Problem 16 - Algebra and Trigonometry with Analytic Geometry

Question

The intensity of illumination \(I\) from a source of light varies inversely as the square of the distance \(d\) from the source. (a) Express \(I\) in terms of \(d\) and a constant of variation \(k\). (b) A searchlight has an intensity of \(1,000,000\) candlepower at a distance of 50 feet. Find the value of \(k\) in part (a). (c) Sketch a graph of the relationship between \(I\) and \(d\) for \(d>0 .\) (d) Approximate the intensity of the searchlight in part (b) at a distance of 1 mile.

Step-by-Step Solution

Answer

(a) \(I = \frac{k}{d^2}\); (b) \(k = 2,500,000,000\); (d) \(I \approx 89.5\) candlepower at 1 mile.

Step 1: Understanding Inverse Variation

The problem states that the intensity of illumination \(I\) varies inversely as the square of the distance \(d\). This means as \(d\) increases, \(I\) decreases proportionally to the square of \(d\). Mathematically, this is expressed as \(I = \frac{k}{d^2}\), where \(k\) is a constant.

Step 2: Express I in terms of d and k

Given the inverse square law, the relationship is given by \(I = \frac{k}{d^2}\). This equation means that the intensity \(I\) will decrease as the distance \(d\) increases since \(d^2\) is in the denominator.

Step 3: Substitute Known Values to Find k

We're given that at \(d = 50\) feet, \(I = 1,000,000\) candlepower. Substitute these values into the equation \(I = \frac{k}{d^2}\) to find \(k\).\[1,000,000 = \frac{k}{50^2}\]Solve for \(k\):\[k = 1,000,000 \times 50^2 = 2,500,000,000\]

Step 4: Graph the Relationship

This graph will show \(I\) on the y-axis and \(d\) on the x-axis. The equation \(I = \frac{2,500,000,000}{d^2}\) represents a hyperbola in the first quadrant since \(I\) is always positive as long as \(d > 0\). As \(d\) becomes larger, \(I\) approaches zero.

Step 5: Calculate Intensity at 1 Mile

We are asked to find the intensity \(I\) at \(d = 1\) mile. Since there are 5280 feet in a mile, \(d = 5280\) feet. Substitute \(d = 5280\) into the formula \(I = \frac{2,500,000,000}{d^2}\):\[I = \frac{2,500,000,000}{5280^2}\]\[I \approx 89.5\]

Key Concepts

intensity of illumination

The intensity of illumination, often denoted by the symbol \( I \), describes how bright a light appears from a particular distance. As an important parameter in lighting, it measures how much visible light is emitted from a source and hits a surface.
The unit for intensity is often measured in candlepower in practical problems, but it may also be referred to in lumens or lux depending on the context. In the given problem, the intensity of the searchlight is given as 1,000,000 candlepower.
Understanding the characteristics of light intensity can help us determine how light behaves over varying distances, which becomes even more fascinating due to the concept of inverse variation.

inverse square law

The inverse square law is a fundamental principle that explains how certain quantities diminish with increasing distance. Specifically for light, it states that the intensity \( I \) inversely varies with the square of the distance \( d \) from the source. This is expressed in the formula:

• \( I = \frac{k}{d^2} \)

where \( k \) is a constant. This means as you move further away from the light source (increase \( d \)), the intensity sharply decreases.
The term 'inverse square' suggests that if you double the distance from a light source, the intensity becomes one-fourth; if you triple the distance, it drops to one-ninth, and so forth. This is because \( d \) is squared in the denominator of the equation.
This law is crucial in various fields like physics, photography, astronomy, and even sound.

constant of variation

The constant of variation \( k \) in the equation \( I = \frac{k}{d^2} \) is crucial because it links light intensity to distance in a specific scenario. Without it, the relationship between the two would be undefined.
For a specific light source, \( k \) must be computed using known values of intensity and distance. In our example, where the intensity is 1,000,000 candlepower at 50 feet, we utilize these values in the formula to solve for \( k \). This reveals that \( k = 2,500,000,000 \).
It's this constant that allows us to predict how bright the light will be at different distances, making it vital for designing lighting systems or evaluating how light travels in its environment.

distance and illumination relationship

The relationship between distance and illumination is elegantly represented in our equation: \( I = \frac{k}{d^2} \). In essence, as the distance \( d \) from the light source increases, the illumination \( I \) falls rapidly.
The graph for this relationship is not linear; instead, it forms a hyperbola, highlighting the steep drop in intensity as distance grows. For instance, as mentioned in the problem, at a distance of one mile (or 5280 feet), the searchlight's brilliance drastically diminishes from 1,000,000 candlepower to approximately 89.5 candlepower.
This stark change exemplifies why it's crucial to understand the distance-illumination relationship, especially in practical scenarios like theatre lighting, road safety, and designing outdoor illumination systems. Comprehending this principle helps in optimizing lighting setups for efficiency and effectiveness.