A spectrometer gives the following reading when used to measure the angle of a prism.
Main scale rea

Q: Why do we subtract 1 VSD from 1 MSD to get the least count?

The least count represents the smallest measurable difference. Because the vernier scale divisions are intentionally made slightly smaller than the main scale divisions, the difference in length between one MSD and one VSD is exactly the tiny extra distance the vernier scale can resolve.

Q: What does '09 divisions' on the vernier scale actually mean?

It means that if you look closely at the two scales, the 9th mark on the vernier scale forms a perfectly straight, continuous line with one of the marks on the main scale. This alignment tells us the exact fractional measurement.

Q: Can I convert the least count to minutes instead of keeping it in degrees?

Absolutely! Since $1^{\circ} = 60'$, a least count of $(1/60)^{\circ}$ is exactly $1'$. You could calculate the reading as $58^{\circ} 30' + 9' = 58^{\circ} 39'$, which is mathematically identical to $58.65^{\circ}$.

Q: What if the zero of the vernier scale exactly aligns with a main scale mark?

If the zero mark aligns perfectly, it means there is no fractional part. The vernier scale reading (VSR) would be 0, and the total reading would simply be the main scale reading.

Q: Why is the main scale reading taken as 58.5 and not 59?

The main scale reading is always the mark that the vernier's zero has *already passed*. Since the vernier zero is between 58.5 and 59.0, it has passed 58.5 but hasn't reached 59.0 yet. Therefore, the base reading is 58.5.

Question

A spectrometer gives the following reading when used to measure the angle of a prism.
Main scale reading : $58.5^{\circ}$
Vernier scale reading : $09$ divisions
Given that $1$ division on main scale corresponds to $0.5^{\circ}$. Total divisions on the vernier scale is $30$ and match with $29$ divisions of the main scale. The angle of the prism from the above data is

Options:

(A) $58.59^{\circ}$

(B) $58.77^{\circ}$

(C) $58.65^{\circ}$

(D) $59^{\circ}$

Sigma Percentile · Accepted Answer

### Mastering the Vernier Spectrometer

Imagine you are in a dark optics lab, carefully adjusting a spectrometer to measure the angle of a glass prism. The beauty of such instruments lies in their ability to measure angles with astonishing precision, far beyond what a simple protractor can achieve. This precision is made possible by the elegant mathematics of the Vernier scale.

In this problem, we are given a spectrometer where the main scale is calibrated such that one division corresponds to $0.5^{\circ}$. This is our **Main Scale Division (MSD)**.

$$1 	ext{ MSD} = 0.5^{\circ}$$

Right below it, the vernier scale is constructed so that $30$ of its divisions perfectly span the same length as $29$ divisions on the main scale. This slight mismatch is the secret to its precision.

### The Master Equation: Least Count

Before we can read any measurement, we must determine the instrument's **Least Count (LC)**. The least count is the smallest value the instrument can reliably measure, and it is defined as the difference between one main scale division and one vernier scale division.

$$LC = 1 	ext{ MSD} - 1 	ext{ VSD}$$

From the given constraint, we know that $30 	ext{ VSD} = 29 	ext{ MSD}$. Therefore, a single vernier scale division is slightly smaller than a main scale division:

$$1 	ext{ VSD} = \frac{29}{30} 	ext{ MSD}$$

Substituting this back into our least count formula, we get:

$$LC = 1 	ext{ MSD} - \frac{29}{30} 	ext{ MSD} = \frac{1}{30} 	ext{ MSD}$$

Since we know that $1 	ext{ MSD} = 0.5^{\circ}$, we can find the absolute value of the least count:

$$LC = \frac{1}{30} 	imes 0.5^{\circ} = \frac{1}{60}^{\circ}$$

This value, $\frac{1}{60}^{\circ}$, is also known as one minute of arc ($1'$). It tells us that our spectrometer can measure angles accurately down to a single minute!

### Decoding the Reading

Now, let's look at the actual measurement taken during the experiment. The total reading of any vernier instrument is the sum of the **Main Scale Reading (MSR)** and the fractional part provided by the vernier scale.

$$	ext{Total Reading} = 	ext{MSR} + (	ext{VSR} 	imes 	ext{LC})$$

We are told that the main scale reading is $58.5^{\circ}$. This means the zero mark of the vernier scale has just passed the $58.5^{\circ}$ line on the main scale but hasn't yet reached the $59.0^{\circ}$ line.

To find out exactly how far past $58.5^{\circ}$ it is, we look for the vernier division that perfectly aligns with *any* main scale mark. The problem states that the $9^{	ext{th}}$ division of the vernier scale matches a main scale division. Thus, our **Vernier Scale Reading (VSR)** is $9$.

### Final Calculation

Let's bring it all together. We substitute our known values into the total reading formula:

$$	ext{Total Reading} = 58.5^{\circ} + \left(9 	imes \frac{1}{60}^{\circ}ight)$$

We simplify the fractional part:

$$9 	imes \frac{1}{60}^{\circ} = \frac{9}{60}^{\circ} = \frac{3}{20}^{\circ} = 0.15^{\circ}$$

Finally, we add this tiny fractional angle to our main scale reading:

$$	ext{Total Reading} = 58.5^{\circ} + 0.15^{\circ} = \mathbf{58.65^{\circ}}$$

And there we have it! The exact angle of the prism is $58.65^{\circ}$. By understanding the geometric relationship between the two scales, we transformed a simple visual alignment into a highly precise numerical measurement.

Select Answer:

Visualized Solution

$Visualizing the Scales$

$Least Count Formula$

$Calculating Least Count$

$Identifying Readings$

$Final Substitution$

$Final Answer$

The Sigma Insight: Vernier Calipers and Screw Gauge

Mastering the Vernier Spectrometer

The Master Equation

Decoding the Reading

Final Calculation

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