Prism and Dispersion: Do you know the white light that we see through our naked eyes is a mixture of seven different colours of light? When a ray of white light is made to pass through a prism, this light gets separated into its component colours; this phenomenon is known as dispersion. And the different colours are red, orange, yellow, green, blue, indigo and violet.  Do you know when a beam of white light is passed through a prism of transparent material, light of different wavelengths are deviated by different amounts? In this article, we will study the dispersion and deviation of light by a prism.  We will also discuss the dispersive power, dispersion without average deviation, and average deviation without dispersion.

Know Everything About Spectrum Here

Dispersion in Prism

When a beam of white light incident on a prism, it splits into a band of colours. The sequence of colours that appears on the screen is Violet, Indigo, Blue, Green, Yellow, Orange, and Red, as shown in the given figure.  To remember this sequence of colours, we use the acronym VIBGYOR. Spectrum is the band of the coloured components of a light beam. Dispersion is the splitting of white light into its component colours as they pass through a prism, different colours of light bend through different angles for the incident ray. In which the red light bends the least while the violet light bends the most.

To obtain the spectrum of sunlight, Isaac Newton used a glass prism and tried to split the colours of the spectrum of white light by using another similar prism, but he could not get any more colours. He then placed a second identical prism in an inverted position to the first prism, as shown in the figure given below. Due to this, all the colours of the spectrum again pass through the second prism. Then, he found a beam of white light emerging from the other side of the second prism. This observation shows that sunlight is made up of seven colours, and any light that gives a spectrum similar to that of sunlight is often referred to as white light.

Deviation and Dispersion of Light by a Prism

White light is a superposition of waves in which the wavelengths extend throughout the visible spectrum. In a vacuum, the speed of light is the same for all wavelengths, but the speed in a material substance is different for different wavelengths. Therefore, the index of refraction of a material depends on wavelength. In most materials, the value of the refractive index decreases with increasing wavelength.

Dispersive Power 

When a beam of white light is made to pass through a prism of transparent material, light of different wavelengths are deviated by different amounts. If \({\delta _r},{\delta _y},\) and \({\delta _v}\) are the deviations for red, yellow, and violet components, then average deviation is measured by \({\delta _y}\) as yellow light falls in between red and violet. \({\delta _v} – {\delta _r}\) is called angular dispersion. The ratio of angular dispersion to the average deviation when a white beam of light is passed through it is called the dispersive power. It is denoted by \omega. As we know that:

\(\delta = (\mu – 1)A\)

We use this equation when \(A\) and \(i\) are small.

Suppose a beam of white light is passed through such a prism as shown in the above figure, the deviation of red, yellow and violet light is given as:

\({\delta _r} = \left( {{\mu _r} – 1} \right)A\)

\({\delta _y} = \left( {{\mu _y} – 1} \right)A\)

And \({\delta _y} = \left( {{\mu _v} – 1} \right)A\)

The angular dispersion is given as :\({\delta _v} – {\delta _r} = \left( {{\mu _v} – {\mu _r}} \right) \cdot A\)

And the average deviation is given as: \({\delta _y} = \left( {{\mu _y} – 1} \right)A\)

Thus, the dispersive power of the medium will be: 

\(\omega = \frac{{{\mu _v} – {\mu _r}}}{{{\mu _y} – 1}}………\left( 1 \right)\)

Dispersion without Average Deviation and Average Deviation without Dispersion

The given figure shows two prisms of refracting angles \(A\) and \(A’\) and dispersive powers \(\omega \) and \({\omega ^\prime }\) respectively. They are placed in contact so that the two refracting angles are reversed for each other. A ray of light passes through the combination, as shown.

By the two prisms, the deviations produced are:

\({\delta _1} = (\mu – 1)A\) and \({\delta _2} = \left( {{\mu ^\prime } – 1} \right){A^\prime }\)

As the two deviations are opposite to each other, then the net deviation may be written as:

\(\delta = {\delta _1} – {\delta _2} = (\mu – 1)A – \left( {{\mu ^\prime } – 1} \right){A^\prime }…….\left( 2 \right)\)

Using this equation, the average deviation produced by the combination of white light is passed is, 

\({\delta _y} = \left( {{\mu _y} – 1} \right)A – \left( {\mu _y^\prime – 1} \right){A^\prime }……..\left( 3 \right)\)

And the net angular dispersion is:

\({\delta _v} – {\delta _r} = \left( {{\mu _v} – {\mu _r}} \right)A – \left( {\mu _v^\prime – \mu _r^\prime } \right){A^\prime }\)

But, as \({\mu _v} – {\mu _r} = \omega \left( {{\mu _y} – 1} \right)\) from Eqn \((1),\) we have \({\delta _v} – {\delta _r} = \left( {{\mu _y} – 1} \right)\omega A – \left( {\mu _y^\prime – 1} \right){\omega ^\prime }{A^\prime }…….\left( 4 \right)\)

Dispersion without Average Deviation

From Eq. \((3),\)     \({\delta _y} = 0\) if

\(\frac{A}{{{A^\prime }}} = \frac{{\mu _y^\prime – 1}}{{{\mu _y} – 1}}……..\left( 5 \right)\)

This is the necessary condition of dispersion without average deviation. Using this in Eqn. \((4),\) the net angular dispersion produced is:

\({\delta _v} – {\delta _r} = \left( {{\mu _y} – 1} \right)A\left( {\omega – {\omega ^\prime }} \right)\)

