Calculating Vernier Scale Division Length from Least Count

Understanding how to calculate the length of a vernier scale division is fundamental when working with a vernier caliper. A vernier caliper is a precision instrument used to measure linear dimensions.

It consists of two main parts:

• The main scale: This is fixed and graduated similar to a regular scale.
• The vernier scale: This is a smaller scale that slides along the main scale.

The least count (LC) of a vernier caliper is the smallest measurement that can be taken accurately using the instrument. It represents the difference between the value of one main scale division (MSD) and the value of one vernier scale division (VSD).

The formula relating these quantities is:

$\text{LC} = \text{MSD} - \text{VSD}$

In this specific question, we are given the following values:

• Least Count (LC) = 0.02 mm
• Least count of the main scale, which is the value of one Main Scale Division (MSD) = 1 mm

We need to find the length of one division of the vernier scale, which is VSD.

We can rearrange the formula to solve for VSD:

$\text{VSD} = \text{MSD} - \text{LC}$

Now, we substitute the given values into the rearranged formula:

$\text{VSD} = 1 \text{ mm} - 0.02 \text{ mm}$

$\text{VSD} = 0.98 \text{ mm}$

Therefore, the length of one division of the vernier scale is 0.98 mm.

Let's summarize the calculation:

Using the formula: $\text{VSD} = \text{MSD} - \text{LC}$

Calculation:

$\text{VSD} = 1 \text{ mm} - 0.02 \text{ mm} = 0.98 \text{ mm}$

The length of one vernier scale division is 0.98 mm.

Revision Table: Vernier Caliper Concepts

Additional Information on Vernier Caliper Calculation

The calculation of VSD using the formula $\text{LC} = \text{MSD} - \text{VSD}$ is applicable when the vernier scale has divisions that are slightly smaller than the main scale divisions. This is the most common type of vernier caliper. In some cases, the vernier scale divisions might be slightly larger, but the formula $\text{LC} = \text{MSD} - \text{VSD}$ holds true, though VSD would be calculated differently based on the total length of the vernier scale divisions matching a certain number of main scale divisions.

Another way to define the least count is $\text{LC} = \frac{\text{MSD}}{N}$, where N is the number of divisions on the vernier scale that coincide with $(N-1)$ divisions on the main scale. From this, if N divisions on the vernier scale equal $(N-1)$ main scale divisions, then $N \times \text{VSD} = (N-1) \times \text{MSD}$. This gives $\text{VSD} = \frac{(N-1)}{N} \times \text{MSD}$. Substituting this back into the LC formula:

$\text{LC} = \text{MSD} - \text{VSD} = \text{MSD} - \frac{(N-1)}{N} \times \text{MSD} = \text{MSD} \left( 1 - \frac{N-1}{N} \right) = \text{MSD} \left( \frac{N - (N-1)}{N} \right) = \text{MSD} \left( \frac{1}{N} \right) = \frac{\text{MSD}}{N}$.

This confirms the relationship. In our problem, with MSD = 1 mm and LC = 0.02 mm, we could find N using $\text{N} = \frac{\text{MSD}}{\text{LC}} = \frac{1 \text{ mm}}{0.02 \text{ mm}} = 50$. This means 50 vernier scale divisions likely match 49 main scale divisions. Then $\text{VSD} = \frac{49}{50} \times 1 \text{ mm} = 0.98 \text{ mm}$, which matches our result.