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The intensity of illumination at any point from a light source is proportional to the square of the reciprocal of the distance between the point and the light source. Two​ lights, one having an intensity nine times that of the​ other, are 11 m apart. On the line between the two light​ sources, how far from the stronger light is the total illumination​ least?

Respuesta :

Answer:

the stronger light 5.5 m apart from the total illumination​

Explanation:

From the problem's statement , the following equation can be deducted:

I= k/r²

where I = intensity of illumination , r= distance between the point and the light source , k = constant of proportionality

denoting 1 as the stronger light and 2 as the weaker light

I₁= k/r₁²

I₂= k/r₂²

dividing both equations

I₂/I₁ = r₁²/r₂²=(r₁/r₂)²

solving for r₁

r₁ = r₂ * √(I₂/I₁)

since we are on the line between the two light​ sources , the distance from the light source to the weaker light is he distance from the light source to the stronger light + distance between the lights . Thus

r₂ = r₁ + d

then

r₁ = (r₁ + d)* √(I₂/I₁)

r₁ = r₁*√(I₂/I₁) + d*√(I₂/I₁)

r₁*(1-√(I₂/I₁)) =  d*√(I₂/I₁)

r₁ = d*√(I₂/I₁)/(1-√(I₂/I₁))  =

r₁ = d/[√(I₁/I₂)-1)]

since the stronger light is 9 times more intense than the weaker

I₁= 9*I₂ → I₁/I₂ = 9 →√(I₁/I₂)= 3

then since d=11 m

r₁ = d/[√(I₁/I₂)-1)] = 11 m / (3-1) = 5.5 m

r₁ = 5.5 m

therefore the stronger light 5.5 m apart from the total illumination​

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