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\\ $\mathrm{MSR}=8.45 \mathrm{~cm}, \mathrm{VC}=26$\\ 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\\ $\mathrm{MSR}=4.05 \mathrm{~cm}, \mathrm{VC}=1$\\ (MSR = Main Scale Reading, VC = Vernier Coincidence)\\ Refractive index of the glass slab is :

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Units & Measurement

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 :

Answer: (B)

The least count is

1 MSD - 1 VSD

with

1 MSD = 0.05

cm and

1 VSD = \frac{49}{50} \times 0.05 = 0.049

cm, giving

LC = 0.001

cm. Reading for the mark on paper:

L_{1} = 8.45 + 26 (0.001) = 8.476

cm. Reading for the mark seen through the slab:

L_{2} = 7.12 + 41 (0.001) = 7.161

cm. Reference reading for the powder particle:

Z = 4.05 + 1 (0.001) = 4.051

cm. Real depth

= L_{1} - Z = 4.425

cm; apparent depth

= L_{2} - Z = 3.110

cm.

μ = \frac{real depth}{apparent depth} = \frac{4.425}{3.110} \approx 1.42

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Single correct option