In an experiment the valno of refractive index of glass was found to be 1.54, 1.53, 1.44,1.54, 1.56 and 1.45 m in successive measurement. Calculate (i) the mean vlaue of refractive index (ii) absolute error of each measurement (iii) mean absolute error (iv) relative error and (v) percentage error.

Explanation

i) Mean value of refractive index,
`mu_("mean") = (1.54 + 1.53 + 1.44+ 1.54 + 1.56 + 1.45)/(6) = 1.51`
ii) Taking `mu_("mean")` as the true value, the errors in the six measurements are
`1.54 - 1.51 = + 0.03 , 1.53 - 1.51 =+ 0.02 , 1.44 - 1.51 =-0.07`
`1.54-1.51 =+0.03 , 1.56 - 1.51 =+0.05 and 1.45 - 1.51 =-0.06`
The absolute errors are
0.03, 0.2,0.07,0.05,and 0.06
iii) Mean absolute error in the value of `mu` is
`Deltamu_("mean") =(0.03+0.02+0.07+0.03+0.05+0.06)/(6) =(0.26)/(6) = 0.04`
(iv) Relative error in the value of `mu`
`=(Delta mu_("mean"))/(Delta mu_("mean"))= (0.04)/(1.51) - 0.02649 = 0.03`
(v) Percentage error in the value of
`mu =0.03 xx100 =3%` s.