The steel wire of Searle's experiment is only \(0.40\,\text{mm}\) thick. Laid against a millimetre scale, it covers less than half of one division. A metre rule cannot tell it from a wire of \(0.3\) or \(0.5\,\text{mm}\). Yet Young's modulus needs its area, and the area goes as the square of the diameter.
Two scales slightly out of step
A vernier calliper carries two scales. The main scale on the beam is marked in millimetres. A short vernier scale slides beside it on the moving jaw. Its \(10\) divisions span \(9\,\text{mm}\), so each vernier division is \(0.9\,\text{mm}\). The two scales are made slightly out of step on purpose.
With the jaws closed, the two zeros coincide. Vernier line 1 is then \(0.1\,\text{mm}\) short of main line 1. Line 2 is \(0.2\,\text{mm}\) short of main line 2, and so on. Now open the jaws by \(0.3\,\text{mm}\). Line 3 moves forward by \(0.3\,\text{mm}\) and meets main line 3 exactly. The number of the coinciding line counts tenths of a millimetre.
To read the calliper, first take the main-scale reading: the last millimetre mark before the vernier zero. Then find the vernier line \(k\) that coincides with some main-scale line. The reading is the main-scale reading plus \(k\) times the least count. In the figure above that is \(3 + 7 \times 0.1 = 3.7\,\text{mm}\).
\[ n \times \text{VSD} = (n - 1) \times s \]
Why. the vernier is made so that its \(n\) divisions span \(n - 1\) main divisions of size \(s\).
\[ \text{VSD} = \dfrac{(n - 1)\,s}{n} \]
Why. divide both sides by \(n\) to find one vernier division.
\[ \text{LC} = s - \text{VSD} = s - \dfrac{(n - 1)\,s}{n} = \dfrac{s}{n} \]
Read it. the least count is the step by which one vernier division falls short of one main division.
On a vernier, \(20\) vernier divisions equal \(19\) main-scale divisions, and one main-scale division is \(0.5\,\text{mm}\). What is the least count, in mm?
A vernier calliper of least count \(0.01\,\text{cm}\) has its vernier zero just past the \(1.8\,\text{cm}\) mark, and vernier line 6 coincides with a main-scale line. What is the reading, in cm?
One turn, one pitch
A screw gauge holds a wire between a fixed anvil and a spindle driven by a fine screw. One full turn moves the spindle by the pitch of the screw, \(p\). The thimble that turns the screw carries a circular scale of \(N\) divisions. One division is \(1/N\) of a turn and moves the spindle by \(p/N\), the least count.
The sleeve carries a reference line with millimetre marks above it and half-millimetre marks below. The sleeve reading is the last mark the thimble has uncovered. Then read the circular division \(k\) on the reference line. The reading is the sleeve reading plus \(k\) times the least count. A gauge that shows \(1.5\,\text{mm}\) on the sleeve and \(27\) on the thimble reads \(1.5 + 27 \times 0.01 = 1.77\,\text{mm}\).
Two habits keep the reading honest. The thimble is turned by its ratchet, which slips once the wire is gripped, so the pressure is the same every time. And the screw is always turned the same way into the final position. A worn screw has a little play in its thread, called backlash. Reversing the turn takes up that play before the spindle moves.
A screw gauge of pitch \(0.5\,\text{mm}\) and \(50\) divisions shows the \(3.5\,\text{mm}\) mark uncovered on its sleeve, and division \(27\) on the reference line. What is the reading, in mm?
Zero errors, with their signs
Close the jaws of a calliper with nothing between them. If the vernier zero lies to the right of the main zero and line \(k\) coincides, the zero error is \(+k \times \text{LC}\). If it lies to the left, the instrument reads below zero. With \(k\) the coinciding line of an \(n\)-division vernier, the zero error is then \(-(n - k) \times \text{LC}\).
A screw gauge is checked the same way, with the anvil and the spindle touching. If the zero of the circular scale stops \(k\) divisions below the reference line, the gauge already reads \(+k\) divisions, a positive zero error. If it stops \(k\) divisions above the line, the zero error is \(-k \times \text{LC}\). In every case the true value is the observed reading minus the zero error.
A \(10\)-division vernier has a least count of \(0.01\,\text{cm}\). With its jaws closed, the vernier zero lies to the left of the main zero and line 6 coincides. A rod then reads \(3.26\,\text{cm}\). What is its true length, in cm?
Back to physics
The wire of Searle's experiment is measured with a screw gauge of the same kind as the figure above. The physics chapter uses its diameter, \(0.40\,\text{mm}\), to find the area and then Young's modulus. Here is where that number comes from.
This is the wire of Elastic Energy, Thermal Stress and Real Materials, lesson 1, done with the tool. A screw gauge of pitch \(0.5\,\text{mm}\) and \(50\) divisions has a zero error of \(+0.02\,\text{mm}\). On the wire the sleeve shows \(0\,\text{mm}\) and the thimble \(42\). Find the diameter, the area, and Young's modulus from the chapter's slope of \(12.5\,\text{N/mm}\) and length \(2.00\,\text{m}\).
