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

Given: 

$$\begin{gathered} 50 VSD=49MSD \\ \Rightarrow 1VSD=\frac{49}{50}MSD \end{gathered}$$

So least count will be

$$\begin{gathered} =1MSD-1VSD \\ =\frac{1}{20} \mathrm{cm}-\frac{49}{50}·\frac{1}{20} \mathrm{cm} \end{gathered}$$

$$\begin{gathered} =\frac{1}{50}\times \frac{1}{20} \mathrm{cm} \\ =0.01 \mathrm{mm} \end{gathered}$$

As we know, $\mu =\frac{h}{h\text{'}}=\frac{5.25}{5.00}$

Applying log on both sides, we get

$ln\mu =lnh–lnh’$

Differentiating both sides, we get

$\frac{d\mu }{\mu }=\frac{dh}{h}-\frac{dh\text{'}}{h\text{'}}$

For calculation, all errors will be added, therefore

$d\mu =\left[\frac{0.01}{5.25}+\frac{0.01}{5.00}\right]\mu$

$=\frac{41}{10}\times {10}^{-3}$

Hence, $x=41$