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Optics Flashcards

(95 cards)

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1
Q

radio wavelength

A

> 1m

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2
Q

microwave wavelength

A

between 1m and 1mm

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3
Q

visible light wavelength

A

400-700nm

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4
Q

UV wavelength

A

100-400nm

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5
Q

IR wavelength

A

between 780 nm and 1 mm

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6
Q

x rays and gamma rays wavelength

A

< 10nm

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7
Q

direction of energy flow is given by

A

the poynting vector

S=1/u0 ExB

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8
Q

wave equation

A

describes its position in space

E(t,z) = Asin(wt-kz+Φ)

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9
Q

polarisation

A

direction of the electric field oscilation of a light beam

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10
Q

circular polarisation

A

E-field direction is processing around circle

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11
Q

elliptical polarisation

A

somewhere in between linear and circular

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12
Q

random polarisation

A

ordinary light source consists of a very large number of randomly orientated atomic emitters

result is unpredictable, rapidly changing polarisation

(unpolarised)

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13
Q

in general, light is neither completely…

A

polarised or unpolarised

(partially polarised)

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14
Q

a polariser is

A

an instrumnet that selects only a specific polarisation from incoming randomly polarised light

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15
Q

4 fundamental mechanisms that polarisers are based on

A

dichroism (by absorption)
reflection
scattering
birefringence

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16
Q

polarisation by absorption (dichroism)

A

selective absorption of one of the two orthogonal E field polarisation components

wire-grid polariser: polarisation parallel to wires is absorbed

wire spacing needs to be much smaller than wavelength

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17
Q

polarised filter, stretched polymer chains

A

stretched sheet of polyvinyl alcohol, get long alligned molecules

sheet dipped in iodine solution, iodine attaches to plastic molecules

conduction of electrons of iodine can then move along those molecules as if they were long wires

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18
Q

dichroism in crystals

A

some crystals have a strong anisotropy in their crystal structure

light that is perpendicular to their optical axis is absorbed

leads to different colours depending on viewing angle

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19
Q

Malus’ law

A

E parallel = EcosΦ

I=Imaxcos^2Φ

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20
Q

polarisation by reflection

A

by reflection from a higher refractive index

unpolarised light incident at polarising angle, reflected light 100% polarised perpendicular to the plane of incidence

