In finding out refractive index of glass slab the following observations were made through travelling microscope 50 vernier scale division $$= 49$$ MSD; 20 divisions on main scale in each cm. For mark on paper MSR $$= 8.45 \text{ cm}$$, VC $$= 26$$. For mark on paper seen through slab MSR $$= 7.12 \text{ cm}$$, VC $$= 41$$. For powder particle on the top surface of the glass slab MSR $$= 4.05 \text{ cm}$$, VC $$= 1$$. (MSR = Main Scale Reading, VC = Vernier Coincidence) Refractive index of the glass slab is :

The microscope has a main scale and a vernier scale.

Length of $$1$$ main scale division (MSD): 20 divisions occupy $$1 \text{ cm}$$, hence
$$1 \text{ MSD} = \frac{1}{20} \text{ cm} = 0.05 \text{ cm}$$ $$-(1)$$

Given: $$50$$ vernier scale divisions (VSD) $$= 49$$ MSD.
Total length of $$50$$ VSD $$= 49 \times 0.05 = 2.45 \text{ cm}$$
Therefore length of $$1$$ VSD $$= \frac{2.45}{50} = 0.049 \text{ cm}$$ $$-(2)$$

Least count (LC) of the microscope is the difference of one MSD and one VSD:
$$\text{LC} = 1 \text{ MSD} - 1 \text{ VSD}$$
Substituting from $$(1)$$ and $$(2)$$:
$$\text{LC} = 0.05 - 0.049 = 0.001 \text{ cm}$$ $$-(3)$$

The reading for any setting of the microscope is
$$\text{Reading} = \text{MSR} + (\text{VC} \times \text{LC})$$ $$-(4)$$

Case 1: Mark on paper (no slab)
$$\text{MSR} = 8.45 \text{ cm}, \; \text{VC} = 26$$
Using $$(4)$$:
$$R_1 = 8.45 + 26 \times 0.001 = 8.476 \text{ cm}$$ $$-(5)$$

Case 2: Same mark seen through the slab (apparent depth)
$$\text{MSR} = 7.12 \text{ cm}, \; \text{VC} = 41$$
$$R_2 = 7.12 + 41 \times 0.001 = 7.161 \text{ cm}$$ $$-(6)$$

Case 3: Powder particle on the top surface of the slab
$$\text{MSR} = 4.05 \text{ cm}, \; \text{VC} = 1$$
$$R_3 = 4.05 + 1 \times 0.001 = 4.051 \text{ cm}$$ $$-(7)$$

The true thickness $$t$$ of the slab is the distance between the top surface and the bottom surface (which is in contact with the paper mark).
Using $$(5)$$ and $$(7)$$:
$$t = |R_3 - R_1| = |4.051 - 8.476| = 4.425 \text{ cm}$$ $$-(8)$$

The apparent thickness $$t'$$ is the distance between the top surface and the apparent position of the mark as seen through the slab.
Using $$(6)$$ and $$(7)$$:
$$t' = |R_3 - R_2| = |4.051 - 7.161| = 3.110 \text{ cm}$$ $$-(9)$$

The refractive index $$\mu$$ of the glass slab (normal incidence) is given by
$$\mu = \frac{t}{t'}$$ $$-(10)$$

Substituting from $$(8)$$ and $$(9)$$:
$$\mu = \frac{4.425}{3.110} \approx 1.42$$ $$-(11)$$

Hence the refractive index of the glass slab is $$1.42$$.
Option C is correct.