added new edits from Irina

1 files changed, 1 insertions, 1 deletions

diff --git a/interferometry_new.tex b/interferometry_new.tex
index 10bbbea..27d40b7 100644
--- a/interferometry_new.tex
+++ b/interferometry_new.tex
@@ -36,7 +36,7 @@ Assuming that the initial optical field is described by $E_0\cos(kz-\omega t)$,
 \begin{equation}
 E_{total} =  \frac12 E_0 \cos(kz-\omega t + kL_1) + \frac12 E_0 \cos(kz-\omega t + kL_1) = E_0 \cos(k\Delta L/2) \cos(kz-\omega t + k(L_1+L_2)),
 \end{equation}
-where $L_{1,2}$ are the round-trip optical path 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.
+where $L_{1,2}$ are the round-trip optical path 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).