Question 75: Units , Dimensions and Error Analysis

Question

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

Accepted Answer

To find the refractive index of the glass slab, we first need to calculate the actual readings using the Main Scale Reading (MSR) and Vernier Coincidence (VC), and understand how to translate these values into a measurement of the refractive index. We will start by calculating the Least Count (LC) of the vernier calipers.
The least count (LC) of the traveling microscope is given by the formula:
$ 	ext{LC} = 	ext{MSD} - 	ext{VSD} $
Where MSD is the value of one main scale division and VSD is the value of one vernier scale division in terms of the main scale. We are told 50 vernier scale divisions equal 49 main scale divisions (MSD), so 1 VSD is $ \frac{49}{50} $ of an MSD.
Given that 20 divisions on the main scale represent 1 cm, each main scale division (MSD) represents:
$ 1 \, 	ext{MSD} = \frac{1 \, 	ext{cm}}{20} = 0.05 \, 	ext{cm} $
Therefore, the least count (LC) of the microscope is:
$ 	ext{LC} = 0.05 \, 	ext{cm} - \left( \frac{49}{50} 	imes 0.05 \, 	ext{cm} ight) = 0.05 \, 	ext{cm} - 0.049 \, 	ext{cm} = 0.001 \, 	ext{cm} $
Now, using the least count to find the total reading (TR) from both supplied observations:
For the mark on paper, the total reading (TR) is:
$ 	ext{TR} = 	ext{MSR} + (	ext{VC} 	imes 	ext{LC}) $
$ 	ext{TR}_{	ext{paper}} = 8.45 \, 	ext{cm} + (26 	imes 0.001 \, 	ext{cm}) = 8.45 \, 	ext{cm} + 0.026 \, 	ext{cm} = 8.476 \, 	ext{cm} $
For the mark on the paper seen through the slab, the total reading is:
$ 	ext{TR}_{	ext{seen through slab}} = 7.12 \, 	ext{cm} + (41 	imes 0.001 \, 	ext{cm}) = 7.12 \, 	ext{cm} + 0.041 \, 	ext{cm} = 7.161 \, 	ext{cm} $
For the powder particle on the top surface of the glass slab:
$ 	ext{TR}_{	ext{top surface}} = 4.05 \, 	ext{cm} + (1 	imes 0.001 \, 	ext{cm}) = 4.05 \, 	ext{cm} + 0.001 \, 	ext{cm} = 4.051 \, 	ext{cm} $
The real depth (RD) observed directly is the difference between the first and third observations (mark on paper and powder particle on top surface):
$ 	ext{RD} = 	ext{TR}_{	ext{paper}} - 	ext{TR}_{	ext{top surface}} = 8.476 \, 	ext{cm} - 4.051 \, 	ext{cm} = 4.425 \, 	ext{cm} $
The apparent depth (AD) when viewed through the slab is the difference between the second and third observations:
$ 	ext{AD} = 	ext{TR}_{	ext{seen through slab}} - 	ext{TR}_{	ext{top surface}} = 7.161 \, 	ext{cm} - 4.051 \, 	ext{cm} = 3.11 \, 	ext{cm} $
The refractive index ($n$) of the glass slab can be found using the formula:
$ n = \frac{	ext{Real Depth (RD)}}{	ext{Apparent Depth (AD)}} = \frac{4.425 \, 	ext{cm}}{3.11 \, 	ext{cm}} $
Calculating this gives:
$ n = \frac{4.425}{3.11} \approx 1.42 $
Therefore, the refractive index of the glass slab is approximately 1.42, which corresponds to Option D.

Answer

1.52

Answer

1.35

Answer

1.24

Answer

1.42

Units , Dimensions and Error Analysis

Choose the correct answer: