chapter 11/12 - waves Flashcards
mechanical waves
energy is transferred by vibration of particles carrying the wave through a medium eg sound seismic water
longitudinal waves
oscillate parallel to the direction of energy transfer
transverse waves
oscillate perpendicular to the direction of energy transfer
amplitude
max displacement from undisturbed position
wavelength
distance between two adjacent points on the same cycle of a wave
frequency
number of complete oscillations / time
displacement position graphs
give a snapshot of what a wave looks like at a point in time
- to find which direction each part is moving draw the wave slightly later
speed
freq * wavelength
phase
is the fraction of a cycle it has traveled since the start of the cycle - always and angle in rad
1 cycle = 360 or 2𝞹
phase difference
difference in degrees/ rad / angle between points on the same wave or similar points on 2 waves
- fraction of the cycle that passes between 2 maximums
one complete cycle
2𝞹 = 360
in phase
two waves at the same point in the cycle
phase difference = 2pi or 360
antiphase
two waves doing opposite things at the same point
phase difference = pi or 180
coherent
waves with constant phase difference
must have the same freq to make phase diff constant
interference
the superposition of coherent waves
principle of superposition
when waves are at the same point there resultant displacement is = to the sum of individual displacements
CRO traces
amplitude - count squares * volts
time period - count squares * time base
frequency - 1/T
progressive wave
transfers energy from one point to another
phase difference equation
phase difference = 2pi d/ wavelength
phase diff/ 2 pi = d/ wavelength
intensity
radiant power per unit area
Power/ Area
- radiates equally in all directions I = P/ 4pi r^2
(inverse square law)
intensity is proportional to
KE
v^2
A^2
v is proportional to A
refraction diagram
- incident ray
- normal
- angles
- emergent ray
- refracted ray
- direction arrows
- change of direction
- title
reflection
wavelength is same because speed doenst change
refraction
wavelength decreases if speed decreases as f stays same
diffraction
- no effect on wavelength or speed
- waves going through a gap
- larger aperture = minimal diffraction
- max diffraction is when wavelength is closed to gap size
refractive index
tells us how much faster the wave travels in a vacuum than the medium
refractive index equations
n2/n1 = c1/ c2 = wavelength1/ wavelength2 = sin i / sin r
snells law
n1 sin i = n2 sin r
TIR
- occurs if i> C
- travelling into a less dense medium
critical angle
C = sin-1 (n2/n1)
in a vacuum/ air
C = sin-1 (1/n)
optical fibres (cladding)
- cladding has lower n than core so light travels as it’s internally reflected along
- if you just used air crossover would occur when fibres touch
- cladding also products core and gives it strength
multi path dispersion
- diff paths with diff lengths and times (eg straight through is fastest)
- leads to pulse broadening which reduces max freq- can be prevented with narrower fibres
spectral dispersion
- speed of light decreases with shorter wavelength - diff colours have diff λ so diff speeds
- use monochromatic light to prevent pulse merging
images carried along optical fibres
formed from light and dark dots
- need a coherent bungle so image appears at the other end
analogue vs digital signals
when waves are sent noise and attenuation can occur
- when you use digital signals it’s easier to see the value of the wave (1 or 0)
hwo signals are sent
carrier wave and signal wave
- can be amplitude modulated - amplitude if the carrier wave shows the message wave
- or freq modulated - means the frequency of the wave shows the peaks and troughs do the message wave
attenuation
further away the weaker the signal becomes
hwo to fix attenuation
repeaters can regenerate digital signals to their full strength agains
noise
signal becomes less sharp which can be prevented by an amplifier or repeater
- easier to fix on digital signals
polarisation
- limits vibrations to one plane
(only transverse not longitudinal - as vibrations are perp to direction of travel)
partial polarisation
- when waves are reflected off transparent surfaces (glare)
metal grille with polarised microwaves
- put through polarising filter
- grilled allows vertically polarised vibrations
- when it’s turned 90° all waves are blocked (absorbed) as it would onyl let through horizontal polarisation
as you tuyrn back to 180 /0 microwaves pass through (max at 180/0 as I = Icos²θ
hwo to create microwaves
- the transmitter has ac passed through it at the same freq as the intended wave
- the electric field created oscillates vertically at the same freq as the current
why does the grille block vertically polarised microwaves
- free electrons in grille oscillate vertically inducing I
- electrons emit an electric field- in all directions
- do very little is transmitted in the direction of the receiver
why does a horizontal grille not block vertical microwaves
-electrons in grille aren’t effected by vertical field
- so continues to transmit wave vertically as no current is induced
grille at 45°
some are recieved as the field is resolved into parallel and perpendicular to the grille
when reciever and emitter are differently orientated
if perp no signal is recieved
- if 180 = 0 so all is recieved
inverse square law
I is proportional to 1/r^2
(also I is prop to a^2)
malus law
intensity of polarised light
I = I0 cos^2(angle)
where I0 = initial intensity
angle = angle of the polariser
polarise sunglasses
- glare is partially polarised light in the horizontal plane by reflection
- glasses only transmit vertically polarised light
- glare isn’t let through so reduced
2 filters
if they are perp no lights will pass through
strain also alters optical activity
brewsters angle ???
