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

Answer 2

To find the refractive index of the glass slab, we need to establish the apparent shift that occurs when viewing through the slab. Let's break this down:

1. Understanding the Observations:
   • The measurements are made using a travelling microscope with certain specifications:
   • \(50\) vernier scale divisions (VSD) coincide with \(49\) main scale divisions (MSD).
   • Each cm of the main scale contains \(20\) divisions.
2. Calculate Least Count of the Vernier:
   • The formula for the least count (LC) of a vernier scale is given by:
   • \(\text{LC} = \text{MSD} - \text{VSD}\)
   • \(\text{MSD} = \frac{1}{20} \, \text{cm} = 0.05 \, \text{cm}\)
   • \(\text{VSD} = \frac{49}{50} \times 0.05 \, \text{cm} = 0.049 \, \text{cm}\)
   • Therefore, \(\text{LC} = 0.05 \, \text{cm} - 0.049 \, \text{cm} = 0.001 \, \text{cm}\)
3. Determine the Actual Measurements:
   • For the mark on paper:
   • Total reading = \(\text{MSR} + \text{VC} \times \text{LC} = 8.45 + 26 \times 0.001 = 8.476 \, \text{cm}\)
   • For the mark seen through the slab:
   • Total reading = \(7.12 + 41 \times 0.001 = 7.161 \, \text{cm}\)
   • For the particle on top of the glass slab (thickness):
   • Total reading = \(4.05 + 1 \times 0.001 = 4.051 \, \text{cm}\)
4. Calculate Real and Apparent Thickness:
   • Real thickness \((t)\) of the slab = mark on paper - top particle reading = \(8.476 - 4.051 = 4.425 \, \text{cm}\)
   • Apparent thickness seen through the slab = mark through slab - top particle reading = \(7.161 - 4.051 = 3.110 \, \text{cm}\)
5. Calculate the Refractive Index:
   • The formula for the refractive index \((\mu)\) is:
   • \(\mu = \frac{\text{Real Thickness}}{\text{Apparent Thickness}} = \frac{4.425}{3.110} \approx 1.42\)
6. Conclusion:
   • Therefore, the refractive index of the glass slab is \(1.42\).

Thus, the correct answer is 1.42.