SHORT EXPERIMENT
Michelson Interferometer
The Michelson interferometer is one of the simplest of a whole set of instruments (interferometers) which can be used to study small differences in thickness, refractive index, wavelength, etc. The purpose of this experiment is:
· to understand how interference fringes are formed in a Michelson,
· to find them in the interferometer (both in "monochromatic" and "white" light), and
· to use these fringes to determine, to an accuracy of 1-2%, the separation of the two spectral lines of the sodium doublet.
The Michelson interferometer is shown schematically in Figure 1. Light from an extended source S is split in two by a beamsplitter B and the two beams then travel to mirrors Ml and M2. C is a "compensator plate" (i.e. a piece of glass of exactly the same thickness as the beamsplitter substrate) which ensures that both beams travel through the same amount of glass: we shall see later why this is necessary. The light reflected by the mirrors is re-combined at the beamsplitter and is viewed as indicated. The two returning beams interfere, and the fringes depend on the positions and orientations of the mirrors.
Figure 1: Schematic of the Michelson interferometer
The analysis of the instrument is simplified if it is recognised that the two arrangements shown in Fig. 2 are equivalent. From Fig 2, it is clear that the fringes are similar to those observed from thin films, the interference occurring between waves reflected from the "front" and "back" surfaces (M2 and M1 in our case). Let us consider in a little more detail the fringes formed when the two mirrors are precisely parallel and perfectly flat. Because the source is extended, rays are incident at different angles on the pair of mirrors M1 and M2. Referring to Fig. 3, it is obvious that the path difference for normal incidence (q = 0), is simply 2d. If this path difference is equal to an integer number of wavelengths, then a bright fringe will be seen at the centre of the field of view[1].
Figure 2: Optically equivalent ray diagrams for the Michelson interferometer.
Figure 32: Fringe formation after reflection by M1 and M2.
Now, for an angle q of incidence, i.e. light from another part of the extended source, the path difference is (referring to Fig. 2), ABC-AD. Show that this is equal to 2dcosq. Thus the condition for a bright fringe is
2dcosq = nl (1)
Can you see that this means the fringes are circular? Suppose that the interferometer is set up so that there is a bright fringe at the centre, i.e. 2d = nl. Then the first bright fringe will occur for an angle given by 2dcosq = (n-1) l, or, substituting,
or for small q
Clearly, the fringes subtend small angles when d is large and subtend larger angles as d decreases. When d is zero, i.e. the two paths in the interferometer are exactly equal, then the circular fringes have infinite diameter (in optical jargon, the fringes are "fluffed out") and a uniform field is seen.
In practice, the two mirrors Ml and M2 will not be exactly parallel or perfectly flat and therefore the pattern of fringes will be some combination of circular fringes, straight-line tilt fringes and any additional distortion due to the imperfections in the flatness of the mirrors. In the interferometer provided, the mirrors are reasonably flat (to about a wavelength or perhaps a bit less) and the dominant perturbation on the circular fringes are the straight-line tilt fringes, which combine to give oval fringes.
List of Equipment
· Ealing-Beck Michelson interferometer
· Dual light source (mercury/tungsten) with power supply
· Sodium lamp
Experiment 1: Finding the Fringes with a Mercury Source
(i) Make a light pencil mark (or, better, use an existing one!) at one edge of the ground glass diffuser. Switch on the Mercury source and look into the instrument: you should see three images of the mark, i.e. three images of the source. Why three? One is to be ignored: which one? Adjust the mirror screws until the other two images are "exactly" superposed: do this as accurately as you can. Mirrors M 1 and M2 are now approximately parallel. Fringes should now be visible. (If they are not, make fine adjustments until you see them or start again, making all adjustments as carefully as you can.) Make further fine adjustments until you see circular fringes in the field of view. Remember to keep an accurate record of your observations in your lab book.
(ii) Turn the micrometer to move M2 (i.e. alter d). Do the fringes collapse towards the centre or expand? Think about the expression for the path difference (= 2dcosq) to decide whether d is increasing or decreasing, Write down your reasoning in your lab book. Turn the micrometer fairly rapidly in the direction of decreasing d: you may need to keep the circular fringes central by adjustment of the two mirror screws. Notice that the size of the pattern (radius of any given ring) is becoming greater (well, it should be: if it is not, you are moving in the wrong direction).
