Streetlamps in a line

Geometry Level 3

There are 5 streetlamps arranged in a line from west to east, spaced 1 meter apart, that radiate the same amount of light. You are 1 meter south of the westernmost streetlamp.

How many times more light do you receive from the line of streetlamps combined than just from the westernmost streetlamp, to 3 decimal places?


The answer is 1.859.

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1 solution

Jake Castillejos
Apr 2, 2018

Let us label each streetlamp Lamp A to E from west to east. You receive x x light from Lamp A, and through the Pythagorean theorem and the inverse square law, you find that you receive x 1 + 1 2 = x 2 \frac{x}{1+1^{2}}=\frac{x}{2} light from Lamp B, x 1 + 2 2 = x 5 \frac{x}{1+2^{2}}=\frac{x}{5} light from Lamp C, x 1 + 3 2 = x 10 \frac{x}{1+3^{2}}=\frac{x}{10} light from Lamp D, and x 1 + 4 2 = x 17 \frac{x}{1+4^{2}}=\frac{x}{17} light from Lamp E. Thus, you receive 158 x 85 \frac{158x}{85} light from the entire line of lamps. Knowing that you receive x x light from Lamp A (the westernmost streetlamp), all that remains is to divide the last two values to obtain a value of 158 85 \frac{158}{85} , or, to 3 decimal places, 1.859 \boxed{1.859} .

why do you receive x/(1+1) light from B. Shouldnt it be radial so that the distance from B is sqrt(2)...and so on in square..it should have been 2.713

Ols Muka - 3 years, 1 month ago

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The distance from B is indeed the square root of 2, so through the inverse square law you receive x/[sqrt(2)^2] or x/2 units of light from Lamp B. The same process follows with the rest of the lamps, where the square root in the distance and the square in the inverse square law cancel out.

Jake Castillejos - 3 years, 1 month ago

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