If nth division of main scale coincides with (n + 1)th division of vernier scale,find the least count of the vernier. Given one main scale division is equal to 'a' units.

The correct option is a/(n+1) units.

Understanding Least Count:
The least count (LC) of a Vernier caliper is defined as the difference between the value of one main scale division (MSD) and one vernier scale division (VSD).
Mathematically, this is expressed as:
$\mathrm{LC}=1\mathrm{MSD}-1\mathrm{VSD}$

Given Data:
1. Value of one main scale division (1 MSD) = $a$ units.
2. The $n$-th division of the main scale coincides with the $(n+1)$-th division of the vernier scale.
This relationship can be written as:
$(n+1)\mathrm{VSD}=n\mathrm{MSD}$

Step-by-Step Derivation:
First, let's find the value of one vernier scale division (1 VSD) in terms of MSD:
$1\mathrm{VSD}=\frac{n}{n+1}\mathrm{MSD}$

Now, substitute this value of 1 VSD into the formula for Least Count:
$\mathrm{LC}=1\mathrm{MSD}-\frac{n}{n+1}\mathrm{MSD}$

Taking 1 MSD common from both terms:
$\mathrm{LC}=(1-\frac{n}{n+1})\mathrm{MSD}$

Simplify the term inside the parentheses:
$\mathrm{LC}=\left(\frac{(n+1)-n}{n+1}\right)\mathrm{MSD}$
$\mathrm{LC}=\frac{1}{n+1}\mathrm{MSD}$

Since we are given that $1\mathrm{MSD}=a$ units, we substitute $a$ for MSD:
$\mathrm{LC}=\frac{a}{n+1}\text{units}$

Thus, the least count of the vernier scale is $\frac{a}{n+1}$ units.