Question

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

• A. 1.42
• B. 1.52
• C. 1.24
• D. 1.35

Answer 1

Answer: (A)

1.42

Answer 2

1. Calculating the Least Count (LC):
$1 \, \text{MSD} = \frac{1 \, \text{cm}}{20} = 0.05 \, \text{cm}$
$1 \, \text{VSD} = \frac{49}{50} \, \text{MSD} = \frac{49}{50} \times 0.05 \, \text{cm} = 0.049 \, \text{cm}$
$\text{LC} = 1 \, \text{MSD} - 1 \, \text{VSD} = 0.05 \, \text{cm} - 0.049 \, \text{cm} = 0.001 \, \text{cm}$
2. For mark on paper, $L_1$:
$L_1 = 8.45 \, \text{cm} + 26 \times 0.001 \, \text{cm} = 8.45 \, \text{cm} + 0.026 \, \text{cm} = 8.476 \, \text{cm}$
3. For mark on paper seen through the slab, $L_2$:
$L_2 = 7.12 \, \text{cm} + 41 \times 0.001 \, \text{cm} = 7.12 \, \text{cm} + 0.041 \, \text{cm} = 7.161 \, \text{cm}$
4. For powder particle on the top surface, $ZE$:
$ZE = 4.05 \, \text{cm} + 1 \times 0.001 \, \text{cm} = 4.051 \, \text{cm}$
5. Calculating the thickness of the slab:
$\text{actual } L_1 = 8.476 - 4.051 = 4.425 \, \text{cm}$
$\text{actual } L_2 = 7.161 - 4.051 = 3.110 \, \text{cm}$
6. Refractive index $\mu$:
$\mu = \frac{L_1}{L_2} = \frac{4.425}{3.110} = 1.42$