imaging + signalling Flashcards

Question

what are longitudinal waves? I + S

Answer 1

longitudinal waves are waves where the oscillations are parallel to the direction of the waves travel

Answer 2

intensity is the rate of flow of energy per unit area perpendicular to the direction of travel of the wave units: Watts per square metre, W m^-2

Answer 3

intensity = power/area I = P/A

Answer 4

the amount of light energy that hits your retina per second

Answer 5

intensity is the measure of how much energy a wave is carrying

Answer 6

refraction happens when a wave changes speed at a boundary between two mediums

Answer 7

when a wave/ray meets a boundary between two mediums, some of it’s energy is reflected back into the first medium and the rest of the energy is transmitted through into the second medium

Answer 8

when light meets the boundary at an angle, the TRANSMITTED light is bent - “refracted” - as it travels at a different speed in each medium

Answer 9

• the more optically dense a material is, the more slowly light travels in it • so if light travels through a very optically dense material, light will slow down A LOT

Answer 10

• the amount of refraction depends on the λ of the light • this means that the focal length for a given lens will change depending on the λ of light passing through it

Answer 11

a converging lens changes the curvature of wavefronts

Answer 12

lenses change the curvature of wavefronts by adding curvature to waves via refraction as they pass through the lens (converging lens adds positive curvature, convex lens adds negative curvature)

Answer 13

if a wave has zero curvature before passing through a converging lens (and the wavefronts are also parallel to the lens axis), the wave will be given a spherical curvature, centred on the focal point of the lens

Answer 14

a converging lens curves the wavefronts by slowing down the light travelling through the middle of the lens more than the light travelling through the lens’ edges

Answer 15

all points on a wave take the same amount of time to get to the focus point of the lens

Answer 16

the focal length is the distance between the lens [axis] and the focus of the lens

Answer 17

you can show lens diagrams in the following two ways: • lens diagram using wavefronts or • lens diagram using rays (aka ray diagram)

Answer 18

the more powerful (thicker) the lens, the more it will curve wavefronts that travel through it, so the shorter the lens' focal length

Answer 19

the focus of a converging lens is the point where all the wavefronts that have passed through the lens converge

Answer 20

lens power = 1/focal length P = 1/f when working out the power of a lens, the focal length HAS to be in METRES

Answer 21

the power of a lens is the amount of curvature added by the lens to a wave passing through the lens units: dioptres

Answer 22

curvature of a wave = 1/radius of wave's curvature curvature = 1/r

Answer 23

the thicker the lens, the more curved its sides and so the more curvature the lens adds to the wave

Answer 24

you can use the lens equation to find where an image will be formed

Answer 25

the distances between the lens, the image and the source of light

Answer 26

the lens equation: 1/v = 1/u + 1/f where: * v = image distance (the distance between a lens and the image) * u = object distance (the distance between a lens and the object (object == "light source") * f = focal length (the distance between a lens and it's focal point)

Answer 27

the lens equation assumes that the lens is thin

Answer 28

* you always measure distances from the lens axis (from the lens basically) and count distances to the right of the lens axis as positive and distances to the left of the lens axis as negative - just like the negative and positive x-axis when you draw graphs * so due to cartesian graphs, u is ALWAYS measured as NEGATIVE and must be used as NEGATIVE in calculations

Answer 29

1/v = 1/u + 1/f is the same as curvature of wavefront after lens = curvature before lens + curvature added by lens

Answer 30

* if you have a distant light source, the wavefronts approaching the lens will be [almost] flat/parallel * this is bc when u = ∞ (signifying very distant light source), 1/u (curvature before lens) = 1/∞ * 1/∞ approaches 0, so the wave has [basically] zero curvature

Answer 31

flat wavefronts are called plane wavefronts

Answer 32

a converging lens gives plane wavefronts a curvature of exactly 1/f

Answer 33

⋅ when the source is at the focus of the lens (there is a focus to both left + right side of the lens obvs bc you can shine light from both directions), wavefronts will start off with negative curvature (bc u is measured as negative) ⋅ this negative curvature is then cancelled out by positive curvature added by converging lens ⋅ so wave fronts will be made flat

Answer 34

⋅ waves coming from a source between the focus and the lens will have curved wavefronts before and after they pass through the lens ⋅ the difference between curvature before and after is 1/f

Answer 35

⋅ you can use ray diagrams bc the lens equation work for both models ⋅ the light rays are just "bent" by the lens instead of being curved like with wavefronts ⋅ [INSERT RAY DIAGRAM EG]

