The circular main scale of a spectrometer is divided into...

The value of one main scale division is $1\text{ MSD} = 0.5^\circ$.

Given that $30$ vernier scale divisions (VSD) coincide with $29$ main scale divisions (MSD), we can write:
$1\text{ VSD} = \dfrac{29}{30}\text{ MSD}$

The least count (LC) of the instrument is given by:
$\text{LC} = 1\text{ MSD} - 1\text{ VSD}$
$\text{LC} = 1\text{ MSD} - \dfrac{29}{30}\text{ MSD} = \dfrac{1}{30}\text{ MSD}$

Substitute the value of $1\text{ MSD}$:
$\text{LC} = \dfrac{1}{30} \times 0.5^\circ = \left(\dfrac{1}{60}\right)^\circ$

Since $1^\circ = 60\text{ minutes}$, we convert the least count into minutes:
$\text{LC} = \dfrac{1}{60} \times 60\text{ minutes} = 1\text{ minute}$

Answer: $1$