Electromagnetic waves are able to travel in a vacuum - they do not need a medium. It is a transverse wave that can be thought of as alternating electric and magnetic fields oscillating at right angles to eachother. the electromagnetic spectrum is a spectrtum of all the types of electromagnetic wave - they are classified according to the wavelength of each wave. We need to learn the frequencies and wavelengths of each wave class. However, we know that v = f λ and v for amm electromagnetic waves is 3x108 ms. This means that if we learn just the frequencies then we can work out the wavelengths (and vice versa), meaning we only hae to learn one (frequencies or wavelegths). That being said, here are the wavelengths (commit these to memory even if s the last thing you do - it is a must!! tbh I'm just willing myself on here as I havetn learnt them yet):
- Radiowaves; 106 - 10-1
- Microwaves: 10-1 - 10-3
- Infrared radiation: 10-3 - 7-7
- Visible light: 7-7 - 4-7
- Red: 7-7
- Blue: 4-7
- Ultraviolet: 4-7 - 10-8
- X-rays: 10-8 - 10-13
- Gamma rays: 10-10 - 10-16
All electromagnetic waves can be reflected, refracted, diffracted, and polarised. I have covered polarisation of electromagnetic waves (microwaves) in post 4.4.1.
The refractive index and the refraction of light
The angle at which light is bent depends on the relative speeds of light through the two materials. Each material has a refractive index which can be calculated using the following equation:
n = c / v
The law of refraction: The product of the refractive index and sinθ (sin of angle between the normal and the incident ray) is constant. In other words:
n₁sinθ₁ = n₂sinθ₂
Total internal reflection occurs at the boundary between two different media provided the angle at which the light strikes the boundary is above the critical angle (depends on the refractive index) and the light is travelling through a medium with a higher refractive index as it strikes a boundary with a lower refractive index. The relationship between C (the critical angle) and n (the refractive index) is as follows:
sin C = 1/n
Optical fibers totally internally reflect very well and such have many uses including the fast transmission of data and keyhole surgery! A simple optic fibre has a fine glass core surrounded by a glass cladding which has a lower refractive index. This ensures that light travelling in the fine glass core reflects at the cladding boundary and stays in the core.
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