Answer 36

a lens can produce a magnified image

Answer 37

linear magnification = image height/object height m = v/u

Answer 38

you can find the power of a converging lens by focusing an image on a screen

Answer 39

1) set up equipment as shown, connecting bulb to low-voltage power supply 2) place bulb exactly 0.200 m away from lens (i.e. u = 0.200 m) + turn on power supply. move screen until you can see clear picture of filament on screen 3) measure DISTANCE between LENS + SCREEN - this is your value for v. record u and v in table, as shown 4) repeat experiment 5 or more times, for range of different values of u (however, don't increase u so much that you can no longer see image of filament though) 5) WORK OUT 1/u and 1/v for each of your readings. ADD them together to get 1/f, which is POWER of lens 6) find AVERAGE of your values for 1/f + DIVIDE 1 by your answer to find FOCAL LENGTH of your lens

Answer 40

a bit is a single binary digit

Answer 41

8 bits = 1 byte

Answer 42

⋅ the binary system is used to store data in the computer memory ⋅ computers store files as a string of bits

Answer 43

the number of bits (b) in a string determines how many alternatives that string can represent

Answer 44

number of alternatives = 2^(number of bits) N = 2^b EG) 1 bit only has 2 alternatives (0 and 1), while 1 byte (8 bits) has 256 alternatives - number of alternatives double with each additional bit - further explanation: bc for 1 bit, 2^1 = 2 alternatives bc the bit can only be either 1 or 0 - whereas for 1 byte (8 bits), 2^8 = 256 alternatives

Answer 45

the number of bits you need depends on how many alternatives you want

Answer 46

number of bits = log[2](number of alternatives] b = log[2](N) EG) if you wanted to represent any letter of alphabet, you'd need 26 alternatives - one for every letter. substituting 26 into equation gives log[2](26) = 4.7, so you'd need five bits (ALWAYS ROUND UP, SO YOU HAVE ENOUGH BITS, NOT TOO FEW)

Answer 47

images are stored as arrays of binary number

Answer 48

a pixel is the smallest single digital unit that can make up an image on a digital screen

Answer 49

⋅ when an image is stored (eg, on a computer or memory card), each pixel is represented by a binary number ⋅ the binary numbers are then stored in an array ⋅ the image is then stored as arrays of binary numbers

Answer 50

⋅ binary numbers are stored in an array ⋅ this array is a grid of numbers where the location of each number in the grid also matches the location of a pixel in a photo (not definition for array, just how it's used to store photos)

Answer 51

the value of the binary number maps to the colour of the corresponding pixel

Answer 52

three binary numbers, 1 for each of the primary colours (R, G + B)

Answer 53

the length of the binary number depends on how many colours are needed EG) on typical PC display, each of numbers for red, green, + blue are 8 bits (1 byte) long, giving 256^3 = 16.8 million possible colours

Answer 54

yes, image resolution can mean different things: 1) image resolution refers to how many bits (length of binary number) are represented in each pixel (eg, if object of 1.0 m is represented by 200 pixels in image, resolution is 1/200 = 0.005 metres per pixel) OR 2) resolution of an image can also mean the number of pixels in an image, in the format width x height (eg, 1920 x 1080) -> this type of resolution can also be given as just the total number of pixels in image without formating (eg, 1920 x 1080 = 2 073 600 pixels, so resolution is 2.1 megapixels)

Answer 55

⋅ the amount of information in an image depends on the number of bits per pixel ⋅ the more bits per pixel, the more information that is held by each pixel ⋅ so therefore, the more pixels there are in an image, the more information is held by an image also

Answer 56

total amount of information in an image = number of pixels x bits per pixel

Answer 57

multiplying the values of the binary numbers by number larger than 1 increases the contrast (by stretching the range of the values of the pixels)

Answer 58

⋅ adding false colour can highlight important features by making them very bright colours ⋅ adding false colours is done by changing what colour the values of the binary numbers map to ⋅ EG) thermal imaging cameras show heat in colour instead of visible light

Answer 59

⋅ you can reduce/get rid of the noise in an image by replacing pixels with the median value of themselves and the 8 pixels surrounding it pixels ⋅ the result is that any 'odd' (i.e. very high or very low) values are removed and the image is made smoother

