The puzzle was as follows: On a clear dark night, you are driving your car with your headlights on along a narrow but straight country road. A stopped bicycle 30 meters away, with no lights of its own, no reflectors, and a rather dull paint job, would be visible with 10 times the minimum brightness for you to notice an object. Suppose you are driving 30 meters per second and need to notice any such stopped bicycle at least two seconds in advance to safely avoid a tragic collision. Can you safely avoid a collision with such a stopped bicycle? If your headlights had totaled 100 watts of power, what would be the minimum headlight power to have four seconds of advance warning?
Since there are no shiny, mirror, or glossy surfaces, all light reflected from the bicycle will be reflected diffusely or scattered. Light emitted from the headlights declines in intensity as 1/d2, where d is the distance from the headlights to the bicycle, and light reflected off the bicycle also declines in intensity as 1/d2. This leads to the famous “radar equation” whereby the intensity of radiation received back at the source of emission declines as 1/d4. Note: This is why long-range radar units tend to consume so much power, often getting really hot and requiring their own cooling equipment!
The intensity of reflection the car driver sees is 1/16 as great at 60 meters as at 30 meters. Since the bicycle was 10 times minimum visible brightness at 30 meters, it will only be 10/16 of minimum visible brightness at 60 meters. So, the driver, driving at 30 meters per second, will have less than two seconds to react to the bicycle and likely not be able to avoid a collision.
At four seconds before collision the car is 120 meters away and the brightness of the bicycle is 1/256 as great as at 30 meters, or 10/256 of minimum visible brightness. If headlight power had been 100 watts, then 2560 watts would be needed for four seconds warning.