The light source located 2.75 m from a surface produces an illumination of 528 lux on that surface. Find the illumination if the distance is changed to 1.55 m.
Well, let's shed some light on this! If we have a light source that produces an illumination of 528 lux at a distance of 2.75 m, and we want to calculate the illumination at a distance of 1.55 m, we can use the inverse square law for illumination.
According to the inverse square law, the illumination is inversely proportional to the square of the distance. So, we can set up a proportion:
(illumination 1)/(distance 1)^2 = (illumination 2)/(distance 2)^2
Substituting the given values, we get:
528 lux /(2.75 m)^2 = x /(1.55 m)^2
Now, let's solve for x, the illumination at a distance of 1.55 m:
x = 528 lux * [(1.55 m)^2 / (2.75 m)^2]
Calculating this, we find that x ≈ 528 lux * (1.55^2 / 2.75^2) ≈ 528 lux * 0.501 ≈ 264.2 lux.
So, the illumination at a distance of 1.55 m would be approximately 264.2 lux. Just remember, when it comes to lighting equations, always keep your distance!
To find the new illumination when the distance is changed to 1.55 m, we can use the inverse square law for light intensity. According to the inverse square law, the intensity of light is inversely proportional to the square of the distance from the source.
The formula for calculating light intensity based on distance is:
I1 / I2 = (D2^2 / D1^2)
Where:
I1 = initial intensity of light
I2 = new intensity of light
D1 = initial distance
D2 = new distance
Given:
I1 = 528 lux
D1 = 2.75 m
D2 = 1.55 m
Plugging these values into the formula, we can solve for I2:
528 / I2 = (1.55^2 / 2.75^2)
Simplifying the equation:
I2 = 528 * (2.75^2 / 1.55^2)
I2 = 528 * (7.5625 / 2.4025)
I2 = 528 * 3.148148
I2 ≈ 1660.02 lux
Therefore, the new illumination when the distance is changed to 1.55 m is approximately 1660.02 lux.
To find the illumination when the distance is changed from 2.75 m to 1.55 m, we can use the inverse square law for light intensity.
The inverse square law states that the intensity of light is inversely proportional to the square of the distance from the source.
Mathematically, this relationship can be expressed as:
I1 / I2 = (D2)^2 / (D1)^2
where I1 and I2 are the initial and final illuminations, and D1 and D2 are the initial and final distances.
We are given I1 = 528 lux and D1 = 2.75 m. Let's use this information to find I2.
(I1 / I2) = (D2)^2 / (D1)^2
Substituting the given values:
(528 lux / I2) = (1.55 m)^2 / (2.75 m)^2
To find I2, we can rearrange the equation:
I2 = (528 lux * (2.75 m)^2) / (1.55 m)^2
Calculating the equation yields:
I2 ≈ 906.08 lux
Therefore, the illumination when the distance is changed to 1.55 m is approximately 906.08 lux.