\(\text{LC} = 0.5/50 = 0.01\,\text{mm}\). The observed reading is \(0 + 42 \times 0.01 = 0.42\,\text{mm}\).
True diameter \(= 0.42 - (+0.02) = 0.40\,\text{mm}\), so \(r = 0.20\,\text{mm}\).
\(A = \pi r^2 = \pi \times (0.20 \times 10^{-3})^2 = 1.257 \times 10^{-7}\,\text{m}^2\).
\(Y = \text{slope} \times L/A = 1.25 \times 10^4 \times 2.00/(1.257 \times 10^{-7}) = 1.99 \times 10^{11}\,\text{Pa}\), the chapter's value for steel.
A student forgets the \(+0.02\,\text{mm}\) zero error and uses \(0.42\,\text{mm}\) for the diameter of Searle's wire. How does her Young's modulus compare with the true value?
The spherometer
A spherometer is a screw gauge on a tripod. Three legs stand at the corners of an equilateral triangle of side \(l\). A central screw with a circular scale moves up and down through the middle. First it is set on a plane glass plate, touching the glass. Then it is placed on the lens, and the screw is turned until it touches the pole. The screw has moved by \(h\), the height of the curved surface above the plane of the feet.
\[ R^2 = a^2 + (R - h)^2 \]
Why. Pythagoras in the right-angled triangle of Fig 4.1, where \(a\) is the distance of each foot from the axis.
\[ 2Rh = a^2 + h^2 \]
Why. expand \((R - h)^2 = R^2 - 2Rh + h^2\) and cancel \(R^2\) from both sides.
\[ R = \dfrac{a^2}{2h} + \dfrac{h}{2} \]
Why. divide both sides by \(2h\).
\[ R = \dfrac{l^2}{6h} + \dfrac{h}{2} \]
Read it. the feet sit at the corners of an equilateral triangle of side \(l\), whose corners are \(a = l/\sqrt{3}\) from its centre, so \(a^2 = l^2/3\).
On a concave surface the screw moves down from the plane-glass position instead of up. The triangle is the same, so the formula is the same. The second term, \(h/2\), is usually small. For a lens it is a correction of less than one per cent, but it costs nothing to keep it.
A spherometer of pitch \(1\,\text{mm}\) with \(100\) divisions has its legs \(4.0\,\text{cm}\) apart. From the plane-glass position to the pole of a convex lens it turns \(1\) full turn and \(34\) divisions. Find the radius of the surface. This laboratory setting is not from a physics chapter; the lens is the one used there.
\(\text{LC} = 1/100 = 0.01\,\text{mm}\), so \(h = 1 + 34 \times 0.01 = 1.34\,\text{mm} = 0.134\,\text{cm}\).
\(R = l^2/6h + h/2\), with \(l = 4.0\,\text{cm}\).
\(R = 16/(6 \times 0.134) + 0.067 = 19.90 + 0.07 = 19.97\,\text{cm}\), that is \(20.0\,\text{cm}\).
Spherical Surfaces and Lenses, lesson 3, finds that an equiconvex glass lens of radii \(20\,\text{cm}\) has \(f = R = 20\,\text{cm}\). The spherometer supplies that radius.
The legs of a spherometer are \(6.0\,\text{cm}\) apart, and on a curved mirror its screw moves by \(h = 0.30\,\text{cm}\). What is the radius of curvature of the mirror, in cm?
An instrument cannot be read better than its least count, and it can easily be read worse. The eye must be straight in front of the scales, or parallax shifts the reading. When no vernier line coincides exactly, take the nearest one; the reading is then uncertain by one least count. The sign of a zero error is the commonest slip. A vernier zero to the left, or a circular zero above the line, is negative. Subtracting it increases the reading.
Pierre Vernier was an official of the castle at Ornans, in the Franche-Comté of eastern France. In 1631 he published in Brussels a short book on a new quadrant for measuring angles. On it, a small sliding scale divided \(n - 1\) main divisions into \(n\) equal parts. A reader could then count the fraction of a division from the line that coincided. The sliding scale soon spread to every precise instrument, and it has carried his name since the eighteenth century.
The screw gauge and the micrometer screw are the instruments of Searle's experiment in Elastic Energy, Thermal Stress and Real Materials. The screw reads the extension to \(0.01\,\text{mm}\), and the gauge reads the diameter that sets the area. The spherometer gives the radii that the lens maker's formula of Spherical Surfaces and Lenses needs. In the practicals of the book, a vernier calliper measures the bore of a capillary tube and the diameter of a pendulum bob.
- Adding the number of the coinciding line instead of that number times the least count.Reading = main-scale reading + \(k \times \text{LC}\). Line 6 of a \(0.01\,\text{cm}\) vernier adds \(0.06\,\text{cm}\).
- Missing the half-millimetre mark below the reference line of a screw gauge.Read the sleeve to the last mark uncovered, above or below the line.
- Taking a negative zero error as \(-k \times \text{LC}\) on a vernier whose zero lies to the left.With the zero to the left, the zero error is \(-(n - k) \times \text{LC}\).
- Using the leg distance itself, not its square, in the spherometer formula.\(R = l^2/6h + h/2\). Square \(l\) first.