from reflecting off the sides

Question 53

What is a wavefront?

Answer 53

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

Question 54

Importance of satellite dish design

Answer 54

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

Question 55

Why do EM waves lose energy as they propogate?

Answer 55

Waves "spread out" and lose intensity

Question 56

Difference between emission and absorption spectra

Answer 56

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

Question 57

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

Answer 57

Wider central fringe, + maxima further apart (as subsequent fringes are also wider)

Question 58

Effect of using a range of wavelengths in single slit experiment

Answer 58

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

Question 59

Effect of using a range of wavelengths in single slit experiment

Answer 59

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

Question 60

Polarization by reflection

Answer 60

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)

Question 61

Diff light sources

Answer 61

Discharge tubes/ vapour lamps - Produce dominant colour Filament/Sun Produces continuous spectrum (white) Laser - monochromatic

Question 61

Pros of diffraction grating

Answer 61

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

Question 62

Applications of diffraction gratings

Answer 62

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

Question 63

Assumptions in diffraction eqs

Answer 63

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

Question 64

First minimum for single slit diff

Answer 64

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

Question 65

Single slit diffraction pattern for white light

Answer 65

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)

Question 66

Effect of slit width + wavelength

Answer 66

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

Question 67

Modulation of two slit interference pattern

Answer 67

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)

Question 68

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

Answer 68

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

Question 69

Which derivations involve sma vs don't

Answer 69

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

Question 70

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.

Answer 70

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

Question 71

Critical angle for diamonds

Answer 71

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

Question 72

Uses of polarisers

Answer 72

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

Question 73

How polarising filters helps to take picture of underwater fish

Answer 73

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

Question 74

Types of waves that can be polarised

Answer 74

EM radiation,Water waves, S(shear) waves

Question 75

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

Answer 75

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

Question 76

Why can't coherent waves produce constant patterns

Answer 76

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

Question 77

Condition for standing waves

Answer 77

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)

Question 78

n=ci/cs derivation

Answer 78

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

Question 79

Angle of incidence for a wavefront vs a ray

Answer 79

Complements of each other, prove to yourself sasageyo

Question 80

Relationship between wavelength and amount of diffraction

Answer 80

n=l/ls Smaller l greater n greater diffraction (capacity to change speed)

Question 81

Why diamonds sparkle in white light

Answer 81

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

Question 82

What can prevent TIR in an optical fibre (bend)

Answer 82

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

Question 83

Pulse broadening and absorption

Answer 83

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

Question 84

Measuring fringe width practically

Answer 84

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

Question 85

What determines the "single slit envelope"

Answer 85

Width of single slit

Question 86

Path difference as

Answer 86

Integer number of wavelengths + path difference (nλ + x) also to convert from radians to fractions of T/λ divide by 2pi