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01 Speed of Light; Foucault-Michelson

Aim

To “measure” speed of light

Subjects

Diagram

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Figure 1:.

Equipment

Presentation

Preparation:

Make the laser beam go as horizontal as possible; careful alignment is essential in this demonstration. (Start at the laser and work step by step working yourself through the light path.)

The +5 m+5 \mathrm{~m}-lens is positioned at about 5 m5 \mathrm{~m} distance from the rotating mirror.

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Figure 2:.

Image-distance bb (see Figure 2A) is chosen such that the image of the first mirror (M1)\left(\mathrm{M}_{1}\right) is sharp at M3(1f=1r+f+1b)M_{3}\left(\frac{1}{f}=\frac{1}{r+f}+\frac{1}{b}\right). (In our assembly this means that bb is around 25 meters.) Lens +5 m+5 \mathrm{~m} and M3M_{3} are carefully adjusted until, in the right postion of the rotating mirror (Mr)\left(M_{r}\right) the laserbeam is reflected to the camera (the camera and M1M_{1} have the same distance to M3\mathrm{M}_{3} ). Making the rotating mirror turn at its highest speed (about 500 rev. per second) the light spot displaces, in our assembly, the whole width of the monitor screen. This displacement is calibrated by placing a plastic ruler between the grey filter and the camera (see Figure 3A).

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Figure 3:.

Shadow projection of the mm-lines on the sensitive layer of the camera show these lines on the monitor (we have 7 mm7 \mathrm{~mm} across the full width of the monitor screen; see Figure 3B). The geometry of the assembly makes it possible to link the rotation of the mirror to the displacement on the monitor screen (we have a displacement of around

2.5 mm2.5 \mathrm{~mm} on the camera, corresponding to ϕ=d2r=2.5×1032×2.6=0.48×103rad\phi=\frac{d}{2 r}=\frac{2.5 \times 10^{-3}}{2 \times 2.6}=0.48 \times 10^{-3} \mathrm{rad} of Mr\mathrm{M}_{\mathrm{r}} ).

Demonstration:

The laser is switched on and the light path is shown to the students. By hand the rotating mirror is turned until a flash is seen on the monitor screen. In this position the light path is as drawn in the Diagram. The principle of operation is explained to the students: In the time it takes the light beam to travel the distance MrM3MrM_{r}-M_{3}-M_{r} the rotating mirror has made a little angle ( φ\varphi ). This is observed on the monitor screen (angle 2φ2 \varphi, see Figure 2B).

Also the calibration is explained to the students.

While the speed of rotation of MrM_{r} increases the students see an increasing displacement of the light spot on the monitor screen. While running at a convenient speed (we use n=500 s1n=500 \mathrm{~s}^{-1} to make calculations easy), this reading and that of the displacement on the monitor screen are used to calculate the speed of light.

Explanation

During the time ( Δt\Delta t ) it takes the light beam to travel the distance MrM3MrM_{r}-M_{3}-M_{r} the rotating mirror has made a little angle (φ)(\varphi) that is read from the monitor screen. Observing the speed of rotation on the frequency meter, the time it took to make this little angle can be calculated. For instance we measure when MrM_{r} runs at 500 s1500 \mathrm{~s}^{-1}, a light spot displacement of d=2.5 mmd=2.5 \mathrm{~mm} on the monitor screen. This means an angle of rotation of that mirror of ϕ=d2r=2.5×1032×2.6=0.48×103rad\phi=\frac{d}{2 r}=\frac{2.5 \times 10^{-3}}{2 \times 2.6}=0.48 \times 10^{-3} \mathrm{rad}. With n=500 s1(500×2πrad/sec)n=500 \mathrm{~s}^{-1}(500 \times 2 \pi \mathrm{rad} / \mathrm{sec})

this means a time-span of Δt=0.48×103500×2π=0.15μsec\Delta t=\frac{0.48 \times 10^{-3}}{500 \times 2 \pi}=0.15 \mu \mathrm{sec}. (So, the displacement measurement on the monitor screen becomes in this way a Δt\Delta t-measurement.) Our distance f+b=23\mathrm{f}+\mathrm{b}=23 meter. This gives us c=2×230.15×106=3.1×108 m/sc=\frac{2 \times 23}{0.15 \times 10^{-6}}=3.1 \times 10^{8} \mathrm{~m} / \mathrm{s}. (The measurement of the light spot displacement on the monitor screen is done very roughly, so our result of c=3.1×108 m/sec\mathrm{c}=3.1 \times 10^{8} \mathrm{~m} / \mathrm{sec} is satisfying.)

Remarks

Sources