Inverse Square Law

Conference rooms have their own set of issues, ranging from poor acoustics to inadequate connectivity. While there are several methods to enhance your conference experience, one method that is sometimes ignored is microphone positioning. The inverse square law is a useful idea to understand the relevance of distance to and between microphones. Any science law stating that a given physical quantity is inversely proportional to the square of its distance from its source is called inverse square law. This article covers the fundamentals of the inverse square law and Coulomb's inverse square law.

Inverse Square Law Definition

If you ask a physicist to define the inverse square law, they would answer something like "the intensity of a force changes in inverse proportion to the square of the distance from that force." This definition may be a little perplexing, however, this physical law is not as difficult as it appears. In simple words, the inverse square rule asserts that when energy or force flows away from a point source, its strength decreases dramatically. The inverse square law applies to any source that expands out in all directions without becoming lost. The following energy and forces can be attributed to this "source".

• Sound

• Gravitation

• Electrostatics

• Light

• Radiation

Let's take the case of light. Consider the following scenario: you're driving down a dark highway with your headlights turned on. As they go, the light beams from the headlights scatter. As a result, the light beams get increasingly spread out as they travel further away from the headlights. As a result, the light beams' power weakens and shallows.

Inverse Square Law of Light

This rule describes the power of light in relation to its distance from the source. According to the inverse square law, the intensity of radiation is inversely proportional to the square of the distance. Or the intensity of light emitted by a source to an observer is inversely proportional to the square of the distance between the observer and the source.

Light loses brightness or luminosity as it goes away from the source, according to this rule. For example, if you turn on a light in one corner of the room and then move away from the source, the light seems dull or less brilliant owing to the increase in distance (away from the source). The diagram below depicts the source of light and the source distance relationship.

(Image Wil be Updated Soon)

Consider the distance to be 3, and the distance squared is 9. Therefore the inverse of 9 is $$\frac{1}{9}$$. When we compare the results, we can see that the outcome is one-quarter the original power. As a result, the light intensity is $$\frac{1}{d^2}$$. When the distance between an object and the light source is doubled, one-fourth of the light from the source is experienced in that region. When the distance between the source and the specified area is quadrupled, the given region receives one-ninth of the light. In a similar way, we can obtain the inverse square law of radiation or inverse square law in radiology by taking an X-ray as the source of light. Some of the inverse square law examples are the universal law of gravitational, electric field, electric force, the intensity of sound, etc.

Inverse Square Law Formula and Its Applications

The inverse square law formula is as follows:

$$I\alpha \frac{1}{d^2}$$

Here, d is the distance and I is the intensity of radiation.

Consider two light sources with intensities $$I_{1}and{I}_2$$ separated by distances $$d_{1}and{d}_2$$. Using the inverse-square law we can write:

$$\frac{I_1}{I_2}\alpha {\left(\frac{d_2}{d_1}\right)}^2$$

Where the distance is measured in metres and the intensity of light is measured in candelas or Lumens.

Applications of the Formula:

• The intensity of any given radiation or distance is calculated using this law.

• In X-ray procedures, the inverse square law is used to compute the distance between the source and the film.

• It also aids in determining the duration of x-ray exposure as well as the x-ray tube's intensity.

• When the brightness of the source is known, the traditional candle technique may be used to compute the distance from the Earth.

• Various astronomical distances are measured using the inverse square law.

Inverse Square Law Graph

A graph depicting the decrease in light intensity caused by the inverse square law demonstrates the inverse square law rapid loss. In this plot, the places are connected by straight lines, although the actual fall is a smooth curve between the points.

(Image Will be Updated Soon)

Coulomb's Inverse Square Law

Coulomb is the basic unit of electric charge. The force between two charges separated by a distance is determined by Coulomb's law. It has been discovered that every system's electric charge is always an integer multiple of the least amount of charge e, where e is the charge of the proton or electron. According to this law, the attraction or repulsion force between two point charges is proportional to the product of charges and inversely proportional to the square of the distance between them. Say two charges $$q_{1}and{q}_2$$ are separated by the distance r than from this law we can write,

$$F\alpha \frac{q_1{q}_2}{r^2}$$

We may write after removing the proportionality indication,

$$F=\frac{(K{q}_1{q}_2)}{r^2}$$

Here k is the proportionality constant and it is given as $$k=\frac{1}{4\pi {ϵ}_0}$$. Here $${ϵ}_0$$ is the permittivity in free space and its value is$$8.854\times {10}^{-12}\frac{C^2}{N{m}^2}$$.

Solved Example on Inverse Square Law

1. What is the intensity of the flashlight 50m from the lens, if a bright flashlight has a light intensity of 10 Candela at a distance of 2m from the lens?

Ans: To solve this question we have to apply the inverse square law equation according to which intensity of the flashlight is inversely proportional to the square of the distance and this can be expressed mathematically as,

$$I\alpha \frac{1}{d^2}\dots \dots .(1)$$

Here, I is the intensity of the flashlight and d is the distance.

In case of different values of intensity for different distances, the equation (1) is written as,

$$\frac{I_1}{I_2}={\left(\frac{d_2}{d_1}\right)}^2$$……….(2)

According to the question$$I_1=10candela,d_1=2m,d_2=50m,and{I}_2$$ we have to determine. So putting these values in equation (2). We can get, 

$$\frac{10}{I_2}={\left(\frac{50m}{2m}\right)}^2$$

So $$I_2$$ is written as,

$$I_2=(\frac{2}{50}{)}^2\times 10candela$$

After solving we get,

$$I_2=(\frac{1}{25}{)}^2\times 10=\frac{10}{625}=0.016candela$$

Hence the intensity of the flashlight is 0.016 candela when the distance is 50m from the lens.