Errors in Measurement

While doing an experiment several errors can enter into the results. Errors may be due to faulty equipment, carelessness of the experimenter or random causes. The first two types of errors can be removed after detecting their cause but the random errors still remain. No specific cause can be assigned to such errors.

When an experiment is repeated many times, the random errors are sometimes positive and sometimes negative. Thus, the average of a large number of the results of repeated experiments is close to the true value. However, there is still some uncertainty about the truth of this average. The uncertainty is estimated by calculating the standard deviation described below.

Let \(x_1, x_2, x_3, \ldots, x_N\) are the results of an experiment repeated \(N\) times. The standard deviation \(\sigma\) is defined as

\(\sigma=\sqrt{\frac{1}{N} \sum_{i=1}^N\left(x_i-\bar{x}\right)^2}\)

where \(\bar{x}=\frac{1}{N} \sum_i x_i\) is the average of all the values of \(x\). The best value of \(x\) derived from these experiments is \(\bar{x}\) and the uncertainty is of the order of \(\pm \sigma\). In fact \(\bar{x} \pm 1.96 \sigma\) is quite often taken as the interval in which the true value should lie. It can be shown that there is a \(95 \%\) chance that the true value lies within \(\bar{x} \pm 1.96 \sigma\).

If one wishes to be more sure, one can use the interval \(\bar{x} \pm 3 \sigma\) as the interval which will contain the true value. The chances that the true value will be within \(\bar{x} \pm 3 \sigma\) is more than \(99 \%\).

All this is true if the number of observations \(N\) is large. In practice, if \(N\) is greater than 8 , the results are reasonably correct.

Example 1: The focal length of a concave mirror obtained by a student in repeated experiments are given below. Find the average focal length with uncertainty in \(\pm \sigma\) limit.
\(\begin{array}{cc}\text { No. of observation } & \text { focal length in cm } \\ 1 & 25 \cdot 4 \\ 2 & 25 \cdot 2 \\ 3 & 25 \cdot 6 \\ 4 & 25 \cdot 1 \\ 5 & 25 \cdot 3 \\ 6 & 25 \cdot 2 \\ 7 & 25 \cdot 5 \\ 8 & 25 \cdot 4 \\ 9 & 25 \cdot 3 \\ 10 & 25 \cdot 7\end{array}\)

Solution:

The average focal length \(\bar{f}=\frac{1}{10} \sum_{i=1}^{10} f_i\) \(=25 \cdot 37 \approx 25 \cdot 4\).
The calculation of \(\sigma\) is shown in the table below:
\(
\begin{array}{rlrlc}
\hline i & \begin{array}{l}
f_i \\
\mathrm{~cm}
\end{array} & \begin{array}{r}
f_i-\bar{f} \\
\mathrm{~cm}
\end{array} & \begin{array}{c}
\left(f_i-\bar{f}\right)^2 \\
\mathrm{~cm}^2
\end{array} & \begin{array}{c}
\Sigma\left(f_i-\bar{f}\right)^2 \\
\mathrm{~cm}^2
\end{array} \\
\hline 1 & 25.4 & 0.0 & 0.00 & \\
2 & 25.2 & -0.2 & 0.04 & \\
3 & 25.6 & 0.2 & 0.04 & \\
4 & 25.1 & -0.3 & 0.09 & \\
5 & 25.3 & -0.1 & 0.01 & 0.33 \\
6 & 25.2 & -0.2 & 0.04 & \\
7 & 25.5 & 0.1 & 0.01 & \\
8 & 25.4 & 0.0 & 0.00 & \\
9 & 25.3 & -0.1 & 0.01 & \\
10 & 25.7 & 0.3 & 0.09 & \\
\hline
\end{array}
\)
\(
\begin{aligned}
\sigma &=\sqrt{\frac{1}{10} \sum_i\left(f_i-\bar{f}\right)^2}=\sqrt{0.033 \mathrm{~cm}^2}=0.18 \mathrm{~cm} \\
& \cong 0.2 \mathrm{~cm}
\end{aligned}
\)
Thus, the focal length is likely to be within (25.4 \(\pm\) \(0.2 \mathrm{~cm}\) ) and we write
\(
f=(25 \cdot 4 \pm 0 \cdot 2) \mathrm{cm} .
\)