Converting source intensity to the maximum reach of beam flux

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 1K views
Qube
Gold Member
Messages
461
Reaction score
1
Homework Statement
Consider a headlamp that is measured to have an intensity of 35,000 candela at FMVSS 108 test point 1.5D, 2R. That is, 1.5 degrees below the horizontal axis and 2 degrees right of the vertical axis with respect to the lamp. On a completely unlit road, what is the maximum distance at which there is 5 lux illumination?
Relevant Equations
E = 10.76*(I/d^2)
We can solve for the maximum 5 lux illumination distance with the above equation.

E = 10.76*(35,000/d^2)

d = 275 feet (approximately).

However, the 5 lux illumination distance is not 275 feet. The 35,000 cd value is not an axial intensity value. It is at a point that is slightly down and to the right with regard to the intersection of the lamp's horizontal and vertical axes (H-V).

Therefore, we have to set up a right triangle as such:

IMG_20190727_171557.jpg


Is this the correct setup? Note that we cannot solve this problem as stated, since no mounting height of the lamp is given.
 
Physics news on Phys.org
No, this is not the correct setup. The equation given is E=10.76*(35000/d^2), where E is the lux illumination and d is the distance from the lamp. There is no mention of mounting height in the equation.