Question

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

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.

Answer 1

Correct Answer: 1.95

Step 1: Recall the principle of reciprocal levelling.
Reciprocal levelling is used to find the true difference of level between two points A and B when the line of sight is long, so that errors due to curvature of the earth, atmospheric refraction, and any collimation error in the instrument are eliminated. Two sets of staff readings are taken: once with the instrument set up very near A, and once with the instrument set up very near B. With the instrument close to a staff, the reading on that near staff is virtually free of curvature, refraction and collimation error, while the reading on the far staff carries the full error.
Step 2: Write down the given readings.
Instrument near A: reading on A (near) \(a_1 = 1.275\) m, reading on B (far) \(b_1 = 2.005\) m. Instrument near B: reading on A (far) \(a_2 = 1.040\) m, reading on B (near) \(b_2 = 1.660\) m.
Step 3: Compute the true difference of level between A and B.
The true difference in elevation, free of the systematic error, is the average of the apparent differences obtained from the two instrument setups: \[ \Delta H = \frac{1}{2}\left[(b_1 - a_1) + (b_2 - a_2)\right] \] Substituting the values: \[ \Delta H = \frac{1}{2}\left[(2.005 - 1.275) + (1.660 - 1.040)\right] = \frac{1}{2}\left[0.730 + 0.620\right] = \frac{1.350}{2} = 0.675 \text{ m} \] So B is lower than A by \(0.675\) m.
Step 4: Find the correct staff reading on B from the setup near A.
Since the near reading \(a_1 = 1.275\) m (instrument near A, staff on A) is essentially free of collimation error, it can be taken as correct. The correct reading on B, from the same instrument position near A, must be consistent with the true difference of level found in Step 3: \[ b_{1,correct} = a_1 + \Delta H = 1.275 + 0.675 = 1.950 \text{ m} \]
Step 5: State the result.
\[ \boxed{b_{1,correct} = 1.950 \text{ m}} \]