Chapter 6: Spectrum

In the year 1800, the astronomer William Herschel was struggling with a small, stubborn problem. To look at the Sun through his telescope he used coloured glass filters, and he had noticed something odd. Some filters let through a comfortable, cool image. Others grew so hot that they cracked. Why should the colour of a filter decide how much heat came through with the light?

To find out, he darkened a room, let a narrow beam of sunlight fall on a glass prism, and caught the band of colours it produced on a table. Then he laid a thermometer with a blackened bulb in one colour after another, and kept two more thermometers to one side as a check.

The readings rose slowly in the violet, and rose faster as he moved towards the red. That was interesting, but not shocking. Then Herschel moved the thermometer just past the red edge of the band, where the table looked completely dark and his eye saw no colour at all.

The temperature climbed higher there than anywhere inside the band.

Something was streaming out of the prism and warming the bulb, yet nothing was visible. Sunlight, it turned out, was wider than sunlight looked. What Herschel had found that day is what we now call infrared radiation, and it was the first hint that the light our eyes catch is only a narrow slice of something far larger.

Diagram

Fig. 6.0-1 — Herschel’s Thermometers Beyond The Red A darkened room scene: a narrow beam of sunlight enters through a small gap in a window shutter and falls on a triangular glass prism resting on a table; the emergent light spreads into a band of seven colours across the table top, labelled violet, indigo, blue, green, yellow, orange, red in that order. Show one thermometer with a blackened bulb lying in the red band and a second thermometer lying just beyond the red edge in the dark region, with a label ‘no visible colour here’. Show two more thermometers lying well away from the band, labelled ‘control thermometers’. Mark the dark region beyond red with an arrow labelled ‘largest temperature rise’.

NoteCuriosity Corner

Q1. A triangular prism spreads white light into a band of colours, but the flat window glass you look through every day does not. What must the shape of the glass have to do with it? Q2. Herschel found heating beyond the red edge of the band. If something can also lie beyond the violet edge, how could anyone ever detect it, since the eye cannot see it and it may not warm a thermometer? Q3. Violet light is bent the most by a glass prism and red light the least. What must be different about violet light while it is travelling inside the glass? Q4. The sky overhead is blue in the afternoon, yet the Sun near the horizon at sunset looks red. Could one single reason explain both of these at once? Q5. Danger signals, stop lights and railway signals are red almost everywhere in the world. Is there a reason from physics for that choice, or is it only a habit?

6.1 Deviation Produced By A Triangular Prism

A prism is a transparent block bounded by two plane surfaces that are inclined to each other, and it is these two surfaces that do all the work. Light enters through one of them and leaves through the other. Because the two surfaces are not parallel, the ray that comes out is not travelling in the direction it came in.

The two inclined surfaces are called the refracting surfaces, and the angle between them is the angle of prism, written as \(A\) and measured in degrees. The surface opposite this angle is called the base. A triangular glass prism used in a laboratory usually has \(A = 60^\circ\), and it is drawn as a triangle in every ray diagram in this chapter.

The parts you will need to name are shown in Fig. 6.1-1.

Diagram

Fig. 6.1-1 — The Parts Of A Triangular Prism A triangle ABC standing with vertex A at the top and side BC horizontal at the bottom, drawn as a solid glass block. Label the two upper sides AB and AC as ‘refracting surfaces’; label the angle at vertex A as ‘angle of prism A’; label the vertex A itself as ‘refracting edge’; label side BC as ‘base’. Add a note beside the figure: for a laboratory glass prism, A is usually 60 degrees.

6.1.1 How A Prism Deviates A Ray Of Light

Follow a single ray of one colour as it goes through the prism, drawn in Fig. 6.1-2. The ray PQ falls on the first surface AB. Since it is passing from air into glass, which is optically denser, it bends towards the normal at Q and travels inside the glass along QR.

That first bend turns the ray through an angle \(\delta_1\) towards the base BC. This angle is called the deviation at the first surface. Its size depends on the angle at which the ray struck the surface, and on the refractive index of glass with respect to air.

The ray QR now reaches the second surface AC and passes from glass out into air. Here it bends away from the normal, and this second bend turns it through a further angle \(\delta_2\), again towards the base. The ray then leaves the prism as the emergent ray RS and travels in a straight line.

