A question regarding on illuminance and photometry...?

Question

A 100 candle and a 64 candle source are 2m apart. At what point between them will the illuminance be the same?

Answer 1

The intensity of a light source drops off as the distance squared. If you double the distance from the light source, the intensity drops to 1/4 its former value.

If we have the following setup:

100candle<------------2 meters--------------->64candle

We need to solve the following equation:

100 / x^2 = 64 / (2 - x)^2

where x is the distance from the 100-candle source, and (2 - x) is the distance from the 64-candle source. Solving, we find...

100 / x^2 = 64 / (2 - x)^2
100*(2 - x)^2 = 64*x^2
100*(x^2 - 4x + 4) = 64*x^2
100*x^2 - 400*x + 400 = 64*x^2
36*x^2 - 400*x + 400 = 0

From here we use the quadratic formula, resulting in
x = 1.111
or
x = 10.00

The second answer places us outside the scene entirely, but it is a valid answer. If you were 10 meters to the right of the 100-candle source, and 8 meters to the right of the 64-candle source, each would have equal intensity. However, we're interested in the first answer.

So, 1.111 meters away from the brighter source, and 0.889 meters from the dimmer source gives you equal intensity.

Answer 2

Since the illuminance of a candle falls off as you go away from it by the distance squared, the ratio of the distances must be sqrt(64)/sqrt(100) or 8:10 or 4:5. The total distance is 2 m, so 8/9 meters from the small candle and 10/9 meters from the large candle.

Answer 3

using cos cube rule (dalumbert's law)
E=i/h(cosx)^3
E1=E2
i1/h(cosx1)^3=i2/h(cosx2)^3
solving third degree equation
distance from 100 cd source x=1.292
and from 64 cd distance =2-1.292=0.708

Answer 4

(I2 /I1) = (d2 /d1) ^2

(d2 /d1) = √ (I2 /I1) = √ (64/100) = 8/10.

d2 / (d1 +d2) = 8/ 18 = 4/9

d2 = 8/9 = 0.88 m Since (d1 +d2) = 2

d1 = 1.11m