The observations of a reciprocal levelling operation carried out from two stations A and B are given in the table. It is known that the instrument has collimation error. The correct staff reading over station B when the instrument is set very near to station A is ______ m (Rounded off to three decimal places). Assume all other errors are negligible.
Instrument set very near to station
Staff reading at A (m)
Staff reading at B (m)
A
1.275
2.005
B
1.040
1.660
Show Hint
Use the reciprocal levelling formula: true difference of level = average of the two apparent differences obtained with the instrument near A and near B, then add it to the accurate near reading at A.
Step 1: Model the collimation error explicitly. Let the instrument, when set up near A, add an error $e_1$ to any far-staff reading over the sight from A to B (the near-staff reading at A is taken over a very short distance, so its error is negligible). Similarly, when set up near B, let it add an error $e_2$ to the far-staff (A) reading. Near-A setup: true reading on A $= a_1 = 1.275$ (no error); observed reading on B $=$ true reading on B $+ e_1$. Near-B setup: true reading on B $= b_2 = 1.660$ (no error); observed reading on A $=$ true reading on A $+ e_2$. Step 2: Write the two equations for the true elevation difference. Let $\Delta H$ be the true fall from A to B. From the near-A setup, the apparent difference is $2.005 - 1.275 = 0.730$ m, equal to $\Delta H + e_1$. From the near-B setup, the apparent difference is $1.660-1.040=0.620$ m, equal to $\Delta H - e_2$ (the error acts in the opposite sense here because A is now the far staff). Step 3: Eliminate the unknown errors by averaging. Since both setups use the same sight length (A to B) and the same instrument, the collimation error contributes equally and oppositely to the two apparent differences, so averaging cancels it: \[ \Delta H = \frac{(2.005-1.275) + (1.660-1.040)}{2} = \frac{0.730+0.620}{2} = 0.675 \text{ m} \] Step 4: Back-calculate the corrected far reading for the near-A setup. With the instrument near A, the near reading on A, $1.275$ m, is accurate. Because the true fall from A to B is $0.675$ m, the correct staff reading that should have been obtained on B from this same instrument position is: \[ 1.275 + 0.675 = 1.950 \text{ m} \] Step 5: Final answer. \[ \boxed{1.950 \text{ m}} \]