Average Deviation without Dispersion

From Eq. \((4),\)  \({\delta _v} – {\delta _r} = 0,\) if

\(\frac{A}{{{A^\prime }}} = \frac{{\left( {\mu _y^\prime – 1} \right){\omega ^\prime }}}{{\left( {{\mu _y} – 1} \right)\omega }} = \frac{{\mu _v^\prime – \mu _r^\prime }}{{{\mu _v} – {\mu _r}}}……..\left( 6 \right)\)

This is the necessary condition of average deviation without dispersion. Using the above condition in Eq. \((3),\) the net average deviation is

\({\delta _y} = \left( {{\mu _y} – 1} \right)A\left( {1 – \frac{\omega }{{{\omega ^\prime }}}} \right)\)

Summary

When a beam of white light is made to pass through the prism, it gets separates into its component colours; this phenomenon is dispersion. In the visible region of the spectrum, the spectral lines are seen in the order from violet to red. The colours are given by the word VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, and Red).

The ratio of angular dispersion for any two wavelengths (colours) to the deviation of the mean wavelength is the dispersive power of the material of a prism. Thus, the dispersive power of the medium will be:

\(\omega = \frac{{{\mu _v} – {\mu _r}}}{{{\mu _y} – 1}}\)

Where \(\mu \) is the refractive index of red, yellow, and violet colours.

The necessary condition of dispersion without average deviation is given as:

\(\frac{A}{{{A^\prime }}} = \frac{{\mu _y^\prime – 1}}{{{\mu _y} – 1}}\)

The necessary condition of average deviation without dispersion is given as:

\(\frac{A}{{{A^\prime }}} = \frac{{\left( {\mu _y^\prime – 1} \right){\omega ^\prime }}}{{\left( {{\mu _y} – 1} \right)\omega }} = \frac{{\mu _v^\prime – \mu _r^\prime }}{{{\mu _v} – {\mu _r}}}………\left( 6 \right)\)

Where for the two prisms, \(A\) and \({{A^\prime }}\) are refracting angles, \(\omega \) and \({\omega ^\prime }\) are the dispersive powers \(\mu \) and \({\mu ^\prime }\) are the refractive index.

Prism and Dispersion – Solved Examples

Q.1. On entering a glass prism, a beam of light of average wavelength \({\rm{600}}\,{\rm{nm}}\) splits into three coloured beams of wavelengths \(384\;{\rm{nm}},589\;{\rm{nm}}\) and \({\rm{760}}\,{\rm{nm,}}\) respectively. Then for these wavelengths, find the refractive indices of the material of the prism.
Ans: The refractive index of the material of the prism is given by:
\(\mu = \frac{c}{v}\)
where \(c\) is the speed of light in a vacuum, and \(v\) is the speed of light in the medium (prism). Since the velocity of a wave is a product of frequency and wavelength, we can write:
\(c = v{\lambda _a}\) and \(u = v{\lambda _m}\) where \({\lambda _a}\) and \({\lambda _m}\) are the wavelengths in air and medium, respectively, and \(v\) is the frequency of light waves. Thus:
\(u = \frac{{v{\lambda _a}}}{{v{\lambda _m}}} = \frac{{{\lambda _a}}}{{{\lambda _m}}}\)
The refractive index for \(384\;{\rm{nm}}\) wavelength will be:
\({u_1} = \frac{{600 \times {{10}^{ – 9}}\;{\rm{m}}}}{{384 \times {{10}^{ – 9}}\;{\rm{m}}}} = 1.56\)
For a wavelength of \(589\;{\rm{nm}}\)
\({u_2} = \frac{{600 \times {{10}^{ – 9}}\;{\rm{m}}}}{{58.9 \times {{10}^{ – 9}}\;{\rm{m}}}} = 1.02\)
For \(760\;{\rm{nm}}\) wavelength:
\({u_3} = \frac{{600 \times {{10}^{ – 9}}\;{\rm{m}}}}{{760 \times {{10}^{ – 9}}\;{\rm{m}}}} = 0.8\)

FAQs on Prism and Dispersion

Q.1. What is the dispersion of light?
Ans: When a beam of white light is made to pass through a transparent glass object like, for example, a glass prism. It gets split into seven different colours (Violet, Indigo, Blue, Green, Yellow, Orange, and Red), known as the dispersion of light.

Q.2. What do you mean by the dispersive power of a prism?
Ans: The dispersive power of prism is defined as the measure of the difference in refraction of the light of the highest wavelength and the lowest wavelength that enters the prism. The ratio of angular dispersion to the average deviation when a white beam of light is passed through a prism is called the dispersive power. It is denoted by \(\omega .\)
\(\omega = \frac{{{\mu _y} – {\mu _r}}}{{{\mu _y} – 1}}\)
Where \(\mu \) is the refractive index of red, yellow, and violet colours.

Q.3. What do you understand by deviation and dispersion?
Ans: When a light slightly diverges from its path due to a change in the medium is known as a deviation of light. In comparison, the splitting up of white light into its constituent colours when white light is allowed to pass through a prism is known as dispersion.

Q.4. Which colour has the minimum and maximum deviation of a prism?
Ans: Since all the colours have different angles of deviation when white light spreads out into a spectrum. The red colour deviates least because it has the maximum wavelength, and the violet light deviates maximum, as it has the least wavelength.

Q.5. Name the factors on which angle of deviation depends.
Ans: The angle of deviation depends on:
1. The angle of incidence
2. The material of the prism
3. The angle of the prism
4. The colour or wavelength of light.