(iii) Keep turning the micrometer until only ~2 circular fringes can be seen in the field of view at any given moment. Turn one of the mirror screws slightly to introduce about 10 "tilt" fringes. These fringes will be slightly curved (they are sections of large circles whose centre is well outside the field of view). Note the direction of curvature. Now turn the micrometer screw in the same direction as before until the fringes are "straight". When the fringes are "straight" (experimentally, this is rather subjective), you have pure tilt fringes - which can in principle be removed to reveal "fluffed out" fringes - and d = 0, i.e. the optical paths in the two arms of the interferometer are the same. Note the micrometer reading in your lab book.
The interferometer is now set up so that the optical path in the two arms is approximately zero: in Experiment 3 you will refine the above adjustments to make it exactly zero.
Experiment 2: Separation of the Sodium Doublet
The bright yellow light you see from a sodium lamp is actually composed of two slightly different shades of yellow, from different transitions in the sodium atom. The aim of this experiment is to measure the difference in wavelength of these two yellow sodium lines. The difference is only = 0.6nm but you should be able to measure this with an accuracy of just 1% - 2% using readings made on a micrometer.
(i) Replace the Mercury source by the Sodium source. Make sure you have noted the nominal "d = 0" reading on the micrometer. Now observe the fringe contrast as you turn the micrometer and move away from the d = 0 position. Record your observations in your lab book. In part (iii), you will make careful recordings of the micrometer readings of the minima of the fringe visibility: for now, you should just get a "feel" for the phenomenon.
(ii) The fringe visibility falls to a minimum when the two intensity patterns produced by the two wavelengths l1 and l2 are displaced by an odd half-number of pattern spacings (read that again).
Let d be the mirror separation at one such position. If there is a bright fringe in the pattern of l1 at the centre of the field of view (where q = 0 and cos q = 1), we can write:
2d = m l1,
where m is the order of interference. It follows that:
(2)
(3)
For minimum visibility, there must be a dark fringe in the pattern of l2 at the centre of the field. It follows that:
where m' is the order of interference and hence
(4)
and
(5)
Now, combining Eqs. (3) and (5),
(6)
and combining Eqs. (2) and (4),
(7)
Now substituting Dl = l1 - l2 and l1l2 , where Dl is the separation we are seeking and is the mean wavelength (strictly, the geometrical mean), Eqs. (6) and (7) give
(8)
Thus, starting at d = 0, the minimum visibility occurs at spacings d which correspond to
,…
There is no need to repeat this derivation but at least the final result, Eq. (8), with an explanation of the symbols, should be entered in your lab book.
(iii) Take a series of readings of d versus (m' - m). The absolute value of (m' - m) is not required: it changes by 1 between adjacent minima of visibility. For each minimum position d take ~ 2 readings and calculate the mean and standard error.
Note that the lever action of the micrometer reduces the movement by a factor of 5.000.05, i.e. moving the micrometer through 1mm moves the mirror through approximately 0.2mm.
Record your results in your lab book and summarise them in a table with the following headings:
Order (m' - m) / Mean Value of d / Std Error of d(iv) Plot the mean value of d versus (m' - m) in your lab book, determine the slope by hand and hence obtain an estimate for Dl using Eq. (8). You many assume that = 589.3 0.3nm.
(v) Use computer program CurveExpert to find the slope and hence Dl and its standard error.
Experiment 3: Finding the White Light Fringe
This part is fun or frustrating depending both on your attitude, patience and a certain amount of luck.
The aim is to find the micrometer position such that d = 0 exactly. At that point, all wavelengths will give a bright (or dark, see earlier footnote) fringe: we refer to this as the "white light fringe".
(i) Replace the Sodium source by the combined Mercury/Tungsten source and set the micrometer to the nominal d = 0 position recorded in part (iii) of Expt. 1.
(ii) Make sure there are about 10 tilt fringes across the field of view and switch on the Mercury source. Turn the micrometer a few turns to the point where you are confident that the, fringes are curved. Switch on the Tungsten lamp, leaving the Mercury lamp on. Turn the micrometer slowly back towards the d = 0 position, at such a speed that each fringe can be seen as it moves across the field. Watch out for a band of a few fringes that have distinctive colours. When you find them - this may be after a few minutes or >15 minutes - switch off the Mercury lamp and immediately note the micrometer reading so that you can find the fringes again quickly if necessary. The "white light" fringes occupy a narrow band only? Why? When the centra1 fringe is in the middle of the field of view, the two beams in the interferometer have zero path difference.
(iii) When you obtain fringes with white light source, note the sequence of colours and try to account for it. What is the colour of the central fringe?
Remember to keep a record of your observations in your lab book.
1
[1] Actually, it's not as simple as this: the beamsplitter introduces different phase changes for air/film/glass and glass/film/air reflections and thus the fringe will almost certainly not be a bright fringe and usually it is closer to a dark fringe.