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 _______.

\(1 \mathrm{VSD}=\frac{49}{50} \mathrm{MSD}=\frac{49}{50} \times 0.05 \mathrm{~cm}=0.049 \mathrm{~cm}\)

\(\mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD}=0.001 \mathrm{~cm}\)

\(\text { For mark on paper, } \mathrm{L}_1=8.45 \mathrm{~cm}+26 \times 0.001 \mathrm{~cm} \)

\(=84.76 \mathrm{~mm}\)

\(\text { For mark on paper through slab, } \mathrm{L} 2=7.12 \mathrm{~cm}+\)

\(41 \times 0.001 \mathrm{~cm}=71.61 \mathrm{~mm}\)

\(\text { For powder particle on top surface, } \mathrm{ZE}=4.05 \mathrm{~cm}\)

\(+1 \times 0.001 \mathrm{~cm}=40.51 \mathrm{~mm}\)

\(\therefore \text { actual } \mathrm{L}_1=84.76-40.51=44.25 \mathrm{~mm}\)

\(\quad \text { actual } \mathrm{L} 2=71.61-40.51=31.10 \mathrm{~mm}\)

\(\quad \mathrm{~L}_2=\frac{\mathrm{L}_1}{\mu}\)

\(\Rightarrow \mu=\frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{44.25}{31.10}=1.42\)