The circular main scale of a spectrometer is divided such that each...

First, find the value of one main scale division (MSD).
Since $1^{\circ}$ is divided into $3$ equal parts, and $1^{\circ} = 60^{\prime}$ (minutes of arc):
$1 \text{ MSD} = \dfrac{60^{\prime}}{3} = 20^{\prime}$

Given the coincidence on the Vernier scale:
$60 \text{ VSD} = 59 \text{ MSD}$
$1 \text{ VSD} = \dfrac{59}{60} \text{ MSD}$

The least count (L.C.) is:
$\text{L.C.} = 1 \text{ MSD} - 1 \text{ VSD}$
$\text{L.C.} = 1 \text{ MSD} - \dfrac{59}{60} \text{ MSD} = \dfrac{1}{60} \text{ MSD}$

Substitute the value of $1 \text{ MSD}$:
$\text{L.C.} = \dfrac{1}{60} \times 20^{\prime} = \dfrac{1}{3}^{\prime}$

Convert the least count into seconds of arc ($1^{\prime} = 60^{\prime\prime}$):
$\text{L.C.} = \dfrac{1}{3} \times 60^{\prime\prime} = 20^{\prime\prime}$

Answer: $20$