deletions | additions       diff --git a/subsection_Direct_Fit_Method_We__.tex b/subsection_Direct_Fit_Method_We__.tex  index 06b995d..c79319f 100644  --- a/subsection_Direct_Fit_Method_We__.tex  +++ b/subsection_Direct_Fit_Method_We__.tex 
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%We fit our data to a function of the form $V=V_{0}sin(\phi)^2$. We could find $\frac{\Delta V}{\delta \phi}$ by taking the derivative and using φ=45 degrees, and we could find ΔV by taking the difference of the voltage read by the photodetector when the laser is on (at maximum voltage read) and off. We could then find Δφ by calculating how much the angle of maximum transmission through the polarizer shifter. With all of this information, we could find dB/dφand use the equation $\frac{\Delta B}{\Delta \phi}=\frac{1}{L}\times\frac{1}{C_{v}}$ to find the Verdet constant of the glass tube.  $$V_{c}=\frac{1}{L}\times %$$V_{c}=\frac{1}{L}\times  \frac{\Delta\theta}{\Delta B}$$ %$$\frac{\Delta\theta}{\Delta B}=\frac{\Delta V}{\Delta B} \times \frac{\Delta \theta}{\Delta V}$$  $$\theta_{B}=104 %$$\theta_{B}=104  degrees; \theta_{0}=108 degrees$$ $$\Delta\theta=\theta_{B}-\theta_{0}=-0.069 %$$\Delta\theta=\theta_{B}-\theta_{0}=-0.069  radians$$ $$\Delta %$$\Delta  B=-3A\times 11.1\frac{mT}{A}=-33.3mT$$ $$\frac{\Delta\theta}{\Delta %$$\frac{\Delta\theta}{\Delta  B}=\frac{-0.069radians}{-33.3mT}=0.00207\frac{radians}{mT}=2.07\frac{radians}{T}$$ $$V_{c}=\frac{1}{L}\times %$$V_{c}=\frac{1}{L}\times  \frac{\Delta\theta}{\Delta B}=\frac{1}{0.1m}\times 2.07\frac{radians}{T}=20.7\frac{radians}{T \cdot m}$$