To find the least count of a measuring instrument that has a vernier scale, you can use the following formula:
\[ \text{Least Count} = \text{Value of one main scale division} - \text{Value of one vernier scale division} \]
Given:
- \( n_1 \): The number of small divisions on the main scale corresponding to \( n_2 \) divisions on the vernier scale.
Let:
- \( L \) = length of one small division on the main scale.
- \( l \) = length of one small division on the vernier scale.
From the problem, since \( n_1 \) small divisions of the main scale correspond to \( n_2 \) divisions on the vernier scale, we can express their values in equations:
- The value of \( n_1 \) divisions on the main scale: \( n_1 \times L \)
- The value of \( n_2 \) divisions on the vernier scale: \( n_2 \times l \)
Since these two lengths represent the same length, we can equate them:
\[ n_1 \cdot L = n_2 \cdot l \Rightarrow l = \frac{n_1 \cdot L}{n_2} \]
Now, substituting that back into the least count formula:
\[ \text{Least Count} = L - l \]
Substituting \( l \):
\[ \text{Least Count} = L - \frac{n_1 \cdot L}{n_2} \]
Factoring out \( L \):
\[ \text{Least Count} = L \left( 1 - \frac{n_1}{n_2} \right) = L \left( \frac{n_2 - n_1}{n_2} \right) \]
So, the least count of the instrument can be expressed as:
\[ \text{Least Count} = \frac{L(n_2 - n_1)}{n_2} \]
This formula gives you the least count based on the relationship between the divisions of the main scale and the vernier scale.
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