Q62 / 90 · JEE Main 2024 (06 Apr Shift 2) · Physics

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 MSR $=7.12 \mathrm{~cm}, V C=41$ For powder particle on the top surface of the glass slab $\mathrm{MSR}=4.05 \mathrm{~cm}, \mathrm{VC}=1$ (MSR $=$ Main Scale Reading, VC $=$ Vernier Coincidence) Refractive index of the glass slab is :

  • A
    1.52
  • B
    1.35
  • Correct answer
    C
    1.42
  • D
    1.24

Solution

$\begin{aligned} & 1 \mathrm{MSD}=\frac{1 \mathrm{~cm}}{20}=0.05 \mathrm{~cm} \\ & 1 \mathrm{VSD}=\frac{49}{50} \mathrm{MSD}=\frac{49}{50} \times 0.05 \mathrm{~cm}=0.049 \mathrm{~cm} \\ & \mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD}=0.001 \mathrm{~cm} \end{aligned}$
For mark on paper, $\mathrm{L}_1=8.45 \mathrm{~cm}+26 \times 0.001 \mathrm{~cm}$ $=84.76 \mathrm{~mm}$ For mark on paper through slab, $\mathrm{L} 2=7.12 \mathrm{~cm}+$ $41 \times 0.001 \mathrm{~cm}=71.61 \mathrm{~mm}$ For powder particle on top surface, $\mathrm{ZE}=4.05 \mathrm{~cm}$ $+1 \times 0.001 \mathrm{~cm}=40.51 \mathrm{~mm}$
$\begin{aligned} \therefore & \text { actual } \mathrm{L}_1=84.76-40.51=44.25 \mathrm{~mm} \\ & \text { actual } \mathrm{L}_2=71.61-40.51=31.10 \mathrm{~mm} \\ & \mathrm{~L}_2=\frac{\mathrm{L}_1}{\mu} \\ \Rightarrow & \mu=\frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{44.25}{31.10}=1.42\end{aligned}$

Hence, the correct answer is option: 'C'.

Relevant Questions