Why do sensors that emit higher frequency signals give more accurate data? I am doing a technical presentation about RADAR and LiDAR. I understand that LiDAR is several times more accurate and capable of producing really detailed 3-D maps of their surroundings, while RADARs tend to lag behind in accuracy. 
Several sources indicate that the shorter wavelength signals from LiDAR contribute to its higher accuracy, but they don't really explain why.
This webpage explains that higher frequency signals yield more accurate data in RADARs, but does not really explain why either:
http://www.radartutorial.eu/07.waves/Waves%20and%20Frequency%20Ranges.en.html
There are some posts that mention the Heisenberg Uncertainty Principle, but I don't think an explanation at the atomic level is really relevant nor required to explain this phenomenon.
Can anybody give an equation or state a scientific concept as to why sensors/apparatuses that use higher frequency signals yield more accurate data?
 A: High frequency sensing gives a better resolution. The resolution is naturally limited by the wavelength. That being said, the propagation of your signal is also very important. A high frequency signal may have a smaller range or may have a smaller penetration depth, for example through clouds or vegetation. You have to pick the frequency that gives you optimal contrast for the signal that you are after. 
A: This has nothing to do with Heisenberg or his uncertainty principle. To measure range both radar and lidar measure time of arrival for both assume that propagation velocity is known and is the same in all directions and all weather, etc. Timing measurement is, therefore, the measurement of rise time of a pulse in noise; measurement accuracy is defined as the statistical dispersion = standard deviation = jitter and is given by: $\sigma_t \approx k_0\frac{t_r}{\sqrt{SNR}}$ where $t_r$ is the rise-time, $\frac{1}{2} < k_0 <2$ is a number that depends on the details of the pulse and signal processing, and SNR is the effective signal to noise ratio. You can think of $\sqrt{SNR}$ in the denominator as representing the result of averaging across the pulse by the noise filter.
The rise time is $t_r \approx \frac{k_1}{B}$ where $B$ is the band width and $k_1$ is a number that depends on the details of the pulse. In other words  $$\sigma_t \approx k_0k_1\frac{1}{B\sqrt{SNR}}$$ As you can see the jitter is inversely proportional with the signal bandwidth and is independent of the carrier frequency. But note that higher the carrier frequency the wider the signal bandwidth can be! In practice a laser pulse can easily be made a nanosec or shorter while radar bands are constrained both by FCC regulations and by electronics.
There is a similar result for angle measurement accuracy $$\sigma_\phi \approx k_2\frac{1}{D\sqrt{SNR}}$$
Here $D$ is the effective diameter of the antenna (lens) and $0.5 < k_2 <2 $ is a number dependent on the illumination details. This is the so-called diffraction limited case.
A: In the article that anna v mentioned:
robotsforroboticists.com/lidar-vs-radar 
there is a segment which states: 
"The down side [of the RADAR] is that if an object is much smaller than the RF wave being used, the object might not reflect back enough energy to be detected. For that reason many RADAR’s in use for obstacle detection will be “high frequency” so that the wavelength is shorter (hence why we often use mm-wave in robotics) and can detect smaller objects." 
This does provide some intuitive understanding as to why devices emitting higher frequency signals provide high-resolution data. Small objects and the small/fine details of large objects (ie. protrusions of a wall, a pedestrian's facial features, and bumps/cracks/curves that make up the texture of any surface) do not reflect enough low-frequency EM wave energy back to the RADAR. Thus, such fine details are not detected by RADAR. 
If the generated signals were of higher frequency, then the EM wave oscillates faster, and a greater percentage of the wave would hit and be reflected by small objects and the fine details of large objects (ie. protrusions/bumps/curves). Thus, a sensor emitting waves at a higher frequency can detect such details.
The above explanation may not be not most technical, but I hope it provides others with a more intuitive understanding of how I reasoned it out.
Thanks everyone for sharing their ideas!
