Answer

Verified

53.7k+ views

**Hint:**First of all find the mean value of refractive indices observed. Then, find the absolute error of each value observed by subtracting them from their mean value and taking only the absolute value. The relative error is calculated by dividing the mean value of absolute error and mean value of refractive indices observed.

**Complete step by step solution:**

Let the mean value of refractive indices observed in the experiment be $\overline n $ and let the refractive indices be ${n_1},{n_2},{n_3} \ldots + {n_6}$

The refractive indices which were observed are $1.45,1.56,1.54,1.44,1.54$ and $1.53$. Now, the average or mean value is calculated by formula –

$\overline n = \dfrac{{{n_1} + {n_2} + \ldots + {n_n}}}{{{n_t}}}$

where, ${n_t}$ is the total number of observations

Putting the values of refractive indices in the above formula –

$

\overline n = \dfrac{{1.45 + 1.56 + 1.54 + 1.44 + 1.54 + 1.53}}{6} \\

\overline n = \dfrac{{9.06}}{6} \\

\overline n = 1.51 \\

$

Now, we have to calculate the absolute error in each measurement of refractive indices observed –

This is calculated by taking only the absolute value by subtracting observed refractive indices with their mean value –

$ \Rightarrow \left| {observed - mean} \right|$

By using this formula for absolute value of each measurement, we get –

$

\Rightarrow \left| {1.45 - 1.51} \right| = 0.06 \\

\Rightarrow \left| {1.56 - 1.51} \right| = 0.05 \\

\Rightarrow \left| {1.54 - 1.51} \right| = 0.03 \\

\Rightarrow \left| {1.44 - 1.51} \right| = 0.07 \\

\Rightarrow \left| {1.53 - 1.51} \right| = 0.03 \\

\Rightarrow \left| {1.53 - 1.51} \right| = 0.02 \\

$

We got all the values of each absolute error of each measurement.

Now, calculating the mean value of absolute error we got above so, let the mean value of absolute error be $\Delta n$ -

$

\Delta n = \dfrac{{0.06 + 0.05 + 0.03 + 0.07 + 0.03 + 0.02}}{6} \\

\Delta n = 0.043 \\

$

Now, the relative error is calculated by dividing the mean value of absolute error and mean value of refractive indices observed which can be represented mathematically by –

$ \Rightarrow \dfrac{{\Delta n}}{{\overline n }}$

Now, putting the values of mean absolute error and mean value of each measurement of refractive indices observed in the experiment –

$ \Rightarrow \dfrac{{0.043}}{{1.51}} = 0.0285$

**Hence, the relative error is $0.0285$.**

**Note:**Relative error is a measure of the uncertainty of measurement compared to the size of the measurement. It is used for putting error into perspective. This error compares the measurement with an exact value.

Recently Updated Pages

Why do we see inverted images in a spoon class 10 physics JEE_Main

A farsighted man who has lost his spectacles reads class 10 physics JEE_Main

State whether true or false The outermost layer of class 10 physics JEE_Main

Which is not the correct advantage of parallel combination class 10 physics JEE_Main

State two factors upon which the heat absorbed by a class 10 physics JEE_Main

What will be the halflife of a first order reaction class 12 chemistry JEE_Main

Other Pages

The nitride ion in lithium nitride is composed of A class 11 chemistry JEE_Main

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

A given ray of light suffers minimum deviation in an class 12 physics JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Oxidation of succinate ion produces ethylene and carbon class 12 chemistry JEE_Main

when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main