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Optical theory.

One of the difficulties with having great distances between the source and the mirror in Fizeau's scheme is that the intensity of the light will decrease with distance. The image is brightened by placing a lens between the source and the mirror. If, as in the diagram below,
  
Figure: S and M are placed at the point-source focus of each other.
\begin{figure}
\centerline{\psfig{figure=point-source.ps,height=.75in,width=5in}}
\end{figure}

the source, S, and the mirror, M are placed so that a point-source light from one is focused precisely on the other, then the return image will be as bright and as crisp as possible.

Note that the distance between L and M is not equal to that between L and S. As M moves farther from the lens, S will need to be moved closer in order for both points to remain at the focus of the other's point source. This is true provided both points are beyond the focal length of the lens (that point where beams of light parallel on one side of the lens would meet on the other side).

By moving S and M farther apart, all the while keeping each at the other's point focus, we increase the distance the light must travel and therefore the time it will take. Even so, the time taken is exceedingly short and difficult to measure.

Instead of Fizeau's wheel, Foucault used a rotating mirror interposed between S and L as in the next diagram.44

  
Figure: Interposing a mirror, R, between the source S and the lens L.
\begin{figure}
\centerline{\psfig{figure=focus.ps,height=1.0in}}
\end{figure}

Light rays from the source that strike R and proceed through the lens L will strike M and return to the source S. If after the light beam first strikes R outbound from S, R can be rotated
  
Figure: Rotating the mirror R causes the returning beam to be deflected.
\begin{figure}
\centerline{\psfig{figure=mirror.ps,height=1.0in}}
\end{figure}

before it is struck again by the beam returning from M, then the returning beam will no longer return exactly to the source S but will instead be deflected away from S in the direction of the rotation.

By rotating the mirror at a constant speed, the amount of deflection will be the same for all light beams that go through L, strike M and return. Then, for a continuous beam of light from S and a constant high speed of rotation of R, an image of the source will appear beside S instead of coincident

  
Figure: The return image I is displaced from the source S by the rotating mirror R.
\begin{figure}
\centerline{\psfig{figure=displacement.ps,height=1.0in,width=2.6in}}
\end{figure}

upon it (as shown in Figure 4). The faster R rotates or the longer is |RS|, the farther the returned image, I, will be displaced from the source, S and the easier it will be to measure the deflection.

By carefully measuring the amount of displacement from S to I (see Figure 4), and the distance from S to R, the angle of deflection can be determined. Together with the known, fixed speed of rotation, this angle can be used to determine the time it took light to travel the distance from R to M and back. Dividing distance by time gives a determination of the speed of light.

Let $\theta$ denote the angle of deflection. Then the angle through which the mirror has rotated is easily shown to be $\theta / 2$. The angle $\theta$ in degrees is arctan(|IS|/|IR|). If the speed of rotation is n measured in cycles per second, then the time taken for the light beam to travel from R to M and back is $\frac{1}{n} \times \frac{\theta /2}{360}$ seconds. The speed of light transmitted under the conditions of the study is therefore

\begin{displaymath}2 \frac{360 n}{arctan(\vert IS\vert/\vert SR\vert)}\times 2\vert RM\vert
\end{displaymath}

In this arrangement, the distances |IS| and |SR| should be as large as possible to reduce the error in measuring $\theta$. The distance |IS| is maximized by maximizing the speed of rotation of R and the distance |RM|. Michelson's principal innovation in Foucault's design allowed |RM| to be very large. In Foucault's setup, M was spherical with centre at R. The greatest distance |RM| achieved by Foucault was 20 metres (page 117 [39]) which produced a displacement |IS|of only 0.7mm (page 118 [42]). Michelson chose to place the rotating mirror at the focal point of the lens which allowed him to use a flat mirror for M. That is, R should be placed at that point where parallel light beams passing through the lens from M meet on the other side as in Figure 5.

  
Figure: R at the focal point of L.
\begin{figure}
\centerline{\psfig{figure=parallel.ps,height=0.50in}}
\end{figure}

Then if the diameter of M was as large as that of L any single beam passing from R through L would necessarily strike M and return through L to R whatever the distance between L and M. This permitted M to be placed very far away. The only difficulty is that the farther away M is from L, the closer the point-source focus S will be to the focal point R which conflicts with maximizing the distance between S and R. This can be remedied somewhat by using a lens of large focal length.

These innovations produced a displacement of more than 100 mm. Such a large displacement solved another difficulty. Originally the eyepiece to observe the displaced image at S was offset using an inclined plate of silvered glass to avoid interference between the observer and the outgoing beam of light. Once the the displacement exceeded 40 mm, it was possible to remove the inclined plate and observe the displaced image directly. Michelson (page 116 [39]) noted ``Thus the eye-piece is much simplified and many possible sources of error are removed.''


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Next: Physical apparatus Up: Michelson's 1879 determinations of Previous: Michelson's 1879 determinations of

2000-05-24