"The inverse-square law is hugely important in physics. It can be pretty easy to understand using the example of a party balloon.
Imagine taking a red balloon and blowing it up. What would you notice about the colour? The larger the balloon gets the paler the colour - eventually it turns pink (or pops, but that's not the point of our example...)
You can think of the surface of the balloon as being like the wave front radiating out from a wave source - say a loudspeaker. The depth of colour of the balloon is like the intensity of the wave. As the power (or total available amount of red rubber) has to spread out over a larger and larger area, it inevitably gets more thinly spread (sound gets quieter / balloon looks paler). We've also heard this explained by thinking about the thickness of jam you could spread on a table -tennis ball, a cricket ball and a football, if you only have the same amount of jam (wave power) available in each case - but this sounds messier than the balloon idea!
The key thing to remember is that the energy from the source is always the same - it is simply distributed over a larger area, the further away from the source you try to measure.
So - what has all this to do with the 'inverse-square' law?
It comes down to the geometry of the area over which the power spreads out. Let's make it easy for ourselves, and assume the power spreads out equally in all directions, in which case the area is the surface of a sphere (and our balloon should be spherical!). Therefore at a distance r from the source, the power P of the source passes through an area 4πr2 - the surface area of a sphere radius r.
I = P/A = P/ 4πr²
so I ∝ 1/r² (∝ = proportional to)
This is the inverse square part - square because the distance is squared, and inverse because intensity is proportional to one over distance squared.
For example: moving 10 metres away from a source will reduce wave intensity by a factor of 10² = 100.
Remember - the inverse-square law applies, where energy spreads out spherically.
Amplitude and intensity
The energy of a wave is proportional to the square of its amplitude. Therefore the intensity of a wave is also proportional to the square of its amplitude.
I ∝ A²
(∝ = proportional to)
that if Intensity drops off at a rate of 1/r² , wave amplitude drops off at a
rate of 1/r. If we move twice as far from a loudspeaker, the sound intensity
will decrease to one-quarter its original value, and the sound pressure
ampitude will go down to one-half." (source unknown)
With the above in mind, can you explain the following: