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?
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
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
The Illuminati (L) received by an object
Well, my friend, if you got 2 as your answer, then I must say the light must be feeling quite shy because it needs a little more distance to shine brighter.
Let's solve this math problem together. The intensity of illumination (I) varies inversely as the square of the distance (d). This means that if we double the distance, the intensity will decrease by a factor of four.
In this case, the initial distance (d1) is 12 meters and the initial intensity (I1) is 2. We want to find a new distance (d2) where the intensity (I2) is twice as much.
Using the inverse square law, we can set up the equation: I1 × d1^2 = I2 × d2^2.
Plugging in the values we have: 2 × 12^2 = 4 × d2^2.
Simplifying this equation, we get 288 = 4 × d2^2.
Dividing both sides by 4, we get 72 = d2^2.
Taking the square root of both sides, we find that the new distance (d2) is approximately 8.5 meters.
So, my friend, the surface must be approximately 8.5 meters away from the light source to receive twice as much intensity of illumination. Keep shining!
To solve this problem, we need to use the inverse square relationship between intensity and distance. The equation for the relationship is:
Intensity ∝ 1/distance²
where "∝" means "is directly proportional to."
Let's represent the initial distance as d1 and the initial intensity as I1. We are given that d1 = 12 meters and I1 = 2.
So, we have:
I1 = k/12²
where k is the constant of proportionality.
To find the constant k, we can rearrange the equation and solve for k:
k = I1 * 12²
= 2 * 144
= 288
Now we can rewrite the equation as:
Intensity = 288/distance²
To find the new distance, let's represent it as d2, and the desired intensity as I2. We want to find d2, when I2 is twice the initial intensity, I1.
I2 = 2 * I1
= 2 * 2
= 4
Now we can substitute I2 = 4 into the equation and solve for d2:
4 = 288/d2²
Multiply both sides by d2²:
4 * d2² = 288
Divide both sides by 4 to isolate d2²:
d2² = 288/4
= 72
To find d2, we take the square root of both sides:
√(d2²) = √72
d2 = ±√72
Since distance cannot be negative in this context, we only consider the positive square root:
d2 ≈ 8.5 meters
Therefore, the surface must be approximately 8.5 meters from the light source to receive twice as much intensity of illumination.