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phys: polarisation

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@ -550,7 +550,7 @@ Note that that is also equal to $\Delta \vec{P}$ above.
- The **amplitude** ($A$) of an oscillation is the greatest displacement from its equilibrium ($\Delta x$).
- An object at an **equilibrium position** is at rest.
### Simple hormonic motion
### Simple harmonic motion
In **simple harmonic motion** (SHM), an object oscillates in a fixed time interval around a central **equilibrium point** with a **linear restoring force** directed toward that equilibrium point.
$$\vec{F} \propto -\Delta \vec{x}$$
@ -636,6 +636,33 @@ The **superposition principle** states that the displacement of a particle of mo
When waves of the same frequency meet, if their crests or troughs overlap, a **supercrest** or **supertrough** is formed, respectively. Waves that result in a net displacement of particles of zero form a **node**.
### Polarisation
**Unpolarised** waves are those that oscillate in every direction **perpendicular** to the direction of energy propagation, while those that are polarised only do so in one plane. Only **transverse** waves can be polarised.
<img src="/resources/images/polarise-head-on.png" width=700>(Source: Kognity)</img>
Waves are polarised by a **polariser**: a material that only allows charged particles to oscillate in one plane. When unpolarised light passes through a polariser, only one plane can progagate fully. The other directions of oscillation have their amplitudes reduced. **Linearly polarised** electromagnetic waves have a single plane of polarisation.
A **polarising filter** can be used to polarise light, and an **analyser** is a second polariser used to determine if light is polarised.
<img src="/resources/images/polariser.png" width=700>(Source: Kognity)</img>
The energy of a wave is proportional to its intensity which is proportional to the square of the amplitude.
$$E\propto I\propto A^2$$
**Malus's law** states that for a polarised wave of energy $E_0$, the amplitude from the second filter, where $\theta$ is the angle between the polariser and the analyser, such that:
$$E=E_0\cos\theta$$
And so:
$$I=I_0\cos\theta$$
When **unpolarised light** passes through a polariser, the average result of $I\cos\theta$ is $\frac{1}{2}$, so the intensity of polarised light is **half** of the intensity of unpolarised light.
When unpolarised light reflects off of a **smooth non-metallic** surface it will be at least partially polarised.
<img src="/resources/images/reflect-polarise.png" width=700>(Source: Kognity)</img>
## 4.4 - Wave behaviour
### Reflection