A survey team uses a laser rangefinder while checking the position of a floating marker near a harbour wall. Fig. 1 shows how the instrument works. The rang...
Question 1 Report
A survey team uses a laser rangefinder while checking the position of a floating marker near a harbour wall. Fig. 1 shows how the instrument works. The rangefinder sends a very short pulse of red laser light towards a reflector on the marker. Some light is reflected back to the instrument. Its electronic circuit measures the total time from emission to detection. For one reading, the displayed time is 160 ns. Use a speed of light in air of 3.0 x 108 m/s.
(a) What process returns some of the laser light to the rangefinder? [1]
(b) Calculate the total distance travelled by the light pulse in 160 ns. [3]
(c) Explain why the distance from the rangefinder to the marker is half the answer to part (b). [2]
(d) Suggest three reasons why a reflector gives a more dependable distance reading than the surface of water. [3]
(e) Explain how an uncertainty of 4 ns in the measured time affects the calculated distance to the marker. [5]
Answer Details
(a) Some of the laser light is returned by reflection from the reflector. [1]
(b) Convert nanoseconds to seconds:
\[160\text{ ns}=160\times10^{-9}\text{ s}\]
Use \(d=vt\):
\[d=(3.0\times10^8)(160\times10^{-9})=48\text{ m}\]
The total distance travelled by the pulse is \(48\text{ m}\). [3]
(c) The 48 m includes both the outgoing path from the rangefinder to the marker and the reflected path back to the rangefinder. The one-way distance to the marker is therefore:
\[48\div2=24\text{ m}\]
[2]
(d) A reflector gives a more dependable reading because it returns more light to the detector, producing a clearer and stronger signal. Water may absorb some light, ripples can direct reflected light away from the detector, and the water surface position is not fixed. Any three of these points gain credit. [3]
(e) First convert the time uncertainty:
\[4\text{ ns}=4\times10^{-9}\text{ s}\]
The uncertainty in the total light path is:
\[\Delta d=(3.0\times10^8)(4\times10^{-9})=1.2\text{ m}\]
This is for the outward and return journey, so the uncertainty in the one-way marker distance is:
\[\frac{1.2}{2}=0.60\text{ m}\]
The calculated distance to the marker has an uncertainty of \(\pm0.60\text{ m}\). [5]
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