The Vernier scale of a travelling microscope | vedclass

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(B) Given,reading of $1$ main scale division $(MSD)$ = $0.5 	ext{ mm}$.According to the given situation,$50$ vernier scale divisions $(VSD)$ = $49$ m

Vedclass · Accepted Answer

$0.01 	ext{ mm}$

Vedclass · Answer

(B) Given,reading of $1$ main scale division $(MSD)$ = $0.5 	ext{ mm}$.According to the given situation,$50$ vernier scale divisions $(VSD)$ = $49$ main scale divisions $(MSD)$.Therefore,$1$ vernier scale division $(VSD)$ = $\frac{49}{50} 	ext{ MSD}$.The least count of the microscope is defined as the difference between $1 	ext{ MSD}$ and $1 	ext{ VSD}$.$	ext{Least count} = 1 	ext{ MSD} - 1 	ext{ VSD} = 1 	ext{ MSD} - \frac{49}{50} 	ext{ MSD} = \frac{1}{50} 	ext{ MSD}$.Substituting the value of $1 	ext{ MSD} = 0.5 	ext{ mm}$:$	ext{Least count} = \frac{1}{50} 	imes 0.5 	ext{ mm} = 0.01 	ext{ mm}$.

The Vernier scale of a travelling microscope has $50$ divisions which coincide with $49$ main scale divisions. If each main scale division is $0.5 \text{ mm}$,then the least count of the microscope is

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