Math

The intensity of Illumination on a surface varies inversely as the square of the distance from the light source. A surface is 12 meters from a light source and has an intensity of 2. How far must the surface be from the light source to receive twice as much intensity of illumination? Approximate to the nearest tenth of meter. When I solved it, I got 2, is it right?

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  1. no

    2 is 1/6 of 12 , so the intensity would be 36 times 2

    I * d^2 = k ... 2 * 12^2 = 288

    for twice the intensity ... (2 * 2) * d^2 = 288 ... d^2 = 72

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  2. L1 = intensity of Illumination 1

    L2 = intensity of Illumination 2

    d1 = distance 1

    d2 = distance 2

    The inverse-square law:

    L2 / L 1 = ( d1 / d2 )²

    You know L2 / L1 = 2 and d1 = 12

    so:

    2 = (12 / d2 )²

    Take square root of both sides

    √2 = 12 / d2

    Multiply both sides by d2

    √2 ∙ d2 = 12

    Divide both sides by √2

    d2 = 12 / √2 = 6 ∙ 2 / √2 =

    6 ∙ √2 ∙ √2 / √2 = 6 ∙ √2 =

    6 ∙ 1.41421 = 8.48526 m

    d2 = 8.5 m rounded to the nearest tenth of meter

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