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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 × λ

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:

c = 3 × 10 8 m . s - 1

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

c = f × λ

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] .

Colours, wavelengths and frequencies of light in the visible spectrum.
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 λ = 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

    c = f × λ

    We know c and we are given that λ = 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.)

    f = c λ = 3 × 10 8 520 × 10 - 9 = 577 × 10 12 Hz

    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

    c = f × λ

    We know c = 3 × 10 8 m . s - 1 and we are given that f = 490 × 10 12 Hz. So we can substitute in these values and solve for the wavelength λ .

    λ = c f = 3 × 10 8 490 × 10 12 = 6 . 122 × 10 - 7 m = 612 × 10 - 9 m = 612 nm

    Therefore the orange light has a wavelength of 612 nm.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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cm
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Can you compute that for me. Ty
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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Source:  OpenStax, Maths test. OpenStax CNX. Feb 09, 2011 Download for free at http://cnx.org/content/col11236/1.2
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