A light source is placed on a table at a distance
d from the center
of a coin.
The illumination of the coin is poor due to the low angle of illumination.
Lifting the light above table's level will improve the angle, but increase the distance.
What is the height needed to get the best illumination of the coin?
N.B. The puzzle was not placed in the "Calculus" category.
I assumed the coin was a disk on the plane of the table--hence the zero illumination when the light source is on the table's surface, and that the coin was small enough that the degree of illumination at its center would serve as a proxy for the total illumination of the face of the coin.
The illumination is directly proportional to the sine of the angle of elevation, but also inversely proportional to the square of the distance (slant distance) from the coin to the light source.
The sine of the angle is h/sqrt(h^2 + d^2), while the square of the distancee is h^2 + d^2, where h is the height of the light source above the surface of the table.
That makes the illumination proportional to h / (h^2 + d^2)^(3/2).
Elevation relative proportional
to d illumination
0 0
.05 .04981308423330897
.1 .09851853368415735
.15 .1450762375845913
.2 .1885732068636385
.25 .2282688235636075
.3 .2636219133636197
.35 .2942988193568874
.4 .3201643761673307
.45 .3412596069262352
.5 .3577708763999664
.55 .3699952013605803
.6 .3783057025202401
.65 .3831201058667234
.7 .3848740566196834
.75 .384
.8 .3809116143624538
.85 .3759932772688188
.9 .3695937743770441
.95 .3620233702573946
In finer detail:
.70706 .3849001783365646
.70707 .3849001787654365
.70708 .3849001790916561
.70709 .3849001793152265
.70710 .3849001794361511
.70711 .3849001794544334
.70712 .3849001793700768
.70713 .3849001791830846
.70714 .3849001788934601
.70715 .384900178501207
And, cutting to the chase, at sqrt(2)/2 times d:
.7071067811865476 .3849001794597505
corrected sqrt(2) to sqrt(2)/2
Edited on July 6, 2016, 8:38 pm
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Posted by Charlie
on 2016-07-06 14:52:48 |
my solution
