A microscope is provided with a main scale graduated with 20 divisions in 1cm and a vernier scale with 50 divisions on it of length same as of 49 divisions of main scale. Find the least count of the microscope.

i) As we know,

$\text{L.C.} = \dfrac{\text{Value of 1 main scale div (x)}}{\text{Total no. of div on vernier(n)}}$

Given,

20 divisions = 1 cm

Therefore,

$\text{1 division} = \dfrac{1}{20} \\[0.5em] \Rightarrow \text{1 division} = 0.05 \text{cm}$

Hence, value of one main scale division = 0.05 cm

Total number of divisions = 50

Substituting the values in the formula above, we get,

$\text{L.C.} = \dfrac{\text{0.05} \text{cm}}{\text{50}} \\[0.5em] \text{L.C.} = 0.001 \text{cm} \\[0.5em]$

Hence, least count of the microscope is 0.001 cm.