I find the functioning of a laser rangefinder confusing.

The explanation usually goes like this: "you shine a laser beam onto the object, the laser beam gets reflected and gets back to the device and time required for that is used to calculate the distance".

Okay. But the object surface can be uneven and not perpendicular to the laser beam so only a tiny fraction of beam energy is reflected back to the device. And there's plenty of other radiation around, sunlight included.

How does a rangefinder manage to "see" that very weak reflected signal in a reliable manner?


so only a tiny fraction of beam energy is reflected back to the device.

This tiny fraction is enough. With respect to ambient light: One can modulate the laser beam, and filter the the voltage of the receiving photodiode for this modulation frequency and phase. Another precaution is to have a light filter in front of the receiving photodiode which only lets the wavelength of the laser pass. I think both precautions are used. And of course the receiving photodiode is focussed to a spot of some centimeters diameter around the laser spot. Try to point the range finder to a mirror, in that case the range finder should fail, exept the mirror happens to reflect precisely back to the range finder (which is rather unlikely). Reason is that from a (clean) mirror You don't get a stray reflection.


Some laser rangefinding uses a retroreflector, which will bounce the laser light back in the direction it came regardless of orientation.

Otherwise, lasers operate at a very specific frequency, so the signal/noise ratio only needs to be strong enough to be detectable at that frequency.

If you shine a normal laser pointer on a wall, even if the wall is pretty far away, you can see the spot it makes. That means your eye can pick out the reflected laser light. The electronics can be made better than your eye, so it's not too hard to see reflected laser light..

  • $\begingroup$ Why the down vote? Is something here wrong? $\endgroup$ – Mark Eichenlaub Jun 2 '11 at 14:34
  • $\begingroup$ I found your answer useful, so a cosmic karma +1. $\endgroup$ – Captain Giraffe Jun 2 '11 at 17:44
  • 1
    $\begingroup$ The most difficult target for a LR is a shiny surface. Unless it is a corner cube or directly normal to the beam the laser light will bounce off at some angle and not return to the receiver. It's very difficult to do laser scans of clean polished cars fro example $\endgroup$ – Martin Beckett Jun 2 '11 at 22:40
  • $\begingroup$ @Martin Yes, but I don't get what you're driving at. I already mentioned the retroreflector (corner cube). This second point was mentioned by Georg in his answer. $\endgroup$ – Mark Eichenlaub Jun 3 '11 at 0:04
  • $\begingroup$ the OP was concerned how a 'non-reflective' surface could reflect light when the most difficult target is actually a highly reflective one. $\endgroup$ – Martin Beckett Jun 3 '11 at 3:13

The amount of laser energy reflected back will be the limiting factor of its effective range. However, since the laser's radiation is of a specific wavelength, it won't be confused by extraneous radiation from ambient sources.


Simply what it means is if you can see the laser dot on the wall the rangefinder can see it 1000s better! Don't think of it as a reflection coming off the fur of a bear at 300 yards. That seems stupid but if you aim a laser at a bear at 300 yards and had a telescope, you would and could see the dot on the bears fur. That's all the rangefinder does. It sees the dot, calculates time of flight to the bear then tells you 300 yards...simple.

  • $\begingroup$ It also means, that the rangefinder has to find a typically not very strong light emitting point 300 yard away? How does it differentiate from the background? $\endgroup$ – user259412 Mar 15 '18 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.