4.2 Interferometers

Question 1

uses of itnerferometers

Answer 1

When more resolution needed than even very large spectrometer
- measuring Y of single mode lasers
- laser linewidth

Question 2

Basic principle of interferometers

Answer 2

-Incident lightwave with intensity I0 is divided into partial beams with amplitudes Ak,
-which pass digerent optical path lengths sk = nxk (where m is the refractive index)
- before they are again superimposed at the exit of the interferometer.
-Partial beams come from the same source, they are coherent
- Total amplitude of the transmitted wave is the superposition of all partial waves, and depends on the amplitudes Ak and on the phases phik = phi0 + 2pisk/Y of the partial waves

Question 3

Partial beams in an interf come from the same source adn they coherent so long as ….

Answer 3

as long as the maximum path diverence does not exceed the coherence length.

Question 4

WHy in interfometer sensitively dep on the wavelength Y

Answer 4

the transmitted wave is the superpos of all partial waves ad dep on their amplitudes and phases (phik(Y))

Question 5

Maxima.minima in interfogram obtained when

Answer 5

All partial waves interefere constrcutively

OPDiff = dsik = si-sk =mY (full wavlenegths apart)

minima when mY/2 (half wavelength apart -> total destrc intef)

Question 6

FSR of interometer

Answer 6

dY = Ym - Ym+1 = deltas/m - deltas/(m+1)

but note that form one terferometric measurement you can only determine Y module m.dY because all Y are equibalent wrt transmission in interf Y = Y0 +mdY

Question 7

what can use use with interf

Answer 7

Need to masure within 1 FSR of interf first using maybea spectrometer
interformeter CANNOT unqieuly det a wavlength on its own

Question 8

Michaelson Intef

Answer 8

Y dependent filter for accurate Y measurement

usual 2 mirors BS assume no losses at mirror
ohse difference between two waves phi = factor due to path diff + additional phase delta phi due to reflections (constnat)

since R and T are for intensities the resultant waves moduled (A not I) both by sqrt(RT), typical 5050 BS => T = R = 1/2

enforce that A1^2 + A2^2 = A0^2

Question 9

What does the detector measure in a michaelson I

Answer 9

The time avged intensity

I = 1/2 nc ep0 AA*
where A is the resultant E wave of summed two beams towgether

I = 1/2I0 (1+cos(theta))
when you sub in teh E = sqrt(RT) A0 exp{i(wt+phi0} (1+esxp(iphi))

where the exp (iphi ) contrib is the shifted wave (addiational phase …)

Question 10

Howdoes interfereogram output look

Answer 10

Intensity vs time
peaks and troughs
peaks at separations of mY -> 2mPi
troughs at sep of mY/2 -> (2m+1) pi -> half wavelgnths (total destr)

Question 11

What is used to move th edelay stage

Answer 11

not a micrometer since that is not nm secnsitive
use piezo electric crystals to adjust OPL and cycle through many many cycles of peaks and troughs in intensity to get a good grip on Y (statistically)

Question 12

Def coherence length

Answer 12

The max path difference delta s that still gives interference finrges in the plane B is limited by the coherence length of the incident radiation

spectral lamps the coherence length is a few cm , usingsingle mode stabilized laser this is several km.

Question 13

what if you go past coherence length

Answer 13

IRL never get infinite coherence length since that requires perfect Y, if you only see a constnat I you are past the conherence length and no interference pattern is visible

Question 14

Use of MI

Answer 14

LIGO
side arms of 1km where OPL can be increases to 100km by using highly relfective mirros and ultra stable laser with very very long coherence length

Question 15

Spectral resolving power of MicInt calc

Answer 15

Y/dY = ds/Y

determine by saying the the number of maxima at Y1 -> N1 and numebr of maxima at Y2 -> N2 must differ by at least 1 to resolve them N2 >= N1+1

note then that you mean Y1 = Y2 +dY rearr to get resolution

Question 16

Spectral resolving power of MicInt in words

Answer 16

the spectral resolving power ( how closely can you distinguish wavlengths)Y/dY of the MicInt = the amximum path difference delta s/Y measyred in units of wavelength

Question 17

hwo does wavlenth relate to number of maxima achineved in the michaelson interf

Answer 17

Shorter wavlength means more cycles can fit in which means more maxima

longer Y -> smaller N (#peaks)

Question 18

in multiple beam interferometer what does F or R determine

and phi

Answer 18

F (not the finesse) determines the peak height (F(R))

phi determines the periodicity of the peaks

in IT and IR expressions

Question 19

Sagnac rundown

Answer 19

1) BS splits incoming beam into a trans and refl beam as usaul
2) two beams circulate in xyplane in OPP directions while the whole interferometer roates around z axis
3) OPL diff for two beams
phase shaift between partial waves -> interference

this change can also be attrib to dopler effect

Question 20

How can you make sagnac more effective

Answer 20

Have optical fibreas circulate N times round cavity such that the effective area become NxA and that increases teh sensitivty significantly

it is already very sensitve to even small angular velocities Omega since that changes path length

Question 21

what is triangle phi usualy here in interferometers

Answer 21

Phase shift between two partial waves that interfere