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 at $$5.00$$ 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 $$\frac{x}{10} \times 10^{-3}$$, where $$x$$ is ______.

Solution :

Refractive index of glass slab is :

$$\mu = \frac{\text{Real thickness}}{\text{Apparent thickness}}$$

Given :

$$t_r = 5.25\text{ mm}$$

$$t_a = 5.00\text{ mm}$$

Therefore,

$$\mu = \frac{5.25}{5.00}$$

$$= 1.05$$

Main scale :

$$20\text{ divisions} = 1\text{ cm}$$

Hence,

$$1\text{ MSD} = \frac{1}{20}\text{ cm}$$

$$= 0.05\text{ cm}$$

$$= 0.5\text{ mm}$$

Given,

$$50\text{ VSD} = 49\text{ MSD}$$

Least count :

$$LC = 1\text{ MSD} - 1\text{ VSD}$$

$$= \frac{1}{50}\text{ MSD}$$

$$= \frac{0.5}{50}\text{ mm}$$

$$= 0.01\text{ mm}$$

Fractional error in refractive index :

$$\frac{\Delta \mu}{\mu} = \frac{\Delta t_r}{t_r} + \frac{\Delta t_a}{t_a}$$

Substituting values :

$$\frac{\Delta \mu}{1.05} = \frac{0.01}{5.25} + \frac{0.01}{5.00}$$

$$\approx 0.0039$$

Therefore,

$$\Delta \mu = 1.05 \times 0.0039$$

$$\approx 0.0041$$

$$= 4.1 \times 10^{-3}$$

Given uncertainty is of the form :

$$\frac{x}{10} \times 10^{-3}$$

Comparing,

$$\frac{x}{10} = 4.1$$

$$x = 41$$

Final Answer :

$$41$$