Question

In finding out refractive index of glass slab the following observations were made through travelling microscope 50 vernier scale division = 49 MSD; 20 divisions on main scale in each cm For mark on paper

MSR = 8.45 cm, VC = 26 

For mark on paper seen through slab 

MSR = 7.12 cm, VC = 41 

For powder particle on the top surface of the glass slab 

MSR = 4.05 cm, VC = 1 

(MSR = Main Scale Reading, VC = Vernier Coincidence) 

Refractive index of the glass slab is: 

(1) 1.42 

(2) 1.52 

(3) 1.24 

(4) 1.35

Answer 1

Correct option is : (1) 1.42

\( \mathrm{MSD}=\frac{1}{20} \mathrm{cm}=0.05 \mathrm{~cm} \)

\( 50 \ \mathrm{VSD}=49 \ \mathrm{MSD} \)

\(1 \ \mathrm{VSD}=\frac{49}{50} \mathrm{MSD}\)

So, L.C  \(=1 \ \mathrm{MSD}-1 \ \mathrm{VSD}\)

\( =1 \ \mathrm{MSD}-\frac{49}{50} \mathrm{MSD} \)

\(\text { L.C }=\frac{\mathrm{MSD}}{50}=\frac{0.05 \mathrm{~cm}}{50}=0.001 \mathrm{cm}\)

\(\mathrm{t}=8.45+26(0.001)=8.476 \mathrm{~cm} \) 

\(\mathrm{t}^{\prime}=7.12+26(0.001)=7.143 \mathrm{~cm}\) 

width of power particle  \(=4.05+1(0.001)\)

\( =4.051 \mathrm{~cm} \)    

There are dust on the glass surface it width has two be reduce from t and to'

\( \mathrm{t}_{\text {red }}=8.476-4.051=4.425 \mathrm{~cm} \)

\(\mathrm{t}_{\text {app. }}=7.143-4.051=3.092 \mathrm{~cm}\)

\(\mu_{\text {glass }}=\frac{\mathrm{t}_{\text {ral }}}{\mathrm{t}_{\text {app }}}=\frac{4.425}{3.092}=1.43\)