In an experiment of measuring the refractive | vedclass

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Question

(B) The refractive index is given by $\mu = \frac{h}{h'}$,where $h = 5.25\,mm$ and $h' = 5.00\,mm$.The main scale division $(MSD)$ is $\frac{1}{20}\,c

Vedclass · Accepted Answer

$41$

Vedclass · Answer

(B) The refractive index is given by $\mu = \frac{h}{h'}$,where $h = 5.25\,mm$ and $h' = 5.00\,mm$.The main scale division $(MSD)$ is $\frac{1}{20}\,cm = 0.5\,mm$.The least count $(LC)$ of the travelling microscope is $1\,MSD - 1\,VSD = 1\,MSD - \frac{49}{50}\,MSD = \frac{1}{50}\,MSD = \frac{0.5\,mm}{50} = 0.01\,mm$.Taking the natural logarithm of the refractive index formula: $\ln \mu = \ln h - \ln h'$.Differentiating,we get the relative uncertainty: $\frac{d\mu}{\mu} = \frac{dh}{h} + \frac{dh'}{h'}$.Here,$dh = dh' = LC = 0.01\,mm$.Substituting the values:$d\mu = \mu \left( \frac{dh}{h} + \frac{dh'}{h'} ight) = \left( \frac{5.25}{5.00} ight) \left( \frac{0.01}{5.25} + \frac{0.01}{5.00} ight)$.$d\mu = \frac{0.01}{5.00} + \frac{0.01}{5.25} = 0.002 + 0.00190476 = 0.00390476$.$d\mu \approx 3.90476 	imes 10^{-3} = \frac{39.0476}{10} 	imes 10^{-3} \approx \frac{41}{10} 	imes 10^{-3}$ (rounding to the nearest integer for $x$).Thus,$x = 41$.

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