n = tan(angle)
angle of incidence at which light with particular polarisation is perfectly transmitted
arials
- mesh reflects signal back to the front
- circular rod picks up the signal ( 1/2 wavelength tall)
- other rods boost signal (1/2 wavelength apart)
adjacent emitters alternate between horizontal and vertical to avoid interference if they overlap signal areas
superposition
at the point waves overlap we can find the wave position using the principle of superposition
constructive interference
waves arriving at a point in phase causes reinforcement
crest + crest = supercrest
destructive interference
waves in anti phase cause cancellation
crest + trough = nothing
nodes
form where waves are in antiphase (destructive interference)
antinodes
where waves are in phase - constructive interference
path difference
difference in the distance from 2 sources
if an odd multiple of 1/2 wavelengths it’s antiphase
if it’s a whole no. is in phase
change in phase difference
2 pi path difference / wavelength
lines of maxima
connect points where path difference is a whole no. wavelengths
(where displacement is a maximum)
lines of minima
connect points where the path difference is an odd multiple of 1/2 wavelength
(where displacement is zero)
nth order maxima
path difference if n wavelengths
phase difference of 2n pi
central maxima is where path+ phase diff = 0
measuring the wavelength of microwaves
- probe is moved over paper from the central maxima along
- positions of maxima marked and path diff is determined
- each maxima can be used to calc wavelength and then you average
young’s double slit experiment
- pass light through a colour filter (monochromatic light) then two slits
- creates fringes of interference
fringes
uniformly spaced near the centre of the interference pattern - bright and dark patches
how to find wavelength from youngs double split
wavelength = ax/ D
a = slit separation
x = finge separation
D = distance to screen
S2P-S1P (distance from far slit to maxima - distance from close slit to maxima)
is the path difference = n * xa/D =n λ
for first order maxima path difference = 1λ
youngs double slit risks
- don’t look at laser
- remove reflections
- do in dark room
- warning sign of laser
- take into account laser orientation
diffraction grating equation
more slits = narrower bands of brightness
d sin(θ) = n λ
n(max) = d/ λ (as angle can’t exceed 90 (sin(90) = 1))
if x lines per mm
slit serrations = 1 / x*1000
ripple tank
transparent tray with water which is lit up from above - waves can be seen with crests as bright and troughs as dark
if you use a strobe light the waves will appear to nto move
plane waves
travel the same direction at the same speed
equally spaced wavefronts
energy perp to wavefronts
circular waves
wavefronts travel in all directions outwards at the same speed
circular wave reflections at plane barriers
image of the source is formed on the other side of the reflector as waves appear to originate from there
reflection at a concave barrier
circular waves produce plane waves
plane waves form a circular wave
diffraction
plane incident on a gap will produce a circular wave pattern - bend through the gap
the larger the gap the less the diffraction
gap = wavelength means max diffraction
EM waves
- transverse
- travel in vacuum
- at speed of light
- electrics field travels perp to magnetic field
progressive waves
transfer energy
stationary waves
when two waves moving in opossite directions with the same freq, v and a are supperposed
where a = zero = node
where a = max - antinode
hwo stationary waves are formed
waves are transmitted
waves refelct of the wall
waves meed transmitted waves and superpose - the two waves are moving in opossite directions with the same freq, v and a so when supperposed form a stationary wave
stationary waves - antinodes / nodes
where a = zero = node
where a = max - antinode
the distance between two nodes/ antinodes is λ/2
fundamental frequency - streched string
when the freuqency is = to the first harmonic - there is one antinode and 2 nodes
L = λ/2
nth harmoic - on a string
there are n antinodes and n+1 nodes
L = nλ/2
f = nc/2L = nf₀
pipe open both ends
f₀- antinode at both ends and node in middle
fₙ - n+1 antinodes and n nodes
L = nλ/2
f = nc/2L = nf₀
pipe closed a both ends
f₀- node at both ends antinode in middle
fₙ - n+1 nodes and n antinodes
L = nλ/2
f = nc/2L = nf₀
pipe - open one end closed other
f₀ - one node one antinode
fₙ - n nodes n andtinodes
L = (2n+1)λ/4
f = (2n+1)c/4L = (2n+1)f₀
open end
antinode
closed end
node
