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Inverse square law
The Inverse Square Law is a mathematical concept that gives the relationship between intensity and the distance from an energy source.
\({intensity} \ \propto \ \frac{1}{{distance}^2} \)
This is relevant to the energy of wave phenomena whether it be sound, light waves or other forms of radiation. The video will introduce you to the concept and the formula.
Inverse Square Law
Let's talk about the inverse Square law.
In maths it relates to gravity, electricity, the intensity of sound, light and other types of radiation.
Let's look at an example.
Rita runs a deli where she spends a lot of time buttering bread for sandwiches. She needed a way to do this more efficiently, so she bought a butter gun. It was great.
Rita could spray a single serve of butter with a pull of a trigger, but it was still a lot of work. After a few attempts Rita noticed the pattern.
Rita could spray a set amount of butter onto one slice at a certain distance. At twice the distance she could butter four slices with a thickness of one quarter of the original amount.
At three times the distance she could butter nine slices with a thickness of one ninth the original amount.
Let's explain this pattern mathematically. If 'i' is equal to the number of serves of butter per slice, when we're looking at a distance of one 'd' away from the butter gun, the 'i' equals one serve per slice. When we're looking at twice the distance we're looking at 'i' equals one quarter.
One quarter is the same as two one on two squared.
If we're looking at three times the distance we're looking at 'i' equals one ninth. Which equals one on three squared.
Now when you're having a look at this, let's have a look at a pattern. If I also included this in the pattern we can see that intensity is directly proportional to one on 'd' squared.
This translates to the thickness of the butter is inversely proportional to the square of the distance from the butter gun.
This is called the inverse Square law and is a universally recognised rule
Questions
Use simple division and the information from the video to solve the following equations.
Sound
The intensity of sound is measured in Watts per square metre - \( W/m^{2}\)
The sound intensity of a loud siren at \(5m\), is \(36 \times 10^{-2} W/m^{2}\)
What is the sound intensity at:
\(9 \times 10^{-2} W/m^{2}\)
\(4 \times 10^{-2} W/m^{2}\)
\(2.25 \times 10^{-2} W/m^{2}\)
\(144 \times 10^{-2} W/m^{2}\)
Light
The intensity of light is measured in 'candela' - \( cp\)
A bright spotlight has a light intensity of 120.0 candela at a distance 2.00 m.
At what distance is the intensity at:
\(4m\)
\(6m\)
\(1m\)
\(0.5m\)
Gravity
The intensity of gravity is measured in 'g', relative to the force applied on the earth by a mass. For this example we're going to use comparitive intensity of gravity from the surface of the earth, or the radius from the centre of the earth. On average the radius of the Earth is 6371 km.
How far away from the centre of earth do you need to be to reduce your weight by half? (You may give your answer in terms of the Earth radius.)
\(I=1/d{^2}\)
\(d{^2}=2\)
\(d=\sqrt{2}=1.41 \times \)the radius of the earth
(If weight is half, gravity intensity is half.)
\(0.5=1/d{^2}\)(transpose the equation)
\(d{^2}=1/0.5\)\(d{^2}=2\)
\(d=\sqrt{2}=1.41 \times \)the radius of the earth
On average the radius of the Earth is 6371 km.
1.41 x 6371 km = 8983 km from the centre.
i.e. 2612 km above the Earths surface. Incidentally, The International Space Station orbits at 408 km from the surface