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| -rw-r--r-- | interferometry_new.tex | 7 |
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diff --git a/interferometry_new.tex b/interferometry_new.tex index 41402cc..55a855d 100644 --- a/interferometry_new.tex +++ b/interferometry_new.tex @@ -43,10 +43,11 @@ and back. Since this beam travels through the glass beamsplitter plate 3 times, its optical path length is longer than the first beam which only passes through the beam splitter one time. To compensate for that, the first beam passes twice through a clear glass plate called the compensator plate, that has the same thickness. At the beam splitter one-half of this light is transmitted to an observer, overlapping with the beam reflected by $M_2$, and the total amplitude of the light at the screen is a combination of amplitude of the two beams. Assuming that the initial optical field is described by $E_0\cos(kz-\omega t)$, where $E_0$ is its amplitude, $k=2\pi \lambda$ is the wave vector, and $\lambda$ and $\omega$ are correspondingly the wavelength and the frequency of the laser . After splitting and recombining at the beam splitter, the total field at the viewing screen beam splitter is a combination of the electric fields from the two arms of the interferometer : -\begin{align*} - E_{total} &= \frac12 E_0 \cos(kz-\omega t + kL_1) + \frac12 E_0 \cos(kz-\omega t + kL_2) \\ +\begin{align} + E_{total} &= \frac12 E_0 \cos(kz-\omega t + kL_1) + \frac12 E_0 + \cos(kz-\omega t + kL_2) \nonumber \\ &= E_0 \cos(k\Delta L/2) \cos(kz-\omega t + k(L_1+L_2)/2), -\end{align*} +\end{align} where $L_{1,2}$ are the round-trip optical paths for each interferometer arm, and $\Delta L = L_1-L_2$ is the optical path difference for the two beams. You may notice that the final expression describes a laser fields with the amplitude $E_0 \cos(k\Delta L/2)$. As a result, the intensity we are going to see on the screen will be $I_0 \cos^2(k\Delta L/2)$, where $I_0$ is the total intensity of the initial laser beam. Mirror $M_1$ is mounted on a precision traveling platform, and thus we can adjust its position (by turning the micrometer screw) with great precision to make the distances traversed on both arms exactly the same. Because the thickness of the compensator plate and the beamsplitter are the same, both wavefronts pass through the same amount of glass and air, so the path length of the light beams in both interferometer arms is exactly the same. Therefore, the two fields will arrive in phase to the observer, their amplitudes will add up constructively, and we should see all the original laser beam to go toward the viewing screen. However, if now you turn the micrometer to offset the length of one arm by a quarter of the wavelength, $\Delta L = \lambda/2$, the two beams arriving to the beam splitter will be out of phase $k L_1-kL_2 = \pm \pi$, resulting in the total cancellation (destructive interference). |