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Physics 2 > Optics > Flashcards

Optics Flashcards

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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
Q

Brewster’s angle

A

angle of incidnece where reflected polarised ray and refracted (slightly polarised) ray are perpendicular

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26
Q

polarisation by birefringence

A

splitting the light as the parallel and perpendicular polarisations have different refractive indices

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27
Q

ordinary ray

A

linear polarisation that is perpendicular to the optical axis which sees ‘n0’

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28
Q

extraordinary ray

A

linear polarisation at 90 degrees that is parallel to the optical axis, this polarisation sees ‘ne’

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29
Q

waveplates

A

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

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30
Q

half waveplate (HWP)

A

lambda/2

the thickness of the crystal produces a half-wave shift between the two polarisations
phase change = pi

rotates the polarisation

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31
Q

quarter waveplate (QWP)

A

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

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32
Q

3d glasses

A

viewer wears glasses with opposite polarising filters for each eye

each filter passes only light similarly polarised and blocks the opposite polarisation

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33
Q

two different ways of making 3d glasses

A
  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
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34
Q

polarisation by scattering

A

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

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35
Q

phasor length

A

amplitude

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36
Q

phasor angle

A

phase

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37
Q

wave intensity

A

I = |E|^2 = Ae^iwt . Ae^-iwt

multiplied by its complex conjugate

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38
Q

superposition of waves derivation

A

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)

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39
Q

superposition of waves derivation - if A1=A2

A

use 2cos^2theta = 1 + cos2theta

get I0=cos^2(delta/2)

40
Q

superposition - if w1t=w2t

A

no phase difference
get max value

41
Q

superposition - if w1t = -w2t

A

4A^2cos^2wt

42
Q

superposition - if w1t = w2t + pi

A

zero

43
Q

constructive interference

A

when two waves are in phase

for equal intensities I=A^2, the resulting intensity is (2A^2)=4I

44
Q

destructive interference

A

when two waves are out of phase

for equal intensities, I=A^2, the resulting intensity is zero

45
Q

location of minima in intensity

A

delta = (2m+1)pi

path diff = (m+1/2)lambda

46
Q

location of maxima in intensity

A

delta = 2mpi

path diff = mlambda

47
Q

Huygens principle

A

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
Q

applications of Huygens principle

A

refraction

young’s double slits

49
Q

young’s double slits

A

each slit becomes a new emitter producin a new wavefront

the interference pattern of the beams produces a sinusoidal pattern

50
Q

two waves E1=A1e^itheta1 and E2=A2e^itheta2 are said to be coherent if

A

their phase difference (theta 1 - theta 2) is constant

oscillating in step

51
Q

temporal coherence

A

measure of purity of the wave

different wavelength = low coherence
same wavelength = high coherence

52
Q

coherence time tc

A

the time over which the phase of a wave at a given point remains predictable

53
Q

coherence length lc

A

=tc c
length over which the phase of the wave remains predictable

54
Q

coherence length of lasers

A

lasers have very narrow linewidth leading to lc»10km

55
Q

spatial coherence

A

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
Q

temporal coherence can be quantified by

A

measuring variations in phase difference between points on a line radiating out from the source

57
Q

spatial coherence can be quantified by

A

measuring variations in phase difference between points at the same distance from the source, but separated laterally

58
Q

coherent light

A

light that has properties of temporal and spatial coherence

59
Q

interferometer

A

device using interference to measure lengths and/or indices of refraction

60
Q

Mach Zehnder inferometer

A

laser (or other light source)

goes to square of 2 beam splitters and 2 mirrors

ends at screen

61
Q

simple inferometer

A

source goes to beam splitter in middle

straight on or up to mirror

back to middle and down to detector

62
Q

inferometer delta=

A

2|L1-L2|

63
Q

beam splitter

A

usually a semi-silvered mirror

some light reflecting, some transmitted

64
Q

Michelson inferometer

A

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
Q

purpose of the compensation plate

A

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
Q

why is mirror 2 movable

A

can be moved an accurately measurable distance to allow changes in the path difference to be measured

67
Q

inferometer - ether

A

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
Q

mechanical wave on rope - heavy rope tied to light rope

A

transmitted wave continues

reflected wave moves back and undergoes no phase change

waves travel slower on thick ropes than on thin ones

69
Q

mechanical wave on rope - light rope tied to heavier rope or a rigid support

A

reflected wave undergoes a half-cycle phase shift

70
Q

EM wave propagating in optical materials - if transmitted wave moves faster than the incident wave

A

reflected wave undergoes no phase change

71
Q

EM wave propagating in optical materials - if the incident and transmitted waves have the same speed

A

there is no reflection

72
Q

EM wave propagating in optical materials - if the transmitted wave moves slower than the incident wave

A

the reflected wave undergoes a half-cycle phase shift

73
Q

if n1<n2

A

reflected light has phase change of pi

74
Q

if n1=n2

A

no interface so no reflection

75
Q

if n1>n2

A

no phase change

76
Q

Lloyds mirror

A

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
Q

Lloyds mirror phase difference

A

delta = 2pi/lambda dsintheta - pi

78
Q

michelson inferometer fringe pattern

A

circles

going out d

d prop to theta

79
Q

fringe formation for parallel mirrors - the path difference between beams 1 and 2

A

delta = 2dcostheta

80
Q

fringe formation for parallel mirrors - why is there a pi phase difference between beams

A

counting phase flips during reflections

beam 1 undergoes two reflections with n1<n2

beam 2 only 1

overall pi phase change

81
Q

fringe formation with phase flip - although total path difference is 2dcostheta the total phase difference is

A

delta = 2pi/lambda(2dcostheta) - pi

=2pi/lambda (2dcos theta-lambda/2)

82
Q

fringe formation with phase flip - looking at superposition of two electric field

A

I12=I0cos^2(delta/2)

83
Q

fringe formation - condition for constructive interference

A

2dcostheta=lambda(m+1/2)

84
Q

fringe formation - condition for destructive interference

cocentric dark rings

A

2dcostheta=lambda m

85
Q

fringe formation - the physical path difference ‘d’ can changed by

A

moving the mirror

86
Q

dark port condition

A

if d=0 then I12=0 regardless of theta or m

corresponds to darkness across the whole field

87
Q

misalignment fringes - fringe spacing

A

D=lambda / sin thi

approx lambda / thi at small angles

88
Q

misalignment fringes - tilt angle

A

theta = 2 theta_mirror

89
Q

if a source of white light is used, fringes will only be seen if

A

the path length difference is smaller than a few wavelengths

(coherence length approx 1 micro metre)

90
Q

compensation plate is necessary to get

A

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
Q

An optical thin film can be engineered to make

A

a coating that reduces reflection from a surface, filters
wavelength of light or makes highly reflecting surfaces

92
Q

The optical thin films will have a thickness
similar to

A

a single wavelength of light for which the coating is being designed.

93
Q

To calculate the interference effects of a thin film we consider

A

r beam of light from a source that is
incident on to the surface of a thin film with refractive index n

94
Q

thin films - Part of the beam is reflected at the
angle of incidence θ, and a part is

A

refracted at an angle θ
′ according to Snell’s law

95
Q
A

Physics 2 (8 decks)

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