Page 28 - Physics - XI

zero error in the instrument. But this condition is normally not satisfi ed as such and most of the instruments
have zero error. There are two types of zero errors: (i) Positive zero error (ii) Negative zero error.
(i) Positive zero error: When the tips of all the four legs lie in the same horizontal plane and the zero
mark on the circular scale lies above the zero mark on the vertical scale (pitch scale), the zero error is
positive. To fi nd the zero error, note the number of divisions on the circular scale which is coinciding
with the vertical scale and multiply this number by the least count of the spherometer. For example, if

8th division of circular scale coincides with vertical scale and 0.001 cm is the least count, then
Zero error = + 8 × L.C.
= + 8 × 0.001 cm

= 0.008 cm
Zero correction = Negative of zero error

= – 0.008 cm.
(ii) Negative zero error: When the tips of all the four legs lie in the same horizontal plane and the zero
mark on the circular scale lies a little below the zero mark on the vertical scale, then the zero error is
negative. In case of negative zero error, note down which number of divisions on the circular scale is
coinciding with the vertical scale and this number is subtracted from the total number of divisions on
the circular scale and the resultant number is multiplied by the least count. For example, let 0.001 cm
be the least count and there are 100 divisions on the circular scale; if 94th division is coinciding with
the vertical scale, the zero error will be
Negative zero error = – [100 – 94] × L.C. = – 6 × 0.001 cm

= – 0.006 cm
Zero correction = Negative of zero error

= – [– 0.006 cm]
= + 0.006 cm

Principle/Theory
Let a spherometer be placed on the spherical convex surface whose radius of curvature is to be measured
(Fig. 4.2). The central screw tip O and only two fi xed legs A and B are shown in Fig. 4.2. OM gives the
distance h through which the central leg must be lowered so that it may just touch the plane surface. This

distance OM is also called Sagitta. Let r be the distance AM between one of the outer legs and the central
screw when they are resting on the plane surface.
If R is the radius of curvature, we have
OM × MZ = AM × MB

Here, M is the midpoint of AB, also M is the centre of the equilateral triangle ABC formed by the tips of
the legs of the spherometer as shown in Fig. 4.2.
We have
OM × (OZ – OM) = AM × MB

or h × (2R – h) = r 2
r 2 h
or R = +
2 h 2
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