In an experiment of measuring the refractive index of a glass slab...

Given:

$50VSD = 49MSD ⇒ 1VSD = \dfrac{49}{50}MSD$

So least count will be

$= 1MSD - 1VSD = \dfrac{1}{20}cm - \dfrac{49}{50} \cdot \dfrac{1}{20}cm$

$= \dfrac{1}{50} \times \dfrac{1}{20}cm = 0 . 01mm$

As we know, $\mu = \dfrac{h}{h '} = \dfrac{5 . 25}{5 . 00}$

Applying log on both sides, we get

$ln\mu = lnh – lnh ’$

Differentiating both sides, we get

$\dfrac{d\mu}{\mu} = \dfrac{dh}{h} - \dfrac{dh '}{h '}$

For calculation, all errors will be added, therefore

$d\mu = \left[\dfrac{0 . 01}{5 . 25} + \dfrac{0 . 01}{5 . 00}\right]\mu$

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

Hence, $x = 41$