Answer 60

⋅ you use the 'Laplace rule': 1) multiply the value of the pixel by four and then subtract the values of the pixels immediately above, below, to the left and to the right of the pixel 2) if the answer is negative, the pixel is treated as if its value is zero (and so the value of the pixel maps to black) ⋅ the result is that any pixel not on an edge goes white - so you're just left with the edges

Answer 61

polarised waves only oscillate in one direction

Answer 62

EM radiation is made up of two transverse waves (the electric and magnetic fields) oscillating perpendicular to each other

Answer 63

they only oscillate in one direction

Answer 64

if light tries to pass through two polarising filters that are perpendicular to each other, no light will get through as all directions of oscillation have been blocked

Answer 65

only transverse waves can be polarised, you cannot polarise longitudinal waves like sound (the fact that you can polarise visible light is an indication that it is a transverse wave)

Answer 66

you can observe the polarisation of light by shining unpolarised light through 2 polarising filters

Answer 67

1) align transmission axes of 2 polarising filters so they are both vertical. shine unpolarised light on first filter. keep position of first filter fixed + rotate second one 2) light that passes through first filter will always be vertically polarised 3) when transmission axes of 2 filters are aligned, all of light that passes through first filter passes through second 4) as you rotate second filter, amount of light that passes through second filter varies (just like vectors, you can think of transmission axis of rotating filter as having vertical + horizontal component. larger the vertical component, the more vertically polarised light will pass through filter) 5) as second filter is rotated, less light will get through it as vertical component of second filter's transmission axis decreases. this means intensity of light getting through second filter will gradually decrease 6) when two transmission axes are 45 degrees to each other, intensity will be half that getting through first filter. when they're perpendicular to each other, no light will pass through - intensity is 0 7) as you continue turning, intensity should then begin to increase once again 8) when two axes realign (after 180 degree rotation), all light will be able to pass through second filter again

Answer 68

⋅ 3D films use polarised light to create depth - filters in each lens are perpendicular to each other so each eye gets a slightly different picture ⋅ polaroid sunglasses also use polarising filters - reflected light is partially polarised so polaroid sunglasses block this out to help prevent glare

Answer 69

you cannot use polarising filters to polarise microwaves bc their λ is too long, so instead you use a metal grille

Answer 70

you can investigate the polarisation of microwaves using a metal grille, a microwave transmitter and a microwave receiver connected to a voltmeter

Answer 71

1) place metal GRILLE between microwave TRANSMITTER + microwave RECEIVER, as shown below 2) intensity of microwaves passing through grille is at MAXIMUM when direction of oscillation of microwaves + wires on grille are PERPENDICULAR to each other 3) as you rotate grille, INTENSITY of polarised microwaves able to pass through metal grille DECREASES, so reading on voltmeter DECREASES 4) when wires of metal grille are ALIGNED with direction of oscillation of polarised microwaves, NO SIGNAL will be shown on voltmeter

Answer 72

the intensity shown on the voltmeter drops to zero when the wires of the metal grille are aligned with the direction of oscillation of the microwaves bc the electrons in the wires of the grille are absorbing the microwave's energy

Answer 73

microwave transmitters already transmit polarised microwaves, so you only need 1 grille

Answer 74

1) vibrating electric field of microwave EXCITES electrons in metal grille 2) energy of incoming microwave is ABSORBED by electrons in grille + RE-EMITTED in ALL DIRECTIONS 3) only few of those re-emitted waves are vibrating in DIRECTION of microwave receiver 4) microwave RECIEVER only receives microwaves in ONE PLANE, so even if RE-EMITTED wave travels towards receiver, it might not be picked up 5) when wires and oscillations are ALIGNED, MORE electrons are excited than when they're perpendicular to each other - all energy is absorbed + INTENSITY reading drops to ZERO 6) when wires + oscillations are PERPENDICULAR to each other, some electrons in grille are still excited + so there is still SMALL DROP in INTENSITY

Answer 75

analogue signals vary continuously and are not limited in the values they can take

Answer 76

digital signals are represented by binary numbers

Answer 77

yes, the values that a digital signal can take depends on the number of bits used for the signal

Answer 78

⋅ when you transmit an electronic signal, it will pick up noise (interference) from electrical disturbances or other signals ⋅ so the receiver needs to be able to reconstruct the original signal from the noisy signal if they're to get an accurate representation of what was sent ⋅ this is much easier with digital signals than analogue signals bc the number of values a digital signal can take is limited

Answer 79

yes, analogue signals can be digitsed and the process is called digitising digitising an analogue signal means converting an analogue signal into a digital signal

Answer 80

1) take VALUE of analogue signal at REGULAR TIME INTERVALS, then find NEAREST DIGITAL VALUE 2) each DIGITAL VALUE is represented by BINARY NUMBER, so you can CONVERT ANALOGUE values to BINARY numbers

Answer 81

the DIGITAL SIGNAL you end up with won't be EXACTLY the same as the original ANALOGUE SIGNAL, bc some information is always lost (but it's usually quite CLOSE!)

