JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
In finding out refractive index of glass slab the following observations were made through travelling microscope \(50\) vernier scale division \(=\) \(49 \mathrm{MSD} ; 20\) divisions on main scale in each \(\mathrm{cm}\) For mark on paper \(\mathrm{MSR}=8.45 \mathrm{~cm}, \mathrm{VC}=26\) For mark on paper seen through slab \(\mathrm{MSR}=7.12 \mathrm{~cm}, \mathrm{VC}=41\) For powder particle on the top surface of the glass slab \(\mathrm{MSR}=4.05 \mathrm{~cm}, \mathrm{VC}=1\) \((\mathrm{MSR}=\) Main Scale Reading, \(\mathrm{VC}=\) Vernier Coincidence) Refractive index of the glass slab is _______.
- A \(1.42\)
- B \(1.52\)
- C \(1.24\)
- D \(1.35\)
Answer & Solution
Correct Answer
(A) \(1.42\)
Step-by-step Solution
Detailed explanation
\(1 \mathrm{MSD}=\frac{1 \mathrm{~cm}}{20}=0.05 \mathrm{~cm}\) \(1 \mathrm{VSD}=\frac{49}{50} \mathrm{MSD}=\frac{49}{50} \times 0.05 \mathrm{~cm}=0.049 \mathrm{~cm}\) \(\mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD}=0.001 \mathrm{~cm}\)…
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