Diagram

Fig. 6.1-2 — Deviation Of A Ray By A Prism Triangle ABC with vertex A at the top, base BC horizontal. A ray PQ from a slit on the left strikes face AB at point Q; draw the normal at Q as a dashed line and mark the angle of incidence i between PQ and this normal. Inside the glass draw the refracted ray QR going down towards face AC and mark the small angle of deviation delta-1 between the direction of PQ produced and QR. At point R on face AC draw the normal as a dashed line and the emergent ray RS bending away from the normal towards the base, marking the deviation delta-2 there. Produce PQ forward as a dashed line PL, and produce RS backwards to meet it, marking the angle between PL and RS as the total angle of deviation delta. Label the incident ray, the emergent ray, the base BC, and the light source with its slit.

Why Both Bends Go The Same Way

At the first surface the light enters a denser medium and turns towards the normal. At the second it enters a rarer medium and turns away from the normal. These sound like opposite turns, and in a slab with parallel faces they would indeed cancel.

In a prism they do not cancel, because the two faces are inclined. The two normals point in different directions, and both turns end up carrying the ray towards the base. So the deviations add rather than subtract:

\[\delta = \delta_1 + \delta_2\]

where \(\delta_1\) is the deviation at the first surface in degrees, \(\delta_2\) the deviation at the second surface in degrees, and \(\delta\) the total deviation in degrees. This addition holds because both deviations are in the same sense, towards the base of the prism.

The total deviation \(\delta\) is measured between two directions you can see in Fig. 6.1-2: the direction PL in which the incident ray was originally travelling, and the direction RS of the ray that finally emerges.

NoteDefinition

Angle Of Deviation The angle between the direction of the incident ray and the direction of the emergent ray, when light passes through a prism; it is measured in degrees. For a prism, the emergent ray is always turned towards the base.

Angle of prism and angle of deviation. The angle of prism A is a fixed property of the glass block itself, the angle between its two refracting faces. The angle of deviation is a property of the light passing through it, and changes when the incident ray or the colour changes.

6.1.2 Factors On Which The Angle Of Deviation Depends

The total deviation \(\delta\) produced by a given prism is not a fixed number. It depends on three things.

  1. The angle of incidence \(i\) at the first surface, measured in degrees. Tilt the prism so that the light strikes the first face at a different angle, and the whole path through the glass changes.
  2. The angle of prism \(A\), measured in degrees. A prism with a sharper refracting edge presents its second face at a different slant, so the deviation there changes.
  3. The refractive index \(\mu\) of the material of the prism, which is a ratio of two speeds and so has no unit. A denser material bends the ray more sharply at both faces.

The third factor is the one this chapter is built on. The refractive index of a material is not a single number for all light. It depends on the colour, that is, on the wavelength \(\lambda\) of the light being used. So the angle of deviation must depend on colour too.

μ = speed of light in air ÷ speed of light in glass Refractive index compares how fast light travels in air with how slowly it is forced to travel inside the glass. The more the glass slows the light down, the larger μ is, and the more sharply that light is bent. It is a ratio of two speeds, so it carries no unit.

6.1.3 Dependence Of Deviation On The Colour Of Light

In air, light of every colour travels at the same speed. That is why a beam of white light stays white as it crosses a room. Inside a transparent material such as glass, this is no longer true: different colours travel at different speeds.

The pattern is a definite one. In a transparent medium, the speed of light decreases as the wavelength of the light decreases. Violet light, at the short-wavelength end of the visible range, travels the slowest inside glass; red light, at the long-wavelength end, travels the fastest.

Now bring in the refractive index. It is defined as

\[\mu_{\text{glass}} = \frac{\text{speed of light in air}}{\text{speed of light in glass}}\]

where both speeds are in m s⁻¹ and \(\mu\) has no unit. The numerator is the same for every colour, so the colour that moves slowest in glass has the largest refractive index. Hence

\[\mu_{\text{violet}} > \mu_{\text{red}}\]

Because the refractive index of glass increases as the wavelength of light decreases, the deviation caused by a prism also increases as the wavelength decreases.

From Refractive Index To Deviation

It is worth seeing exactly why a larger refractive index means a larger bend. At the first surface, Snell’s law gives

\[\mu = \frac{\sin i}{\sin r} \qquad \text{so} \qquad \sin r = \frac{\sin i}{\mu}\]

where \(i\) is the angle of incidence and \(r\) the angle of refraction, both in degrees. All the colours in a single beam arrive at the same angle \(i\), so \(\sin i\) is fixed. Dividing by a larger \(\mu\) gives a smaller \(\sin r\), and therefore a smaller angle \(r\).

Violet light, having the largest \(\mu\), is refracted through the smallest angle inside the glass, so \(r_{\text{red}} > r_{\text{violet}}\). The deviation at this surface is the difference between the two angles,

\[\delta_1 = i - r\]

which is larger when \(r\) is smaller. So \(\delta_{\text{violet}} > \delta_{\text{red}}\), and the same reasoning at the second surface pushes them further apart.

