Q.Which of the following is the most precise device for measuring length:

Concept: The precision of a measuring instrument is determined by its least count — the smallest length it can reliably measure. The device with the smallest least count is the most precise.

Step 1. For the vernier callipers with 20 divisions on the sliding scale (assuming standard 1 cm main scale divisions):

Step 2. For the screw gauge:

Step 3. For the optical instrument measuring to within a wavelength of visible light (taking $\lambda \sim 500\,\text{nm}$ as typical): …

Precision in measurement means the ability to distinguish between two very close values. A device is more precise when it can detect smaller differences. This ability is quantified by the least count — the smallest division the instrument can reliably measure.

The device with the smallest least count is the most precise. Let's calculate the least count for each instrument.

Step-by-step comparison

1. Vernier callipers with 20 divisions on the sliding scale

The least count of a vernier callipers is given by:

where MSD is the main scale division and VSD is the vernier scale division.

For a standard vernier callipers, 1 MSD = 1 mm. The 20 vernier divisions coincide with 19 main scale divisions (standard construction), so:

Therefore:

2. Screw gauge with pitch 1 mm and 100 divisions on the circular scale

The least count of a screw gauge is:

The pitch is the distance the screw advances in one complete rotation. Here:

This is already five times more precise than the vernier callipers.

3. Optical instrument measuring to within a wavelength of light

Visible light has wavelengths ranging from about 400 nm (violet) to 700 nm (red). Taking a typical value:

$\lambda \approx 500 \text{ nm} = 500 \times 10^{-9} \text{ m} = 0.0005 \text{ mm}$ …

Method: Least Count Comparison

The Least Count of a measuring instrument is the smallest measurement it can accurately read. The device with the smallest least count is the most precise.

Steps

1. Find Least Count of Vernier Callipers

   • Formula: $\text{LC} = \frac{1 \text{ MSD}}{\text{Number of VSD}}$
   • Here, 1 Main Scale Division (MSD) = 1 mm (standard), 20 Vernier Scale Divisions (VSD)
   • $\text{LC} = \frac{1 \text{ mm}}{20} = 0.05 \text{ mm}$

2. Find Least Count of Screw Gauge

   • Formula: $\text{LC} = \frac{\text{Pitch}}{\text{Number of circular divisions}}$
   • Pitch = 1 mm, Circular divisions = 100
   • $\text{LC} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}$

3. Find Least Count of Optical Instrument

   • It can measure to within one wavelength of light …

Here are the common mistakes students make on this "Least Count Precision" question, along with how to avoid each.

Mistake 1: Confusing "Number of Divisions" with "Precision"

The Mistake:

Students see "20 divisions" on the vernier and think "more divisions = more precise." They then compare this directly to "100 divisions" on the screw gauge and pick the screw gauge without calculating the actual least count.

Why it happens:

They ignore the pitch (the main scale distance per rotation). A screw gauge with 100 divisions on a 1 mm pitch is very precise, but a screw gauge with 100 divisions on a 2 mm pitch is less precise.

How to Avoid:

Always calculate the Least Count (LC) for every instrument before comparing. Never compare the number of divisions alone.

• Vernier Callipers: $LC = \frac{\text{1 Main Scale Division (MSD)}}{\text{Number of Vernier Divisions (VSD)}}$
  • Here, 1 MSD is typically 1 mm. So, $LC = \frac{1 \text{ mm}}{20} = 0.05 \text{ mm}$.
• Screw Gauge: $LC = \frac{\text{Pitch}}{\text{Number of Circular Scale Divisions}}$
  • Here, $LC = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}$.
• Optical Instrument: Measures to within one wavelength of light. Visible light wavelength is roughly $500 \text{ nm} = 0.0005 \text{ mm}$.

Result: The optical instrument has the smallest LC ($0.0005 \text{ mm}$), making it the most precise.

Mistake 2: Forgetting to Convert Units Before Comparing

The Mistake:

A student calculates the screw gauge LC as $0.01 \text{ mm}$ and the optical instrument LC as $500 \text{ nm}$, then incorrectly says the screw gauge is more precise because $0.01$ is smaller than $500$.

Why it happens:

They compare the raw numbers without converting to the same unit (e.g., both in mm or both in nm).

How to Avoid:

Convert all LCs to the same unit (usually mm or meters) before comparing.

• Screw Gauge: $0.01 \text{ mm}$
• Optical Instrument: $500 \text{ nm} = 500 \times 10^{-6} \text{ mm} = 0.0005 \text{ mm}$

Now it's clear: $0.0005 \text{ mm} < 0.01 \text{ mm} < 0.05 \text{ mm}$.

Mistake 3: Misinterpreting "Wavelength of Light"

The Mistake:

Students think "wavelength of light" is a large value (like 1 cm or 1 m) because they confuse it with radio waves or sound waves.

Why it happens:

They don't recall the typical order of magnitude for visible light.

How to Avoid:

Memorize the scale: Visible light wavelength is on the order of $10^{-7}$ m (or $10^{-4}$ mm). Specifically, it's about $500 \text{ nm}$.

• $1 \text{ nm} = 10^{-9} \text{ m}$
• $500 \text{ nm} = 5 \times 10^{-7} \text{ m} = 0.0005 \text{ mm}$ …