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Wavelength of Light

Ever wonder how we able to see the color and how we interpret them. The reason for the interpretation is that the colors are nothing but the electromagnetic radiations with different wavelengths. The visible region of the electromagnetic spectrum is also known as Visible Light. In the below section we will discuss light and the characteristics of different wavelengths of the visible spectrum. Before we proceed forward let us understand the concept of light.

It is an electromagnetic wave which is visible to human eyes. It extends from the infrared region in EM spectrum to the UV region. Other than these regions humans are unable to see. Although we won’t be able to see the region other than the visible region still the infrared and ultraviolet region both are useful for us.

The colour of the object which we see is due to the fact that the object emits that color and it absorbs all the other colours. 

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Wavelength of Light Formula

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The light exhibits particle as well as wave nature and hence it can be understood by two different equations:

$v$ = $\lambda f$.........(1)

$E$ = $hf$ .........(2)

where,
v = velocity of the light wave;
E = Energy of the light wave;
$\lambda$  = wavelength of the light wave;
f = frequency of the light wave;
h = Planck’s constant = 6.62$\times$10-34 J-sec

The equation (1) represents wave nature of light while the particle nature is understood by the equation (2).

The visible light ranges from 400 nanometer to 700 nanometer. The colour ranges from violet to red. The sun is the only source which can emit entire electromagnetic spectrum.

Wavelength of Visible Light

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As already discussed, the Wavelength of Visible Light ranges from 400 – 700 nm. In this section let us discuss wavelength of different colours of the visible light spectrum.

Visible Light Spectrum

The visible spectrum of the electromagnetic spectrum is known as visible light. The visible spectrum has various different color with different wavelengths. The violet color has shortest wavelength while the red color has the longest wavelength. It can be seen in the above diagram.

The wavelength of various colours of the visible spectrum is given in the table below, but we will discuss wavelength of each color, in detail, one by one.

Colour
Wavelength (nm)
Red 622 - 780
Orange
597 - 622
Yellow
577 - 597
Green
492 - 577
Blue
455 - 492
Violet
390 - 455

Wavelength of White Light : The white light extends from the 400 nm to 750 nm. The white colour when passed through the prism it gets diffracted into all the other colours.

Wavelength of Ultraviolet Light : Ultraviolet light belongs to electromagnetic spectrum and it extends from end of visible region and x-rays. Ultraviolet light is in the range of 10 nm to 400 nm with energies from 3eV to 124 eV. Ultraviolet light gets its name because it is the light closest to the violet portion of visible light.

Wavelength of Red Light : The red light of the visible spectrum has a wavelength of about 650 nm. The best place to see natural red colour is at sunrise and sunset when red or orange colours are present. This is because at the sunrise and sunset the wavelengths associated with red and orange colours are not properly scattered by the atmosphere than the wavelength of other colours (like blue and purple).

Wavelength of Yellow Light : The yellow light has a wavelength of about 570 nm. Low-pressure sodium lamps, like those used in parking lots, emit a yellow (wavelength 589 nm) light.

Wavelength of Green Light : The green light has a wavelength of about 510 nm. Grass appears green because all of the colours in the visible part of the spectrum are absorbed by the grass except green. The grass reflects green wavelength, therefore the grass appears green.

Wavelength of Blue Light : The blue light which we see has a wavelength of about 475 nm. The atmosphere scatter shorter wavelength efficiently and hence the wavelength associated with blue colour is scattered more efficiently by the atmosphere. This is the reason why we see sky to be blue.

Wavelength of Violet Light : The violet light has a wavelength of about 400 nm. As already discussed the violet and blue which belongs to short wavelength region are more efficiently scattered than other wavelengths. Our eyes are more sensitive to the blue colour and hence we see sky blue and not violet.

Wavelength of Indigo Light : The indigo lies between the blue color and the violet color and hence the wavelength of indigo lies between 420 nm to 450 nm.

Wavelength of Infrared Light : Infrared (IR) radiation belongs to electromagnetic radiation spectrum. Infrared has a longer wavelength than visible light. Infrared is close to red colour in visible spectrum and hence it is sort of "redder-than-red" light or "beyond red" light, that is why the name infrared. Infrared radiation cannot be seen but can only be feel as a heat. The best example of feeling the infrared by yourself is feeling the heat after the burner is turned off.

Calculate the Wavelength of Light

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The following problems help us to learn more about wavelength.

Solved Examples

Question 1:
A visible light with a frequency of 7.26 x 1014 Hz is emitted by an object. Calculate the wavelength of light. Also calculate the wave number of this wavelength? 

Solution:
 
For calculating the wavelength of light we will use the following equation.
$\lambda$ = $\frac{v}{f}$
$\lambda$ = $\frac{3\times10^{8}}{7.26\times10^{14}}$
$\lambda$ = $4.13 \times 10^{-7}$

The wave number is related to wavelength as:
$k$ = $\frac{1}{\lambda}$
So,
$k$ = $\frac{1}{4.13\times 10^{-7}}$
$k$ = $2420000$

 

Question 2:
A visible light with a wavelength of 420 nm is emitted by a star constellation. Calculate the frequency of light. Also calculate the wave number of this wavelength?

Solution:
 
For calculating the frequency of light we will use the following equation.
$f$ = $\frac{v}{\lambda}$
$f$ = $\frac{3\times10^{8}}{420\times10^{-9}}$
$f$ = $7.14 \times 10^{14}$

The wave number is related to wavelength as:
$k$ = $\frac{1}{\lambda}$
So,
$k$ = $\frac{1}{420 \times10^{-9}}$
$k$ = $2.38 \times10^{8}$

 

Question 3: A visible light with the energy of 14 $\times$ 10-20 J is emitted by a far star constellation. Calculate the frequency of light. Also calculate the wavelength of the light and its wave number?
Solution:
 
For finding the frequency of visible light we will use the equation 2.
$E$ = $hf$
Hence the frequency of the light is given by;
$f$ = $\frac{E}{h}$
$f$ = $\frac{14\times10^{-20}}{6.62\times10^{-34}}$
$f$ = $2.11\times 10^{14}$

The wavelength is given by following equation;
$\lambda$ = $\frac{3\times10^{8}}{2.11\times10^{14}}$
$\lambda$ = $1.42 \times 10^{-6}$

The wave number is given by;
$k$ = $\frac{1}{\lambda}$ 
Hence,
$k$ = $\frac{1}{1.42 \times10^{-6}}$
$k$ = $0.703 \times10^{6}$ 

 

Question 4: Calculate the wavelength of radiation emitted from radioactive material with a frequency of 2.80 x 1020 s-1. Find in what region of the eletromagnetic spectrum does this frequency lie in?
Solution:
 
We can calculate the wavelength by the following equation:
$\lambda$ = $\frac{v}{f}$
$\lambda$ = $\frac{3\times10^{8}}{2.8\times10^{20}}$
$\lambda$ = $1.07 \times 10^{-12}$

Now if see the EM spectrum we came to know that the above radiation lies in the gamma band since the gamma band extend for the wavelength less than the 10-9 meters.

 

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