transmitted light is partially polarised parallel to plane of incidence

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21
Q

derivation of Brewster’s angle

A

start with snell’s law

require θp+ θt=90 degrees
θt=90- θp

ni sin θp = nt sin θt
ni sin θp = nt cos θp

sin θp/cos θp = nt/ni

tan θp=nt/ni
rearrange for θp

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22
Q

s polarisation

A

not parallel

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23
Q

p polarisation

A

parallel

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24
Q

at Brewster’s angle, only get

A

s polarisation

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25
Brewster's angle
angle of incidnece where reflected polarised ray and refracted (slightly polarised) ray are perpendicular
26
polarisation by birefringence
splitting the light as the parallel and perpendicular polarisations have different refractive indices
27
ordinary ray
linear polarisation that is perpendicular to the optical axis which sees 'n0'
28
extraordinary ray
linear polarisation at 90 degrees that is parallel to the optical axis, this polarisation sees 'ne'
29
waveplates
rotation of the polarisation by a birefringent crystal light is normal incident there is a phase difference as light travels at a different speed *thicker material = more difference in phase*
30
half waveplate (HWP)
lambda/2 the thickness of the crystal produces a half-wave shift between the two polarisations phase change = pi rotates the polarisation
31
quarter waveplate (QWP)
lambda/4 the thickness of the crystal produces a half-wave shift between the two polarisations phase change = pi/2 produces elliptical or circular polarised light
32
3d glasses
viewer wears glasses with opposite polarising filters for each eye each filter passes only light similarly polarised and blocks the opposite polarisation
33
two different ways of making 3d glasses
1. linearly polarised at +/- 45 degrees viewers must keep level head as tilting causes images to bleed over 2. circular polarised right/left (quarter waveplate and linear polariser viewers can tilt head and still maintain left/right separation
34
polarisation by scattering
electric charges in air molecules oscillate in the direction of the E field of the incident light from the sun that produce scattered light scattered light reaches observer air molecules scatter blue light more than red, sky looks blue overhead red sunset because blue has been scattered away
35
phasor length
amplitude
36
phasor angle
phase
37
wave intensity
I = |E|^2 = Ae^iwt . Ae^-iwt multiplied by its complex conjugate
38
superposition of waves derivation
start with E=A1e^iw1t + A2e^iw2t collect the e^iwt together and e^-1wt together use foil to multiply out take A1A2 out as common factor and change so signs are same in brackets sub in 2cos(w1t-w2t)
39
superposition of waves derivation - if A1=A2
use 2cos^2theta = 1 + cos2theta get I0=cos^2(delta/2)
40
superposition - if w1t=w2t
no phase difference get max value
41
superposition - if w1t = -w2t
4A^2cos^2wt
42
superposition - if w1t = w2t + pi
zero
43
constructive interference
when two waves are in phase for equal intensities I=A^2, the resulting intensity is (2A^2)=4I
44
destructive interference
when two waves are out of phase for equal intensities, I=A^2, the resulting intensity is zero
45
location of minima in intensity
delta = (2m+1)pi path diff = (m+1/2)lambda
46
location of maxima in intensity
delta = 2mpi path diff = mlambda
47
Huygens principle
each point on a wave front acts as a point source of new waves the envelope of secondary waves forms new wave front and so on
48
applications of Huygens principle
refraction young's double slits
49
young's double slits
each slit becomes a new emitter producin a new wavefront the interference pattern of the beams produces a sinusoidal pattern
50
two waves E1=A1e^itheta1 and E2=A2e^itheta2 are said to be coherent if
their phase difference (theta 1 - theta 2) is constant oscillating in step
51
temporal coherence
measure of purity of the wave different wavelength = low coherence same wavelength = high coherence
52
coherence time tc
the time over which the phase of a wave at a given point remains predictable
53
coherence length lc
=tc c length over which the phase of the wave remains predictable
54
coherence length of lasers
lasers have very narrow linewidth leading to lc>>10km
55
spatial coherence
choose two points on same wavefront at t=0 for any t>0 the if phase difference between two points stays constant, wave has spatial coherence between those points
56
temporal coherence can be quantified by
measuring variations in phase difference between points on a line radiating out from the source
57
spatial coherence can be quantified by
measuring variations in phase difference between points at the same distance from the source, but separated laterally
58
coherent light
light that has properties of temporal and spatial coherence
59
interferometer
device using interference to measure lengths and/or indices of refraction
60
Mach Zehnder inferometer
laser (or other light source) goes to square of 2 beam splitters and 2 mirrors ends at screen
61
simple inferometer
source goes to beam splitter in middle straight on or up to mirror back to middle and down to detector
62
inferometer delta=
2|L1-L2|
63
beam splitter
usually a semi-silvered mirror some light reflecting, some transmitted
64
Michelson inferometer
source eg sodium lamp in middle to beam spliiter (half silvered surface) and compensation plate fixed mirror straight on moveable mirror up telescope/eye down
65
purpose of the compensation plate
match the optical path lengths of the two beams beam 2 travels 3 times through BS glass while beam 1 only travels ones compensation plate is identical to the BS glass
66
why is mirror 2 movable
can be moved an accurately measurable distance to allow changes in the path difference to be measured
67
inferometer - ether
at the time thought light travelling through ether as earth orbits sun, should see change as earth orbits no results from rotating in the ehter so ruled out
68
mechanical wave on rope - heavy rope tied to light rope
transmitted wave continues reflected wave moves back and undergoes no phase change waves travel slower on thick ropes than on thin ones
69
mechanical wave on rope - light rope tied to heavier rope or a rigid support
reflected wave undergoes a half-cycle phase shift
70
EM wave propagating in optical materials - if transmitted wave moves faster than the incident wave
reflected wave undergoes no phase change
71
EM wave propagating in optical materials - if the incident and transmitted waves have the same speed
there is no reflection
72
EM wave propagating in optical materials - if the transmitted wave moves slower than the incident wave
the reflected wave undergoes a half-cycle phase shift
73
if n1
reflected light has phase change of pi
74
if n1=n2
no interface so no reflection
75
if n1>n2
no phase change
76
Lloyds mirror
two sources, real (s) and virtual (s') become analogues to the two slits in Young's double slits experiment dark fringe at bottom, bounced off mirror so pi phase change and destructive interference
77
Lloyds mirror phase difference
delta = 2pi/lambda dsintheta - pi
78
michelson inferometer fringe pattern
circles going out d d prop to theta
79
fringe formation for parallel mirrors - the path difference between beams 1 and 2
delta = 2dcostheta
80
fringe formation for parallel mirrors - why is there a pi phase difference between beams
counting phase flips during reflections beam 1 undergoes two reflections with n1
81
fringe formation with phase flip - although total path difference is 2dcostheta the total phase difference is
delta = 2pi/lambda(2dcostheta) - pi =2pi/lambda (2dcos theta-lambda/2)
82
fringe formation with phase flip - looking at superposition of two electric field
I12=I0cos^2(delta/2)
83
fringe formation - condition for constructive interference
2dcostheta=lambda(m+1/2)
84
fringe formation - condition for destructive interference cocentric dark rings
2dcostheta=lambda m
85
fringe formation - the physical path difference 'd' can changed by
moving the mirror
86
dark port condition
if d=0 then I12=0 regardless of theta or m corresponds to darkness across the whole field
87
misalignment fringes - fringe spacing
D=lambda / sin thi approx lambda / thi at small angles
88
misalignment fringes - tilt angle
theta = 2 theta_mirror
89
if a source of white light is used, fringes will only be seen if
the path length difference is smaller than a few wavelengths (coherence length approx 1 micro metre)
90
compensation plate is necessary to get
level of precision in arm length difference the fringes for a given colour are more widely spaced the greater the wavelength and the fringes for different colours only coincide for d=0
91
An optical thin film can be engineered to make
a coating that reduces reflection from a surface, filters wavelength of light or makes highly reflecting surfaces
92
The optical thin films will have a thickness similar to
a single wavelength of light for which the coating is being designed.
93
To calculate the interference effects of a thin film we consider
r beam of light from a source that is incident on to the surface of a thin film with refractive index n
94
thin films - Part of the beam is reflected at the angle of incidence θ, and a part is
refracted at an angle θ ′ according to Snell’s law
95

Physics 2 (8 decks)

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