In an experiment of measuring the refractive index of a glass slab using travelling microscope in physics lab, a student measures real thickness of the glass slab as 5.25 mm and apparent thickness of the glass slab as 5.00 mm. Travelling microscope has 20 divisions in one cm on main scale and 50 divisions on vernier scale is equal to 49 divisions on main scale. The estimated uncertainty in the measurement of refractive index of the slab is x/10×10⁻³, where x is ___________

Solution

Physics sir · 4 steps

Get the least count first, then add the two fractional errors, because μ is a ratio of thicknesses.

1. Least count. 20 divisions in 1 cm means 1 MSD = 0.5 mm. Since 50 VSD = 49 MSD, 1 VSD = 49/50 × 0.5 = 0.49 mm. So the least count is 0.5 - 0.49 = 0.01 mm, and this is the uncertainty in each thickness.
2. Refractive index. μ = tᵣₑₐₗ/tₐₚₚ = 5.25/5.00 = 1.05. For a ratio, the fractional errors add.

   Δμ = μ((Δ tᵣₑₐₗ)/tᵣₑₐₗ + (Δ tₐₚₚ)/tₐₚₚ) = 1.05(0.01/5.25 + 0.01/5.00)

3. Put in numbers. 1.05 × 0.01/5.25 = 0.002 and 1.05 × 0.01/5.00 = 0.0021. Adding them gives Δμ = 0.0041 = 4.1 × 10⁻³.
4. So. Δμ = 41/10 × 10⁻³, so x = 41.

The common slip: 39. Stopping at the fractional error Δμ/μ ≈ 3.9 × 10⁻³ without multiplying by μ = 1.05 gives 39, not 41.

The idea to keep. For any vernier instrument, least count is 1 MSD minus 1 VSD. When a result is a ratio, add the fractional errors, then multiply by the value to get the absolute error.

Answer from the NTA final answer key. Solution by PrepWiser.