Question

The least count of spherometer is given by:
A.pitch$\times$ no of circular divisions

B.$\dfrac{{{\text{pitch}}}}{{{\text{no}}{\text{.}}\,{\text{of}}\,{\text{circular}}\,{\text{divisions}}}}$

C.$\dfrac{{{\text{no}}{\text{.}}\,{\text{of}}\,{\text{circular}}\,{\text{divisions}}}}{{{\text{pitch}}}}$

D.$\dfrac{{{\text{pitch}}}}{{{\text{mean}}\,{\text{distance}}\,{\text{between}}\,{\text{two}}\,{\text{consecutive}}\,{\text{legs}}\,{\text{of}}\,{\text{the}}\,{\text{spherometer}}}}$

Answer 1

Hint:First let us see what a spherometer is: A spherometer is a device for precisely measuring the curvature radius of a sphere or a curved surface. These methods were initially used mainly by opticians to calculate the curvature of a lens surface.

Complete step by step answer:
The spherometer is an instrument that measures the radius of curvature of a spherical surface using the micrometre screw principle. This consists of a graduated disc in $500$ divisions, say, which is turned above by a milled head.
The screw is rotated precisely such that for every turn of the disc, say $0.5\,mm$ , it progresses. At the centre of three symmetrically arranged fixed legs is the blunt end of the bolt. In use, the four points are performed before a reading is taken to contact the area to be weighed. The tool is zeroed onto an optically flat plate and then the travel of the screw can be precisely determined to touch a circular surface, say a mirror or prism.
The majority of spherometers were small instruments, typically from four or five centimetres to less than twenty centimetres in diameter, and were primarily used by lens manufacturers to reliably measure the lens curvature. They can easily bend considering the fragile nature of these instruments, so they often come with a glass layer, called a flat, which can be used to zero the micrometre. Three legs have spherometers, which form an equilateral triangle.In one complete rotation of the circular disc scale, pitch is known as the distance moved by the central screw. The screw's pitch is $1\,mm$ .
The least count of a spherometer is given by: $\dfrac{{{\text{pitch}}}}{{{\text{no}}{\text{.}}\,{\text{of}}\,{\text{circular}}\,{\text{divisions}}}}$
Hence option B is correct.

Note:We have to be careful while taking the reading on a spherometer as the screw might have friction or divisions may not be of equivalent size between circular scales.A spherometer works on the principle of the micrometer screw. It is used to measure either very small thickness of flat materials like glass or the radius of curvature of a spherical surface thus getting its name.