In an experiment of measuring the refractive index of a glass slab using travelling microscope in ph

Question

In an experiment of measuring the refractive index of a glass slab using travelling microscope in physics lab, a student measures real thickness of the glass slab as 5.25 mm and apparent thickness of the glass slab as 5.00 mm . Travelling microscope has 20 divisions in one cm on main scale and 50 divisions on vernier scale is equal to 49 divisions on main scale. The estimated uncertainty in the measurement of refractive index of the slab is $\frac{x}{10} 	imes 10^{-3}$, where $x$ is ________ .\

Professor PC Thomas Classes · Answer

(41) : Real thickness, $y_{0}=5.25 \mathrm{~mm}$ Apparent thickness, $y=5 \mathrm{~mm}$\ 20 divisions in one cm and 50 divisions on vernier scale is equal to 49 divisions on main scale.\ So, $\mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD}=1 \mathrm{MSD}-\frac{49}{50} \mathrm{MSD}$\ $\mathrm{LC}=\frac{1}{50} \mathrm{MSD}=\frac{1}{50} 	imes \frac{1}{20}=\frac{1}{1000} \mathrm{~cm}=\frac{1}{100} \mathrm{~mm}$\ Refractive index, $\mu=\frac{	ext { Real depth }}{	ext { App. depth }}=\frac{5.25}{5}$\ Now, $\mu=\frac{y_{0}}{y}$\ $\ln \mu=\ln y_{0}-\ln y$\ $\frac{d \mu}{\mu}=\frac{d y_{0}}{y_{0}}+\frac{d y}{y}$\ (errors are added)\ $d \mu=\left[\frac{0.01}{5.25}+\frac{0.01}{5}ight] 	imes \frac{5.25}{5}=\frac{41}{10} 	imes 10^{-3}$\ On comparing with the given value we get $x=41$