A travelling microscope is used to determine the refractive index of a glass slab. If $$40$$ divisions are there in $$1$$ cm on main scale and $$50$$ Vernier scale divisions are equal to $$49$$ main scale divisions, then least count of the travelling microscope is ______ $$\times 10^{-6}$$ m.

We need to find the least count of a travelling microscope.

Main scale: 40 divisions in 1 cm

50 Vernier scale divisions = 49 main scale divisions

$$1 \text{ MSD} = \frac{1 \text{ cm}}{40} = \frac{1}{40}$$ cm = $$0.025$$ cm

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

$$1 \text{ VSD} = \frac{49}{50} \text{ MSD}$$

$$\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = 1 \text{ MSD} - \frac{49}{50} \text{ MSD} = \frac{1}{50} \text{ MSD}$$

$$= \frac{1}{50} \times 0.025 \text{ cm} = 0.0005 \text{ cm}$$

$$= 0.0005 \times 10^{-2} \text{ m} = 5 \times 10^{-6}$$ m

Comparing with $$\_\_\_ \times 10^{-6}$$ m, the answer is $$5$$.

The least count is $$5 \times 10^{-6}$$ m.