Q.The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, calculate the minimum inaccuracy in the measurement of distance.

The minimum inaccuracy in any vernier instrument is its least count — the smallest distance it can resolve.

Finding the least count:

The value of 1 main scale division (MSD) is $0.5 \text{ mm}$.

Since 50 vernier scale divisions (VSD) coincide with 49 MSD:

So 1 VSD $= \frac{24.5}{50} = 0.49 \text{ mm}$.

The least count is: …

Why Least Count Matters

The least count of any measuring instrument tells you the smallest distance it can reliably distinguish. It sets the fundamental limit on precision: you cannot measure more accurately than this value, no matter how carefully you read the scale. For a vernier system, the clever design allows you to read fractions of a main-scale division by exploiting the slight mismatch between vernier and main-scale spacings.

The key insight: when $n$ vernier divisions coincide with $(n-1)$ main-scale divisions, each vernier division is slightly smaller than a main-scale division. The difference between one main-scale division and one vernier division is precisely the least count.

Step-by-Step Calculation

1. Identify the main-scale division (MSD) value.

   Each main-scale division is given as $0.5 \text{ mm}$.

2. Find the total length covered by the vernier divisions.

   The problem states that $50$ vernier divisions coincide with $49$ main-scale divisions. Therefore, the total length spanned by $50$ vernier divisions is:

3. Calculate the value of one vernier-scale division (VSD). Dividing the total length by the number of vernier divisions:

4. Determine the least count (LC). The least count is the difference between one main-scale division and one vernier-scale division: $\text{LC} = \text{MSD} - \text{VSD} = 0.5 \text{ mm} - 0.49 \text{ mm} = 0.01 \text{ mm}$ …

Concept: Vernier Caliper Least Count

The minimum inaccuracy (or least count) of a vernier instrument is the smallest distance it can measure reliably. It depends on the value of one main scale division (MSD) and the number of vernier divisions (VSD).

Method: Least Count Formula

Formula

Why this works

• 50 vernier divisions cover the same length as 49 main scale divisions.
• So, 1 VSD = $\frac{49}{50}$ MSD.
• The difference between 1 MSD and 1 VSD gives the smallest measurable length:

This $\frac{1}{50}$ of an MSD is the least count.

Step-by-Step Calculation

1. Identify value of 1 MSD …

Here’s a breakdown of the common mistakes students make on this specific vernier caliper / travelling microscope problem, and how to avoid each one.

The Core Concept

The minimum inaccuracy (or least count) of a vernier instrument is the smallest measurement it can reliably detect. It is not the smallest division on the main scale.

The formula is:

Mistake #1: Using the Wrong Formula (Confusing Coincidence with Division Size)

The Mistake:

Students often see “50 divisions coincide with 49 main scale divisions” and incorrectly assume the least count is simply $0.5 \text{ mm} / 50 = 0.01 \text{ mm}$ without checking the coincidence relationship. Or, they use the formula:

and then incorrectly calculate $1 \text{ VSD}$ as $0.5 \text{ mm} / 50 = 0.01 \text{ mm}$.

Why it’s wrong:

The coincidence tells you the total length of the vernier scale, not the size of one division. If 50 VSD = 49 MSD, then:

So the least count is:

The result is the same in this case, but the reasoning is flawed. If the numbers were different (e.g., 50 VSD = 48 MSD), the shortcut would give a wrong answer.

How to Avoid:

Always write the two steps explicitly:

1. Find the value of 1 VSD using the coincidence relation.
2. Then apply: $\text{LC} = 1 \text{ MSD} - 1 \text{ VSD}$.

Mistake #2: Forgetting to Convert Units

The Mistake:

The problem gives MSD = 0.5 mm. Students calculate LC = 0.01 mm and leave it as is, or they mistakenly convert it to cm incorrectly (e.g., writing 0.01 mm = 0.1 cm).

Why it’s wrong:

Exams often ask for the answer in a specific unit (mm, cm, or m). A unit error costs marks even if the calculation is correct. Also, the minimum inaccuracy is the least count itself, so the unit must match the context.

How to Avoid:

• Check the question: Does it ask for “mm” or “cm”?
• Convert carefully: $1 \text{ mm} = 0.1 \text{ cm}$. So $0.01 \text{ mm} = 0.001 \text{ cm}$.
• Write the unit next to every numerical answer in your rough work.

Mistake #3: Confusing “Minimum Inaccuracy” with “Maximum Possible Error”

The Mistake:

Students think the minimum inaccuracy is half the least count (i.e., $\pm 0.005 \text{ mm}$) or that it’s the total error from multiple readings.

Why it’s wrong: …