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Chapter 4: Problem 41

The intensity of illumination at any point from a light source is proportional to the square of the reciprocal of the distance between the point and the light source. Two lights, one having an intensity eight times that of the other, are 6 m apart. How far from the stronger light is the total illumination least?

Short Answer

Expert verified
The minimum illumination occurs 2 meters from the stronger light.

Step by step solution

01

Understanding the Problem

The intensity of illumination at a point is inversely proportional to the square of the distance from a light source. If we denote the intensity of light 1 as \( I_1 \) and light 2 as \( I_2 \), where \( I_1 = 8I_2 \), we need to find the point on the line between the lights where the total illumination is least.
02

Modeling Equations

Let's place light 1 at point 0 and light 2 at point 6 on a 1D line. Let \( x \) be the distance from light 1 to the point P where we want to find minimum intensity. The intensity at point P due to light 1 is \( \frac{I_1}{x^2} \), and due to light 2 is \( \frac{I_2}{(6-x)^2} \). Total intensity \( T \) is given by the sum of these intensities, so \( T = \frac{8I_2}{x^2} + \frac{I_2}{(6-x)^2} \).
03

Simplifying the Expression

We can simplify the total intensity expression by factoring out \( I_2 \), since it's a common term: \( T = I_2 \left( \frac{8}{x^2} + \frac{1}{(6-x)^2} \right) \). To minimize \( T \), we need to minimize the expression inside the brackets: \( \frac{8}{x^2} + \frac{1}{(6-x)^2} \).
04

Finding the Minimum Point by Differentiation

Take the derivative of \( f(x) = \frac{8}{x^2} + \frac{1}{(6-x)^2} \) with respect to \( x \) and set it equal to zero. Differentiating gives \( f'(x) = -\frac{16}{x^3} - \frac{2}{(6-x)^3} \). Setting \( f'(x) = 0 \) leads to solving \( \frac{16}{x^3} = \frac{2}{(6-x)^3} \).
05

Solving the Equation

Simplify the equality \( \frac{16}{x^3} = \frac{2}{(6-x)^3} \) to get \( 8(6-x)^3 = x^3 \). Expanding \( (6-x)^3 \) gives \( 216 - 108x + 18x^2 - x^3 \). Solve the equation \( x^3 = 8(216 - 108x + 18x^2 - x^3) \).
06

Final Solution for x

Solving the equation \( x^3 = 8(216 - 108x + 18x^2 - x^3) \) results in a polynomial equation. Simplifying and solving this gives the value of \( x \). After simplification, we find that at \( x = 2 \), the total illumination is minimized.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Intensity of Illumination
The intensity of illumination refers to the brightness of light that reaches a certain point from a source. It's crucial in various applications, such as photography, lighting design, and even plant growth. In physics, intensity is typically expressed as power per unit area. When talking about a point in relation to a light source, the intensity diminishes as you move further away. This is due to the Inverse Square Law.

According to the Inverse Square Law, the intensity is inversely proportional to the square of the distance from the light source. This mathematical relationship can be expressed as:
  • If the distance from the light source doubles, the area covered is four times larger, meaning the intensity is a quarter of what it was.
  • Likewise, if you halve the distance, the area shrinks, and the intensity is four times stronger.
This concept helps explain why light from a nearby lamp seems much brighter than when observed from afar. By understanding how intensity decreases with distance, one can effectively harness or shield light for desired effects in practical applications.
Proportional Relationships
Proportional relationships are fundamental in understanding how different quantities relate to each other. In the context of illumination, the intensity of light is directly influenced by distance, following an inversely proportional relationship.

This means that as one value increases, the other decreases, and vice versa. Specifically for illumination:
  • The intensity is proportional to the reciprocal of the square of the distance.
  • Effectively, this means that small changes in distance can lead to significant changes in the perceived brightness of light.
This inverse proportionality helps in calculating the needed brightness levels or adjusting setups in lighting design to achieve desired illuminance. Having a firm grasp of proportional relationships helps simplify complex real-world problems into manageable equations, allowing accurate predictions of outcomes.
Differentiation and Derivatives
Differentiation and derivatives are key concepts in calculus that aid in understanding and analyzing changes in functions. They are extremely useful in determining minimum or maximum values of functions, which is a common requirement in physics and engineering problems.

When dealing with light intensity, differentiation allows us to find the minimum point of total illumination from two lights. By taking the derivative of the intensity function with respect to distance, setting it to zero, and solving for the distance, we can identify where the total intensity is minimized.

In this example:
  • The derivative of the intensity function reveals the rate of change of illumination intensity over distance.
  • By solving the derivative equation, one finds critical points where the intensity might be at an extremum (maximum or minimum).
  • Verifying through second derivative tests confirms whether these points are indeed the minimum.
Using differentiation not only aids in solving for optimal intensity levels but also helps understand how changes in distance affect lighting dynamics effectively.

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