In finding the refractive index of a glass slab, the following observations were made through a travelling microscope: 50 vernier scale divisions $= 49$ MSD; 20 divisions on the main scale in each cm.
For a mark on paper: MSR $= 8.45\ \text{cm}$, VC $= 26$.
For the mark on paper seen through the slab: MSR $= 7.12\ \text{cm}$, VC $= 41$.
For a powder particle on the top surface of the glass slab: MSR $= 4.05\ \text{cm}$, VC $= 1$.
(MSR = main scale reading, VC = vernier coincidence.) The refractive index of the glass slab is
Answer: (C) $1.42$
$1\ \text{MSD} = \dfrac{1}{20}\ \text{cm} = 0.05\ \text{cm}$; least count $= \dfrac{1\ \text{MSD}}{50} = 0.001\ \text{cm}$.
Readings: $R_1 = 8.45 + 0.026 = 8.476\ \text{cm}$, $R_2 = 7.12 + 0.041 = 7.161\ \text{cm}$, $R_3 = 4.05 + 0.001 = 4.051\ \text{cm}$.
Real thickness $= R_1 - R_3 = 4.425\ \text{cm}$; apparent thickness $= R_2 - R_3 = 3.110\ \text{cm}$.
$$\mu = \frac{4.425}{3.110} \approx 1.42$$
Solution by Sreeraj P, M.Sc Physics