deletions | additions       diff --git a/subsection_Slope_Method_The_direct__.tex b/subsection_Slope_Method_The_direct__.tex  index d0ea84b..4c0b827 100644  --- a/subsection_Slope_Method_The_direct__.tex  +++ b/subsection_Slope_Method_The_direct__.tex 
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%This value is consistent with the value we got when we set the second derivative of $\frac{V}{\theta}$ to zero (the inflection point).   To verify the value of the Verdet constant, we used a second method where the polarizer stayed at the angle of greatest sensitivity, which is 45 degrees after we applied trigonometry to our previous fits. We did this because we used $cos$ function instead of $cos^2$ to fit our data in Figure \ref{fig:directmethoddata}, but we will be expressing $V(\theta)$ in the form of $cos^2$ in the rest of our report.   %We got the angle by setting the second derivative of $\frac{dV}{d\theta}$ to zero, which corresponds to the inflection point.  We then varied the current of the solenoid, going in 0.5A steps from -3A to 3A, thereby varying the magnetic field within the solenoid between -33.3 mT and +33.3 mT. We could then graph voltage vs magnetic field, which results in a linear graph as shown in Figure \ref{fig:VB}. The slope of the graph is $\frac{dV}{dB}$. We calculated $\frac{dV}{d\theta}$ by taking the first derivative and setting angle to $45$ degrees as mentioned above, so we can find $\frac{dB}{d\theta}$ and therefore the Verdet constant. In the direct fit method, we estimated that we have an uncertainty of $0.05$ degrees for d$\theta$, and this uncertainty is associated with 45 degrees here since we applied trigonometry as aforementioned. By plugging in the maximum ($45.5$ degrees) and minimum ($44.5$ degrees) into our calculation, we are able to get the uncertainty for Verdet constant.  %Initially, we fit our data to the form $V=V_{0}(sin(\theta_{1}+\theta_{2}))^2$, which using trigonometry, is $V=V_{0}\frac{(1-cos(\theta_{1}+\theta_{2})^2)}{2}$ (the actual equation that we fit to). However, we should have been fitting to the form $V=V_{0}cos(\theta_{1}-\theta_{2})^2$, which is $V=V_{0}\frac{(1+cos(\theta_{1}-\theta_{2})^2)}{2}$. Now,   %$$sin(\theta_{1}+\theta_{2})^2=sin2(θ1-(-θ2))$$ 
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$$V_{c}=\frac{1}{L}\left(\frac{d\theta}{dV_{pd}}\right)_{\theta=45}\frac{dV_{pd}}{dB}$$  In this way, we get:  $$V_{c}=21.095\pm4.12\frac{radians}{T $$V_{c}=21.095\pm0.0025\frac{radians}{T  \cdot m}$$ %As shown in Figure 2:  %$$V=0.000443B+0.127$$  %Therefore: