Question

The intensity, \(I\), of light, in watts per square metre \(\left(\mathrm{W} / \mathrm{m}^{2}\right),\) at a distance, \(d,\) in metres, from the point source is given by the formula \(I=\frac{P}{4 \pi d^{2}},\) where \(P\) is the average power of the source, in watts. How far away from a 500 -W light source is intensity \(5 \mathrm{W} / \mathrm{m}^{2} ?\)

Short answer

d ≈ 1.414 meters.

Step-by-step solution

Identify the given variables

Write down the values provided in the problem. The power of the light source, P, is 500 W. The intensity of light, I, is 5 W/m^2. We need to find the distance, d.

Substitute given values into the formula

Start with the formula for intensity: \( I = \frac{P}{4 \pi d^2} \). Substitute the known values:\( 5 = \frac{500}{4 \pi d^2} \).

Isolate the term with d

Rearrange the equation to solve for d:\( 5 = \frac{500}{4 \pi d^2} \). First, multiply both sides by 4πd² to eliminate the fraction:\( 5 \cdot 4 \pi d^2 = 500 \).

Simplify the equation

Simplify the expression on the left side:\( 20 \pi d^2 = 500 \). Divide both sides by 20π to solve for d²:\( d^2 = \frac{500}{20 \pi} \).

Calculate the value of d

Compute the value:\( d^2 = \frac{500}{20 \pi} = \frac{25}{\pi} \). Take the square root of both sides to find d:\( d = \sqrt{\frac{25}{\pi}} \).

Final computation

Calculate the numerical value of d:\( d = \sqrt{\frac{25}{\pi}} \). Using a calculator,\( d \approx 1.414 \; \text{meters} \).

Key concepts

inverse square law

The inverse square law is a fundamental principle in physics that describes how a certain physical quantity decreases with distance. It applies to various phenomena, including light, sound, and gravitational forces.
For light intensity, this law tells us that the intensity of light diminishes in proportion to the square of the distance from the light source.
Mathematically, the law is expressed as: \[ I = \frac{P}{4 \pi \ d^2} \]
where:

• I represents the light intensity
• P is the power of the light source
• d is the distance from the source

Understanding this relationship helps us grasp why objects appear dimmer as they move further from a light source. This can be observed in daily life - the farther you are from a lamp, the less light you receive.

light source power

The light source power, denoted as P in the formula, is the amount of energy emitted by the light source per second, measured in watts (W).
It represents the total energy output of the light source.
In practical terms, the power of a light source can range from small (like a flashlight) to very large (such as industrial lamps or the sun).
For this exercise, the light source has a power of 500 W.
More power generally means more brightness, but the light's intensity at a particular point also depends on distance, as governed by the inverse square law.
For instance, a 100-W light bulb is less powerful and illuminates less area compared to a 500-W bulb.
Understanding these power values helps in calculating and predicting how much illumination an area will receive from a specific light source.

distance calculation

To find the distance from a light source where the intensity is a given value, we use the light intensity formula, which is derived from the inverse square law.
Here's how you calculate it step-by-step:

• Write down the given variables. In this exercise, the intensity I is 5 W/m², and the power P is 500 W.
• Substitute these values into the formula: \[ I = \frac{P}{4 \pi d^2} \] so \[ 5 = \frac{500}{4 \pi d^2} \]
• Rearrange the formula to isolate d: \[5 = \frac{500}{4 \pi d^2} \]
  Multiply both sides by 4πd²:\[5 \cdot 4 \pi d^2 = 500 \]
• Simplify and solve for d²:\[ 20 \pi d^2 = 500 \] Divide both sides by 20π:\[ d^2 = \frac{500}{20 \pi} \]
• Calculate the value of d² and take the square root: \[ d^2 = \frac{25}{\pi} \] \[d = \sqrt{\frac{25}{\pi}} \approx 1.414 \text{ meters} \]

By following these steps, we determine that the required distance from the light source to achieve an intensity of 5 W/m² is approximately 1.414 meters.
This method can be used for any set of values, as long as you understand the relationship between intensity, power, and distance.