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

The Correct Option is A

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 \( \mu \):
\[\mu = \frac{L_1}{L_2} = \frac{4.425}{3.110} = 1.42\]