Physics 306: Waves Lecture 10 3/31/2011

Interferometers

What is an interferometer? It is any instrument designed to exploit interference of in any number of ways.

Idea: Divide an initial beam into two or more points that travel different path lengths and are reunited to produce an interference pattern.

“Amplitude division” interferometers use a beam-splitter that divides the beam into 2 parts which then combine.

The most famous example of such an interferometer is the Michelson, first constructed in 1881 [Used to establish the evidence of special relativity, to detect and measure hyperfine structure, and much more…]

Basic schematic of a :

(Note: the beams are actually on top of each other but are shown separated for illustration purposes only)

1. Source emits waves, part of which travel to the right. 2. Beam splitter divides the beam into 2 3. The 2 waves are reflected by M1 and M2 4. They return to beam splitter 5. Part of wave coming from M2 goes through the beam splitter and part of wave coming from M1 gets deflected by beam splitter to detector/screen 6. Waves reunite and interfere

Notice that beam (1)-(2)-(4) goes through the beam-splitter only once but beam (1)-(3)- (4) goes through the beam splitter 3 times. To compensate for this, a “compensator”

Page 1 of 4 306: Waves Lecture 10 3/31/2011 plate is typically inserted in beam (2), made of same material as beam splitter (but no reflective coating) to account for the different thickness of glass traveled by each beam. Then any optical path differences arise from actual path differences between the two beams.

To understand more simply the optical path difference between the two beams and how we get fringes, let us envision an equivalent linear optical system, having a single optical axis (instead of the two shown in the schematic above), and by working with virtual images of the source and M1 that occur from in the beam splitter mirror.

So, if we rotate assembly that includes source, M1, and beams (1) and (3) CCW by 90 degrees about the beam splitter interface, we get the following equivalent linear assembly:

(Note, here it is assumed that the path along beams (3)-(4) is shorter than the (1)-(2) path)

Light from the source, emanating from pt. Q reflects from both M2 and M1’. The source appears to come from Q1’ and Q2’. Because light has to travel the round trip distance when traveling from M1 and M2, then this means that the separation between Q1’ and Q2’ has to be twice the mirror separation.

So, the physical optical path difference, Δp between the two emerging beams is:

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"P = 2d cos!

Where θ is the angle between the incoming beam and the optical axis. So, if we had normal incidence, θ = 0 and cosθ=1 and :

! P = 2d

Which we would expect since if M2 is farther from the beam splitter than M1 by d, the extra distance traveled by the beam going from M2 includes the distance d twice (once before and once after reflection).

So, the physical path difference is "P = 2d cos! , but there is also reflection occurring ! which introduces a phase shift of π or " = , so the total path difference is: r 2

! # = # + # = 2d cos" + P r 2 For dark fringes (destructive interference), the total optical path difference has to be half integer or:

! 1 2d cos" + = (m + )! , or 2 2 2d cos! = m" where m=0,1,2,3,…(1)

If d is such that normal rays (θ=0) yield a dark fringe at the center, then its order (m) is given by:

2d m = , max ! where mmax is some large number. Neighboring dark fringes decrease in order outward, since cosθ < 1, and as you can see: 2d cos! m = (so m gets less for θ>0) max " So you get circular fringes :

m= 99

mmax=100(for example)=

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You can see that as d changes, a particular point on the fringe pattern (which corresponds to a particular θ) will correspond to a different m. As d varies, fringes appear to move inward toward center (or outward, depending on whether Δ is increasing or decreasing).

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