Exercise 1.6

Transcript

Exercise 1.6 Which of the following is the most precise device for measuring length : (a) a vernier callipers with 20 divisions on the sliding scale (b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale (c) an optical instrument that can measure length to within a wavelength of light ? We compare the least count of all three Least count is the smallest value that a measuring instrument can accurately read on its scale Usually, formula for Least count is Least count = (𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 𝑜𝑛𝑒 𝑀𝑎𝑖𝑛 𝑆𝑐𝑎𝑙𝑒 𝐷𝑖𝑣𝑖𝑠𝑖𝑜𝑛)/(𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠 𝑖𝑛 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑠𝑐𝑎𝑙𝑒) (a) Vernier callipers Assume the standard arrangement: 1 main−scale division =1 mm 20 vernier divisions =19 main−scale divisions [; ] Then: 1 vernier division =(19)/(20) mm The least count is: LC=1−(19)/(20) LC=(1)/(20) 𝐋𝐂=𝟎.𝟎𝟓 𝐦𝐦 [, ; , ] The number of vernier divisions alone does not fully specify the least count; the scale relationship above is assumed. (b) Screw gauge [LC=( pitch )/( number of circular−scale divisions ); LC=(1 mm)/(100); LC=0.01 mm] (c) Optical instrument A typical visible-light wavelength is approximately: 𝜆∼5×10^(−7) m=0.0005 mm The number of vernier divisions alone does not fully specify the least count; the scale relationship above is assumed. (b) Screw gauge LC=( pitch )/( number of circular−scale divisions ) LC=(1 mm)/(100) LC = 0.01 mm (c) Optical instrument A typical visible-light wavelength is approximately: 𝜆∼5×10^(−7) m=0.0005 mm A smaller resolving interval means finer measurement capability. Thus, The optical instrument is the most precise of the three So, option (c) is the correct answer Device | Approximate resolving interval Vernier callipers | 0.05 mm Screw gauge | 0.01 mm Optical instrument | 0.0005 mm