Intensity of a wave

​​In a nutshell

Intensity is defined as the power passing through a surface per unit area. Intensity and distance have an inverse square relationship. Intensity is proportional to the square of the amplitude of the wave.

Equations

Variable definitions

Intensity

Intensity is the power passing through a surface per unit area. The units of intensity are watts per square metre ($$Wm^{-2}$$).

​$$I = \frac{P}{A}$$​​

A progressive wave spreads out as the distance from the source increases. The energy and power transferred by the wave is also spread out over this increasing area.

Example

A torch light used at night doesn't carry on going all the way into space. The light gets dimmer and spreads out as the distance from the torch increases.

Intensity and distance

A spherical wave is emitted from a point source and spreads out equally in every direction.

1. Point source emitting spherical waves 2. Surface area of a sphere, $$A = 4\pi r^2$$​​

The power and energy radiated by the source spreads out over the surface of the sphere. The surface area of the sphere increases as the distance from the source increases. This means the same power is spread over a larger area.

This means that at a distance $$r$$ from the source, the total power $$P$$ is spread over the surface area of the sphere $$A$$. The surface area of a sphere is $$4\pi r^{2}$$. This gives a new equation for intensity.

$$I = \frac{P}{4\pi r^{2}}$$​​

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From this equation it can be seen that there is an inverse square relationship between intensity $$I$$ and the distance from the source $$r$$. This is called the inverse square law.

$$I \propto \frac{1}{r^{2}}$$​​

Example

The total power output of the Sun is $$3.8 \times 10^{26} \,W$$​. The intensity of the radiation measured on Mars is $$590 \,Wm^{-2}$$. Calculate the distance between the Sun and Mars. Give your answer in $$km$$.

First, write out the quantities given by the question and check they are in the correct form:

$$P = 3.8\times 10^{26}\,W \newline I = 590 \,Wm^{-2}$$​​

Next, write out the equation needed:

$$I = \frac{P}{4\pi r^{2}} \newline \ \newline r = \sqrt{\frac{P}{4\pi I}}$$

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Then, substitute the values into the equation:

$$r = \sqrt{\frac{3.8\times 10^{26}}{4\pi \times 590}} = 2.2639...\times10^{11}$$​​

Make sure to include the correct units and round to the lowest number of significant figures of the values given:

$$r = 2.3 \times 10^{11} \,m = \frac{2.3\times10^{11}}{1000} = 2.3\times10^8 \,km$$​​

$$\underline{The \space distance \space between \space the \space Sun \space and \space Mars \space is \space \,2.3 \times 10^{8} \,km}$$.

Investigating the inverse square law

A light dependent resistor (LDR) can be used to investigate the inverse square law. A lamp can be used as the light source.

An LDR decreases in resistance as the intensity detected decreases. Each LDR has a calibration curve that can be used to match the resistance measured to the intensity of the light.

This can be plotted graphically. 

Example

Intensity and amplitude

As the energy of a wave spreads out, the amplitude of the wave decreases. 

The amplitude of the wave is proportional to the average speed of the oscillating particles. Decreasing the amplitude means decreasing this speed and therefore decrease the kinetic energy of the particles ($$E_k=\frac{1}{2}mv^{2}$$).

Kinetic energy is proportional to the square of the speed and therefore the square of the amplitude. This means that intensity is proportional to the square to the amplitude.

​$$I \propto(amplitude)^2$$​​