1. Find the least count of the travelling microscope

Given:

• $50$ vernier scale divisions $= 49$ main scale divisions
• Main scale has $20$ divisions in $1\,\text{cm}$

So, $1\,\text{MSD} = \frac{1}{20}\,\text{cm} = 0.05\,\text{cm}$

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

Least count: $LC = 1\,\text{MSD} - 1\,\text{VSD}$ $= \left(1 - \frac{49}{50}\right)\text{MSD} = \frac{1}{50}\text{MSD}$ $LC = \frac{1}{50}\times 0.05 = 0.001\,\text{cm}$

2. Calculate the actual readings

Reading formula: $\text{Reading} = \text{MSR} + (\text{VC})\times LC$

(i) Mark on paper

$R_1 = 8.45 + 26\times 0.001 = 8.476\,\text{cm}$

(ii) Mark on paper seen through slab

$R_2 = 7.12 + 41\times 0.001 = 7.161\,\text{cm}$

(iii) Powder particle on top surface of slab

$R_3 = 4.05 + 1\times 0.001 = 4.051\,\text{cm}$

3. Find real thickness of slab

The top surface reading is $R_3$ and bottom mark reading is $R_1$. Thus real thickness, $t = R_1 - R_3 = 8.476 - 4.051 = 4.425\,\text{cm}$

4. Find apparent thickness

Bottom mark seen through slab gives apparent bottom position $R_2$. So apparent thickness, $t' = R_2 - R_3 = 7.161 - 4.051 = 3.110\,\text{cm}$

5. Compute refractive index

For a glass slab, $\mu = \frac{\text{real thickness}}{\text{apparent thickness}}$

So, $\mu = \frac{4.425}{3.110} \approx 1.423$

Hence, $\mu \approx 1.42$

6. Match with options

The correct option is: $\boxed{\text{D: }1.42}$