Problem 19
Question
The inverse square law states that for a surface illuminated by a light source, the intensity of illumination on the surface is inversely proportional to the square of the distance between the source and the surface. A certain light source produces an illumination of 800 lux (a lux is 1 lumen per square meter) on a surface. Find the illumination on that surface if the distance to the light source is doubled.
Step-by-Step Solution
Verified Answer
The illumination on the surface when the distance to the light source is doubled will be 200 lux.
1Step 1: Understand the Inverse Square Law
The Inverse Square Law in this context states that the illumination intensity (I) is inversely proportional to the square of the distance (d) from the light source. Mathematically, this can be expressed as: \( I \propto \frac{1}{d^2} \).
2Step 2: Calculate Initial Intensity
The illumination on the surface at the initial distance (we can call it \(d_1\)) is already given as 800 lux. We can use this value later to find the new illumination.
3Step 3: Determine the New Distance
If the distance is doubled, the new distance \(d_2\) will be \(d_2 = 2d_1\). This means that the distance to the light source is now twice the initial distance.
4Step 4: Apply the Inverse Square Law to Find New Illumination
Using the inverse square law: \( I_1 \propto \frac{1}{d_1^2} \) and \( I_2 \propto \frac{1}{d_2^2} \). By the proportionality we can say that \( \frac{I_1}{I_2} = \frac{d_2^2}{d_1^2} \). Plugging in \( d_2 = 2d_1 \), we get \( \frac{I_1}{I_2} = (2d_1)^2 / d_1^2 = 4 \). Using the initial intensity \( I_1 = 800 \) lux, we solve for \( I_2 \) by rearranging to \( I_2 = \frac{I_1}{4} \).
5Step 5: Calculate the New Illumination Intensity
Substitute the value of the initial illumination into the equation to get \( I_2 = \frac{800 \text{ lux}}{4} = 200 \text{ lux} \), which is the intensity of illumination on the surface when the distance to the light source is doubled.
Key Concepts
Illumination IntensityProportionality in PhysicsLux (Unit of Illumination)
Illumination Intensity
Illumination intensity is a measure of the amount of light that falls on a surface per unit area. In practical terms, it describes how brightly a surface is lit by a light source. The higher the illumination intensity, the brighter the surface appears.
To understand how this intensity changes with distance, it is essential to grasp the inverse square law. This law reveals that if you move away from a light source, the illumination intensity decreases rapidly. In fact, doubling the distance from the light source results in only one-quarter of the original illumination intensity, because the light spreads out over a larger area.
For any student grappling with this concept, it's helpful to picture illumination intensity as rain falling on a sidewalk. If you have a certain amount of rain (light) falling on a small section of the sidewalk (surface area), it will be wetter (brighter) than if the same amount of rain was spread over a broader section. This visual can make the concept of illumination intensity much more relatable and easier to remember when solving related physics problems.
To understand how this intensity changes with distance, it is essential to grasp the inverse square law. This law reveals that if you move away from a light source, the illumination intensity decreases rapidly. In fact, doubling the distance from the light source results in only one-quarter of the original illumination intensity, because the light spreads out over a larger area.
For any student grappling with this concept, it's helpful to picture illumination intensity as rain falling on a sidewalk. If you have a certain amount of rain (light) falling on a small section of the sidewalk (surface area), it will be wetter (brighter) than if the same amount of rain was spread over a broader section. This visual can make the concept of illumination intensity much more relatable and easier to remember when solving related physics problems.
Proportionality in Physics
Proportionality is a foundational concept in physics that encapsulates the relationship between two quantities. When two variables are proportional, as one changes, the other changes in a predictable manner. In the case of the inverse square law, the illumination intensity is inversely proportional to the square of the distance from the light source.
This proportionality is depicted mathematically as: \( I \propto \frac{1}{d^2} \) where \(I\) is the illumination intensity and \(d\) is the distance. If you double the distance (\(d_2 = 2d_1\)), according to the inverse square law, the new intensity will be one-fourth, because \( (2d_1)^2 = 4d_1^2 \) as highlighted in the problem's solution.
Understanding proportionality helps students see the 'big picture' of how different physics concepts are interrelated. It encourages a deeper comprehension of equations and fosters an ability to predict changes in one quantity based on changes in another. This core principle aids not just in physics, but in other areas of science and mathematics as well.
This proportionality is depicted mathematically as: \( I \propto \frac{1}{d^2} \) where \(I\) is the illumination intensity and \(d\) is the distance. If you double the distance (\(d_2 = 2d_1\)), according to the inverse square law, the new intensity will be one-fourth, because \( (2d_1)^2 = 4d_1^2 \) as highlighted in the problem's solution.
Understanding proportionality helps students see the 'big picture' of how different physics concepts are interrelated. It encourages a deeper comprehension of equations and fosters an ability to predict changes in one quantity based on changes in another. This core principle aids not just in physics, but in other areas of science and mathematics as well.
Lux (Unit of Illumination)
Lux is the unit of measurement for illumination intensity. It quantifies how much luminous flux is spread over a given area. One lux is equal to one lumen per square meter. A lumen measures the total amount of visible light emitted by a source, so lux tells us how that light is distributed across a surface.
In the context of the example problem, an illumination intensity of 800 lux means that the light source is emitting enough lumens to provide 800 units of light for every square meter of the surface. When the distance is doubled, the light has to illuminate four times the area (since the area of a circle increases with the square of the radius), resulting in a quarter of the initial illumination intensity, or 200 lux, consistent with the inverse square law.
The concept of lux as a unit may seem abstract, but if you imagine a standard light bulb overhead: the closer you are to the bulb, the brighter and more concentrated the light—this is a high lux level. As you move away, light covers more area but becomes less intense, demonstrating lower lux. This real-world application helps solidify the understanding of the lux unit in practice.
In the context of the example problem, an illumination intensity of 800 lux means that the light source is emitting enough lumens to provide 800 units of light for every square meter of the surface. When the distance is doubled, the light has to illuminate four times the area (since the area of a circle increases with the square of the radius), resulting in a quarter of the initial illumination intensity, or 200 lux, consistent with the inverse square law.
The concept of lux as a unit may seem abstract, but if you imagine a standard light bulb overhead: the closer you are to the bulb, the brighter and more concentrated the light—this is a high lux level. As you move away, light covers more area but becomes less intense, demonstrating lower lux. This real-world application helps solidify the understanding of the lux unit in practice.
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