Answer 82

1) the resolution (the number of possible quantisation levels) 2) the sampling frequency

Answer 83

1) the DIFFERENCE between possible DIGITAL VALUES [aka quantisation levels] (this is just RESOLUTION) ⋅ bc if signal is digitised using only FEW, WIDELY SPACED digital values, it's likely that lot of analogue signal will be FAR from NEAREST DIGITAL VALUE ⋅ but, if LARGE number of CLOSELY SPACED digital values are used, most of analogue values will be VERY CLOSE to digital value, so will only change SLIGHTLY ⋅ this means that HIGHER THE RESOLUTION (i.e. MORE POSSIBLE DIGITAL VALUES there are), MORE CLOSELY the digital signal will MATCH original 2) the TIME from one SAMPLE to the next (SAMPLING FREQUENCY aka sampling rate)

Answer 84

the RESOLUTION in digital signals is determined by the NUMBER OF BITS in BINARY NUMBERS representing digital values - so the GREATER the number of BITS in the binary numbers, the GREATER the RESOLUTION

Answer 85

you can lower the rate of transmission used to send information in real-time by using a lower resolution and a lower sampling frequency

Answer 86

⋅ digital signals can often be sent, received and reproduced more easily than analogue signals bc they can only take a limited number of values ⋅ digital files can be compressed to reduce their size + manipulated easily for artistic effect ⋅ digital signals are more resistant to the effect of noise than analogue signals ⋅ digital signals can be used to represent different kinds of information in the same way (eg, images + sounds can both be represented as a string of bits) ⋅ computers can be used to easily process digital signals since they are digital devices too

Answer 87

⋅ digital signals can never reproduce analogue signals exactly - some information will always be lost ⋅ bc digital signals can be copied more easily, digital information can then easily be reproduced illegally an unlimited number of times (so confidential information may be stolen + copied without the owner's knowledge or consent more easily)

Answer 88

⋅ no, a signal can be made up of lots of different frequencies ⋅ in fact, most signals are made up of several waves (not just one), all with different frequencies, added together

Answer 89

when waves are added together, the amplitude of the final signal is the sum of the amplitudes of the individual waves at each point in time

Answer 90

the fundamental frequency is the lowest frequency wave that can make up a sound wave

Answer 91

you can calculate the fundamental frequency of a sound wave by finding the shortest repeating part of that sound wave and calculating the inverse of its period

Answer 92

noise limits the number of bits that can be used when digitising signals

Answer 93

the higher the resolution used to sample a signal, the better a digitised signal matches the original

Answer 94

if the original signal contains noise, then a really fine resolution will reproduce the variation caused by the noise to higher degree of detail also (which is not useful)

Answer 95

resolution is limited by the ratio of the total variation in the signal to the variation caused by noise

Answer 96

maximum number of [useful] bits = log[2] (total variation/noise signal) b = log[2] (Vtotal/Vnoise)

Answer 97

when you digitise a signal, the minimum sampling frequency must be twice the maximum frequency in the original signal

Answer 98

the sampling frequency must be twice the maximum frequency of the og signal so that the sampling frequency is high enough to record all the high-frequency detail of the og signal

Answer 99

⋅ detail can be lost if the sampling rate is too low ⋅ a low sampling frequency can create false low-frequency signals (called aliases) that weren't in the original signal at all

Answer 100

rate of transmission = samples per second x bits per sample OR rate of transmission = sampling frequency x bits per sample units: bits per second

Answer 101

rate of transmission depends on these two factors: 1) sampling frequency - sampling frequency must be at least twice the highest frequency in the original signal to ensure that all frequencies within the spectrum are transmitted correctly 2) the number of bits per sample - the number of bits per sample must be high enough that the transmitted signal closely matches the original signal, but not so high that signal is negatively affected by noise

Answer 102

time taken to transmit a signal = number of bits to transmit / rate of transmission