How does a sampling rate of 1ns equate to 15cm for an optical instrument? I’m currently working with open source data captured by the NASA GEDI LiDAR instrument and I am applying a math model to it. GEDI’s sampling rate is said to be 1 nanosecond or 15 centimeter resolution. Is the sampling rate due to a 1ns difference in round-trip time translating to a 15cm difference in distance?
The following is a link to the page that has the instrument details: Instrument Overview - GEDI
I’m primarily referring to the following line from the page: “ These photons are then directed towards detectors, converting the brightness of the light to an electronic voltage which is then recorded as a function of time in 1 ns (15 cm) intervals.”
 A: We can easily check this by calculating how far light travels in 1 ns.
$$
x = c \cdot t = 299792458\,\text{m}/\text{s} \cdot 10^{-9}\,\text{s} \approx 0.3\,\text{m} = 30\,\text{cm}
$$
By how a LIDAR works, as you correctly say, it measures the round trip distance from the source to the object and back to the detector, so the 30 cm travel distance resolution means a resolution of 15 cm for the distance between the source/detector and the reflecting object.

Example for clarity
Say the instrument is 400 meters above the ground. We send a laser pulse towards the ground and measure the time to its return. We find a round trip time of $2669\pm 1\,\text{ns}$. To get the distance, we must multiply by $c$:
$$ x_\text{rt} = c \cdot t = 299792458\,\text{m}/\text{s} \cdot (2669\pm 1)\cdot 10^{-9}\,\text{s} = (800.15 \pm 0.30)\,\text{m}$$
Remember this is the round trip time, and thus also the round trip distance. To get the one-way distance from the instrument to the ground, we must divide by 2.
$$ x = \frac{x_\text{rt}}{2} = (400.08\pm0.15)\,\text{m}$$
You see how an uncertainty of 1 ns due to the sampling rate translates to 30 cm uncertainty of the round trip distance (which I above called travel distance), and thus a 15 cm uncertainty of the distance between the instrument and the ground.
