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 :

Solution

Physics sir · 3 steps

Find the least count first, then μ is real depth over apparent depth, both measured from the top surface.

1. Least count. 20 divisions per cm means 1 MSD = 0.05 cm. Since 50 VSD = 49 MSD, the least count is 1 MSD - 1 VSD = 0.05/50 = 0.001 cm.

   Position focused	MSR + VC × LC	Reading
   Mark on paper	8.45 + 26 × 0.001	8.476 cm
   Mark seen through slab	7.12 + 41 × 0.001	7.161 cm
   Powder on top surface	4.05 + 1 × 0.001	4.051 cm

2. Real and apparent. Real depth is from the top surface to the actual mark: 8.476 - 4.051 = 4.425 cm. Apparent depth is from the top surface to the image: 7.161 - 4.051 = 3.110 cm.

   μ = (real depth)/(apparent depth) = 4.425/3.110 ≈ 1.42

3. So. The refractive index of the glass slab is about 1.42, which is option (d).

The idea to keep. In the travelling microscope method, measure both depths from the top-surface reading. Real depth uses the bare mark and apparent depth uses the mark seen through the slab.

Answer from the NTA final answer key. Solution by PrepWiser.