How much refraction correction (in m) is required for a distance of 500 m?
Correct Answer: A. 0.0028 m
📚 Detailed Explanation: Refraction Correction for 500 m = 0.0028 m
Why A (0.0028 m) is correct: The standard formula for atmospheric refraction correction in levelling uses D in kilometres. Converting 500 m to 0.5 km and substituting gives 0.0028 m.
Formula: Refraction correction Cr = 0.0112 × D² (D in kilometres)
D = 500 m = 0.5 km
Cr = 0.0112 × (0.5)²
Cr = 0.0112 × 0.25
Cr = 0.0028 m
Check Against Other Corrections
| Correction Type | Formula (D in km) | Value at 0.5 km |
|---|---|---|
| Curvature (Cc) | 0.0785 × D² | 0.0785 × 0.25 = 0.0196 m |
| Refraction (Cr) | 0.0112 × D² | 0.0112 × 0.25 = 0.0028 m |
| Combined (Cc + Cr) | 0.0673 × D² | 0.0673 × 0.25 = 0.0168 m |
Why Other Options Are Wrong
| Option | Value | Error |
|---|---|---|
| A. 0.0028 | 0.0028 m | CORRECT — refraction correction at 0.5 km |
| B. 0.0056 | 0.0056 m | 0.0028 × 2 — uses D = 0.5 without squaring (wrong) |
| C. 5.61 | 5.61 m | Forgot to convert m to km; used D = 500 km (wrong) |
| D. 2850 | 2850 m | Completely wrong; no valid derivation |
- Refraction correction: Cr = 0.0112 × D² (D in km).
- For D = 500 m = 0.5 km: Cr = 0.0112 × 0.25 = 0.0028 m.
- Refraction is exactly 1/7 of the curvature correction: 0.0196/7 = 0.0028 m ✓.