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In an experiment the refractive index of glass was observed to be `1.45 , 1.56 , 1.54 , 1.44 , 1.54 , and 1.53`. Calculate (a). Mean value of refractive index (b). Mean absolute error ( c ) Fractional error (d) Percentage error (e) Express the result in terms of absolute error and percentage error

Answer» Correct Answer - (i) `1.51 (ii) +- 0.4 ()iii) +- 0.03 (iv) +- 3% ;
mu =1.51 +- 0.04 ; mu =1.51 +- 3%.`
Mean value of
`mu = (1.45 +1.56 +1.54 +1.44 +1.54 +1.53)/(6) =1.51`
Absolute error are :
`1.51 -1.45 = 0.06, 1.51 - 1.56 = - 0.05`
`1.51 -1.54 = 0.03, 1.51 - 1.44 = - 0.07`
`1.51 -1.54 = 0.03, 1.51 - 1.53 = - 0.02`
Mean absolute error
` = +-(0.06 + 0.05 +0.03 +0.07 +0.03 + 0.02)/(6)`
`= +- 0.26//6 = +- 0.04`
`Fractional error = (+- 0.4)/(1.51) =+- 3%`
`mu =1.51 +- 0.04 or mu= 1.51 +- 3%`


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