# 0.5 Colour

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## Introduction

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:

$v=f×\lambda$

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:

$\begin{array}{ccc}\hfill c& =& 3×{10}^{8}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}.{\mathrm{s}}^{-1}\hfill \end{array}$

Since the speed of light is $c$ , we can then say:

$c=f×\lambda$

## Colour and light

Our eyes are sensitive to visible light over a range of wavelengths from 390 nm to 780 nm (1 nm = $1×{10}^{-9}$ 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 $×{10}^{12}$ blue 455 - 492 659 - 610 $×{10}^{12}$ green 492 - 577 610 - 520 $×{10}^{12}$ yellow 577 - 597 520 - 503 $×{10}^{12}$ orange 597 - 622 503 - 482 $×{10}^{12}$ red 622 - 780 482 - 385 $×{10}^{12}$

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.

1. What colour is the light? (Use [link] to determine the colour)
2. What is the frequency of the light?
1. We need to determine the colour and frequency of light with a wavelength of $\lambda =520$ nm = $520×{10}^{-9}$ m.

2. 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.

3. We know that

$\begin{array}{ccc}\hfill c& =& f×\lambda \hfill \end{array}$

We know $c$ and we are given that $\lambda =520×{10}^{-9}$ 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 = $1×{10}^{-9}$ m.)

$\begin{array}{ccc}\hfill f& =& \frac{c}{\lambda }\hfill \\ & =& \frac{3×{10}^{8}}{520×{10}^{-9}}\hfill \\ & =& 577×{10}^{12}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\mathrm{Hz}\hfill \end{array}$

The frequency of the green light is $577×{10}^{12}$ Hz

A streetlight also emits light with a frequency of 490 $×{10}^{12}$ Hz.

1. What colour is the light? (Use [link] to determine the colour)
2. What is the wavelength of the light?
1. We need to find the colour and wavelength of light which has a frequency of 490 $×{10}^{12}$ Hz and which is emitted by the streetlight.

2. We can see from [link] that orange light has frequencies between 503 - 482 $×{10}^{12}$ Hz. The light from the streetlight has $f=490×{10}^{12}$ Hz which fits into this range. Therefore the light must be orange in colour.

3. We know that

$\begin{array}{ccc}\hfill c& =& f×\lambda \hfill \end{array}$

We know $c=3×{10}^{8}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}.{\mathrm{s}}^{-1}$ and we are given that $f=490×{10}^{12}$ Hz. So we can substitute in these values and solve for the wavelength $\lambda$ .

$\begin{array}{ccc}\hfill \lambda & =& \frac{c}{f}\hfill \\ & =& \frac{3×{10}^{8}}{490×{10}^{12}}\hfill \\ & =& 6.122×10-7\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}\hfill \\ & =& 612×{10}^{-9}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}\hfill \\ & =& 612\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\mathrm{nm}\hfill \end{array}$

Therefore the orange light has a wavelength of 612 nm.

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