In visible light, violet colour (wavelength \(\lambda \simeq 4000\ \text{Å}\)) is deviated the most and red colour (wavelength \(\lambda \simeq 8000\ \text{Å}\)) is deviated the least, because in glass the speed of violet light is the least and that of red light is the most. The other five colours are deviated by amounts lying in between, in order of their wavelengths. The result is drawn in Fig. 6.1-3.

Diagram

Fig. 6.1-3 — Violet Is Deviated More Than Red Triangle ABC, vertex A at top, base BC horizontal. A single incident ray labelled ‘violet + red’ strikes face AB at point Q at an angle of incidence i, with the normal drawn dashed. Inside the glass draw two separate refracted rays from Q: the violet one bent closer to the normal with its angle of refraction marked r-violet, and the red one bent less with a larger angle marked r-red. Both rays travel to face AC and emerge, with the violet emergent ray clearly closer to the base BC than the red emergent ray. Label both emergent rays ‘violet’ and ‘red’, and mark the two total deviations delta-violet and delta-red from the direction of the incident ray produced forward as a dashed line. Add the note: angles exaggerated for clarity.

Example 6.1 — Comparing The Refractive Index For Violet And Red

  1. Given: speed of violet light in a certain glass \(= 1.94 \times 10^{8}\ \text{m s}^{-1}\); speed of red light in the same glass \(= 1.99 \times 10^{8}\ \text{m s}^{-1}\); speed of light in air \(= 3 \times 10^{8}\ \text{m s}^{-1}\). To find: the refractive index of the glass for each colour, and which colour the prism deviates more.

  2. Relation used: \(\mu = \dfrac{\text{speed of light in air}}{\text{speed of light in glass}}\). This is the definition of refractive index, and it is the right relation because the question gives speeds and asks for the bending.

  3. For violet: \(\mu_{\text{violet}} = \dfrac{3 \times 10^{8}\ \text{m s}^{-1}}{1.94 \times 10^{8}\ \text{m s}^{-1}} = 1.55\). For red: \(\mu_{\text{red}} = \dfrac{3 \times 10^{8}\ \text{m s}^{-1}}{1.99 \times 10^{8}\ \text{m s}^{-1}} = 1.51\). The units of m s⁻¹ cancel in each case, as they must, since refractive index has no unit.

  4. The refractive index is \(1.55\) for violet and \(1.51\) for red. Since \(\mu_{\text{violet}} > \mu_{\text{red}}\), the prism deviates violet light more than red light. Both values lie close to \(1.5\), which is the expected refractive index for ordinary glass, so the answer is sensible.

Example 6.2 — Deviation Of Two Colours At The First Face

  1. Given: refractive index of the prism material for red light \(\mu_{\text{red}} = 1.51\) and for violet light \(\mu_{\text{violet}} = 1.53\); angle of incidence at the first face \(i = 30^\circ\) for both colours. To find: the angle of refraction and the deviation at this face for each colour.

  2. Relation used: \(\sin r = \dfrac{\sin i}{\mu}\), from Snell’s law at the air–glass surface, and \(\delta_1 = i - r\) for the deviation produced at a single refracting surface.

  3. For red: \(\sin r = \dfrac{\sin 30^\circ}{1.51} = \dfrac{0.500}{1.51} = 0.331\), so \(r_{\text{red}} = 19.3^\circ\) and \(\delta_1 = 30^\circ - 19.3^\circ = 10.7^\circ\). For violet: \(\sin r = \dfrac{0.500}{1.53} = 0.327\), so \(r_{\text{violet}} = 19.1^\circ\) and \(\delta_1 = 30^\circ - 19.1^\circ = 10.9^\circ\).

  4. At the first face the red ray is deviated by \(10.7^\circ\) and the violet ray by \(10.9^\circ\). The violet deviation is the larger, as expected from its larger refractive index, and the difference is small — which is why the colours only separate visibly after the second face.

Hold a thick glass paperweight or the bevelled edge of a mirror in a patch of daylight near a white wall, and turn it slowly. Watch for a narrow fringe of colour appearing at the edges of the bright patch, and note which colour sits at which end of the fringe.

Q3. Violet light is bent the most by a glass prism and red light the least. What must be different about violet light while it is travelling inside the glass?

A. Its speed. All colours travel at the same speed in air, but inside glass violet light is slowed down the most and red light the least. The slower a colour travels in the glass, the larger the refractive index of the glass for that colour, and the more sharply the prism bends it.

