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The light that human beings can see is called visible light . Visible light is actually just a small part of the large spectrum of electromagnetic radiation which you will learn more about in [link] . We can think of electromagnetic radiation and visible light as transverse waves. We know that transverse waves can be described by their amplitude, frequency (or wavelength) and velocity. The velocity of a wave is given by the product of its frequency and wavelength:
However, electromagnetic radiation, including visible light, is special because, no matter what the frequency, it all moves at a constant velocity (in vacuum) which is known as the speed of light. The speed of light has the symbol $c$ and is:
Since the speed of light is $c$ , we can then say:
Our eyes are sensitive to visible light over a range of wavelengths from 390 nm to 780 nm (1 nm = $1\times {10}^{-9}$ m). The different colours of light we see are related to specific frequencies (and wavelengths ) of visible light. The wavelengths and frequencies are listed in [link] .
Colour | Wavelength range (nm) | Frequency range (Hz) |
violet | 390 - 455 | 769 - 659 $\times {10}^{12}$ |
blue | 455 - 492 | 659 - 610 $\times {10}^{12}$ |
green | 492 - 577 | 610 - 520 $\times {10}^{12}$ |
yellow | 577 - 597 | 520 - 503 $\times {10}^{12}$ |
orange | 597 - 622 | 503 - 482 $\times {10}^{12}$ |
red | 622 - 780 | 482 - 385 $\times {10}^{12}$ |
You can see from [link] that violet light has the shortest wavelengths and highest frequencies while red light has the longest wavelengths and lowest frequencies .
A streetlight emits light with a wavelength of 520 nm.
We need to determine the colour and frequency of light with a wavelength of $\lambda =520$ nm = $520\times {10}^{-9}$ m.
We see from [link] that light with wavelengths between 492 - 577 nm is green. 520 nm falls into this range, therefore the colour of the light is green.
We know that
We know $c$ and we are given that $\lambda =520\times {10}^{-9}$ m. So we can substitute in these values and solve for the frequency $f$ . ( NOTE: Don't forget to always change units into S.I. units! 1 nm = $1\times {10}^{-9}$ m.)
The frequency of the green light is $577\times {10}^{12}$ Hz
A streetlight also emits light with a frequency of 490 $\times {10}^{12}$ Hz.
We need to find the colour and wavelength of light which has a frequency of 490 $\times {10}^{12}$ Hz and which is emitted by the streetlight.
We can see from [link] that orange light has frequencies between 503 - 482 $\times {10}^{12}$ Hz. The light from the streetlight has $f=490\times {10}^{12}$ Hz which fits into this range. Therefore the light must be orange in colour.
We know that
We know $c=3\times {10}^{8}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}.{\mathrm{s}}^{-1}$ and we are given that $f=490\times {10}^{12}$ Hz. So we can substitute in these values and solve for the wavelength $\lambda $ .
Therefore the orange light has a wavelength of 612 nm.
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