For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm. The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm) :

The main scale of the travelling microscope has $$300$$ equal divisions spread over a length of $$15 \text{ cm}$$.

Length of one main scale division (MSD) is therefore
$$\text{MSD} = \frac{15 \text{ cm}}{300} = 0.05 \text{ cm}$$

The vernier scale has $$25$$ divisions that coincide with $$24$$ divisions of the main scale.
Thus, length of the entire vernier scale is
$$25 \text{ VSD} = 24 \text{ MSD}$$

Using $$\text{MSD} = 0.05 \text{ cm}$$,
$$25 \text{ VSD} = 24 \times 0.05 \text{ cm} = 1.2 \text{ cm}$$

Length of one vernier scale division (VSD) is then
$$\text{VSD} = \frac{1.2 \text{ cm}}{25} = 0.048 \text{ cm}$$

Formula for least count (LC) of a vernier instrument:
$$\text{LC} = \text{MSD} - \text{VSD}$$

Substituting the values,
$$\text{LC} = 0.05 \text{ cm} - 0.048 \text{ cm} = 0.002 \text{ cm}$$

Hence, the least count of the travelling microscope is $$0.002 \text{ cm}$$.

Option B is correct.