Reciprocal levelling / two-peg test computation: A level set up over peg A has eye-piece height above A = 1.540 m and reads 0.705 m on staff held at peg B. Then set up over peg B with eye-piece height above B = 1.490 m and read 2.195 m on staff at A. What is the difference in level between A and B (magnitude)?

Introduction / Context:
This is a classic reciprocal levelling (or two-peg test style) computation. By observing from each end, collimation and refraction effects are largely canceled, allowing determination of the true difference in level even if the line of sight has a small inclination error.

Given Data / Assumptions:

• Height of eye (instrument axis above peg tops): h_A = 1.540 m at A, h_B = 1.490 m at B.
• Staff reading at B from A: r_B = 0.705 m (upright staff).
• Staff reading at A from B: r_A = 2.195 m (upright staff).
• Sight distances are the same line AB in opposite directions.

Concept / Approach:
Let RL_A and RL_B be elevations of pegs A and B. With the instrument set over A: HI_1 = RL_A + h_A and RL_B = HI_1 − r_B, so RL_B − RL_A = h_A − r_B. With the instrument set over B: HI_2 = RL_B + h_B and RL_A = HI_2 − r_A, so RL_B − RL_A = r_A − h_B. Small collimation/refraction errors make these two values differ slightly; the true difference is taken as their mean.

Step-by-Step Solution:

Verification / Alternative check:
Check sums: h_A + h_B = 3.030 m, r_A + r_B = 2.900 m; the mismatch (0.130 m) indicates collimation/refraction effects, which averaging neutralizes in the final result.

Why Other Options Are Wrong:

• 2.900 m and 3.030 m: these are sums (r_A + r_B) or (h_A + h_B), not a level difference.
• 0.785 m: incorrect rounding; precise mean is 0.770 m.
• 1.770 m: unrelated combination of numbers.

Common Pitfalls:
Subtracting in the wrong order; ignoring the need to average reciprocal values; mixing heights of instrument with staff readings without consistent sign convention.

Final Answer:
0.770 m