Question

Identify the instrument having minimum least count out of the following options.

• A. A vernier calipers with 20 divisions on the vernier scale coinciding with 19 main scale divisions.
• B. A screw gauge of pitch 1 mm and 100 divisions on the circular scale.
• C. A spherometer of pitch 0.1 mm and 100 divisions on the circular scale.
• D. An optical instrument that can measure length to within a wavelength to light.

Answer 1

Answer: (D)

(a) Least count of vernier calipers
= 1MSD – 1 VSD =1 MSD - $\frac{19}{20}$ MSD
= $\frac{1}{20}MSD=\frac{1}{20}mm=\frac{1}{200}cm=0.005 \, cm$
$\left(\right.\because 1 \, MSD=1mm\left.\right)$
(b) Least count of screw gauge= $\frac{P i t c h}{N o . o f \, d i v i s i o n s \, o n \, c i r c u l a r \, s c a l e}$
= $\frac{1}{100}mm=\frac{1}{1000}cm=0.001 \, cm$
(c) Least count of spherometer = $\frac{P i t c h}{N o . o f \, d i v i s i o n s \, o n \, c i r u l a r \, s c a l e}$
= $\frac{0.1 m m}{100}=\frac{1}{1000}mm=0.0001 \, cm$
(d) Wavelength of light, $\lambda \approx10^{- 5}cm=0.00001 \, cm$ Clearly the optical instrument is the most precise.