6.2 Colours In White Light With Their Wavelength And Frequency Range

The white light given out by the Sun, or by a filament lamp, is not a single kind of light at all. It is a mixture of light of many different wavelengths, all travelling together.

The wavelength of light is the distance between two neighbouring crests of the light wave, and its SI unit is the metre (m). The frequency is the number of waves that pass a point each second, and its SI unit is the hertz (Hz). Each of these two quantities has a definite value for each kind of light in the mixture.

6.2.1 Wavelength As The Characteristic Of Colour

When light of a particular wavelength enters your eye and falls on the retina, it produces the sensation of a particular colour. Light of a different wavelength produces the sensation of a different colour. This is why we see the mixture as being made of separate colours at all.

The prominent colours present in white light are violet, indigo, blue, green, yellow, orange and red. What separates one from the next is nothing but wavelength: violet light has the shortest wavelength of the seven, and red light the longest.

This link between wavelength and colour is a fixed one. Wavelength is the characteristic of colour, irrespective of its origin — light of the same colour obtained from different sources will always have the same wavelength. Green light from a leaf, from a laboratory lamp and from a traffic signal all carry the same wavelength.

Notice the direction of the reasoning here. The wavelength is a physical property of the light itself, and can be measured in a laboratory. The colour is the sensation that wavelength produces in an eye and a brain. In other words, colour is the subjective property of light related to its wavelength.

Wavelength and colour. Wavelength is a length, measured in metres, and belongs to the light. Colour is the sensation that light produces when it reaches the retina, and belongs to the observer. The wavelength decides the colour, but the two are not the same thing.

6.2.2 Wavelength And Frequency Ranges Of The Seven Colours

Each of the seven colours is not one single wavelength, but a small range of wavelengths. A range of wavelengths means a range of frequencies as well, because the two are tied together for light travelling in air by

\[c = f\lambda\]

where \(c\) is the speed of light in air, equal to \(3 \times 10^{8}\ \text{m s}^{-1}\), \(f\) is the frequency in Hz, and \(\lambda\) is the wavelength in m. Since \(c\) is the same for every colour in air, a longer wavelength must go with a lower frequency, and a shorter wavelength with a higher frequency.

The ranges for the seven colours are listed in Table 6.1.

Table 6.1 — Wavelength and frequency of different colours in white light

Colour Wavelength (\(\lambda\)) range (nearly) Frequency (\(f = c/\lambda\)) range in \(10^{14}\ \text{Hz}\)
Violet 4000 Å to 4460 Å 7.5 – 6.73
Indigo 4460 Å to 4640 Å 6.73 – 6.47
Blue 4640 Å to 5000 Å 6.47 – 6.01
Green 5000 Å to 5780 Å 6.01 – 5.19
Yellow 5780 Å to 5920 Å 5.19 – 5.07
Orange 5920 Å to 6200 Å 5.07 – 4.84
Red 6200 Å to 8000 Å 4.84 – 3.75

Read the table across a row and you get one colour’s own stretch of the spectrum; read it down a column and you follow the whole visible range from one end to the other. Note that there is no sharp boundary between different colours. Each range shades into the next, and the figures given are approximate.

The Two Ends Of The Visible Range

Two rows of Table 6.1 are worth learning as the limits of the whole visible spectrum, and they are shown on the scale in Fig. 6.2-1.

Red light has the longest wavelength, \(8000\ \text{Å}\), which is \(8 \times 10^{-7}\ \text{m}\) or \(800\ \text{nm}\), and therefore the lowest frequency, \(3.75 \times 10^{14}\ \text{Hz}\). Violet light has the shortest wavelength, \(4000\ \text{Å}\), which is \(4 \times 10^{-7}\ \text{m}\) or \(400\ \text{nm}\), and therefore the highest frequency, \(7.5 \times 10^{14}\ \text{Hz}\).

So as you move from the violet end of the spectrum to the red end, the wavelength increases while the frequency decreases. The two always run in opposite directions, because their product is fixed at the speed of light.

Diagram

Fig. 6.2-1 — The Visible Spectrum With Its Wavelength And Frequency Scale A horizontal band shaded continuously from violet at the left through indigo, blue, green, yellow and orange to red at the right, with the seven colour names printed under their parts of the band and no hard dividing lines between them. Below the band draw a wavelength scale reading 4000 Å at the extreme left and 8000 Å at the extreme right, with 5000 Å, 6000 Å and 7000 Å marked in between, and an arrow along it labelled ