The main scale of a spectrometer is divided such that one main...

Given, $1\text{ MSD} = 0.5^\circ$.
Since $1^\circ = 60'$, we convert the main scale division into arc minutes:
$1\text{ MSD} = 0.5 \times 60' = 30'$

It is given that $60$ vernier scale divisions (VSD) coincide with $56$ main scale divisions (MSD).
$60\text{ VSD} = 56\text{ MSD}$
$1\text{ VSD} = \dfrac{56}{60}\text{ MSD}$

The least count (LC) of the spectrometer is defined as:
$\text{LC} = 1\text{ MSD} - 1\text{ VSD}$
$\text{LC} = 1\text{ MSD} - \dfrac{56}{60}\text{ MSD} = \dfrac{4}{60}\text{ MSD}$

Substituting the value of $1\text{ MSD}$ in arc minutes:
$\text{LC} = \dfrac{4}{60} \times 30' = 2'$

The least count is $2$ arc minutes.

Answer: $2$