8. Wyprowadzenie wzoru na niepewność pomiaru
$\Lambda = \ln\left( \frac{A_{n}}{A_{n + 1}} \right)$
$u_{c}\left( \Lambda \right) = \sqrt{\left( \frac{\partial\Lambda}{\partial A_{n}} \right)^{2}*m_{A_{n}}^{2} + \left( \frac{\partial\Lambda}{\partial A_{n + 1}} \right)^{2}*m_{A_{n} + 1}^{2}}$
$\frac{\partial\Lambda}{\partial A_{n}} = \frac{1}{A_{n}}$
$\frac{\partial\Lambda}{\partial A_{n + 1}} = \frac{- 1}{A_{n + 1}}$
mAn = mAn + 1 = mA = 0, 001[m]
$u_{c}\left( \Lambda \right) = \sqrt{\left( \frac{1}{A_{n}} \right)^{2}*m_{A_{n}}^{2} + \left( \frac{- 1}{A_{n + 1}}\ \right)^{2}*m_{A_{n} + 1}^{2}}$
$u_{c}\left( \Lambda \right) = \sqrt{\left( \frac{1}{A_{n}} \right)^{2}*m_{A}^{2} + \left( \frac{- 1}{A_{n + 1}}\ \right)^{2}*m_{A}^{2}}$
$u_{c}\left( \Lambda \right) = \sqrt{m_{A}^{2}\left( \left( \frac{1}{A_{n}} \right)^{2} + \left( \frac{- 1}{A_{n + 1}}\ \right)^{2} \right)}$
$u_{c}\left( \Lambda \right) = m\sqrt{\left( \frac{1}{A_{n}} \right)^{2} + \left( \frac{- 1}{A_{n + 1}}\ \right)^{2}}$
$u_{c}\left( \Lambda_{1} \right) = m\sqrt{\left( \frac{1}{A_{0}} \right)^{2} + \left( \frac{- 1}{A_{1}}\ \right)^{2}} = 0,001\sqrt{\left( \frac{1}{0,150} \right)^{2} + \left( \frac{- 1}{0,118}\ \right)^{2}} = 0,001\sqrt{44,44 + 71,82}$
uc(Λ1) = 0, 0108
$u_{c}\left( \Lambda_{2} \right) = m\sqrt{\left( \frac{1}{A_{0}} \right)^{2} + \left( \frac{- 1}{A_{1}}\ \right)^{2}} = 0,001\sqrt{\left( \frac{1}{0,118} \right)^{2} + \left( \frac{- 1}{0,103}\ \right)^{2}} = 0,001\sqrt{71,82 + 94,26}$
uc(Λ2) = 0, 0129
$u_{c}\left( \Lambda_{3} \right) = m\sqrt{\left( \frac{1}{A_{0}} \right)^{2} + \left( \frac{- 1}{A_{1}}\ \right)^{2}} = 0,001\sqrt{\left( \frac{1}{0,103} \right)^{2} + \left( \frac{- 1}{0,090}\ \right)^{2}} = 0,001\sqrt{94,26 + 123,46}$
uc(Λ3) = 0, 0148
$u_{c}\left( \Lambda_{4} \right) = m\sqrt{\left( \frac{1}{A_{0}} \right)^{2} + \left( \frac{- 1}{A_{1}}\ \right)^{2}} = 0,001\sqrt{\left( \frac{1}{0,090} \right)^{2} + \left( \frac{- 1}{0,080}\ \right)^{2}} = 0,001\sqrt{123,46 + 156,25}$
uc(Λ4) = 0, 0167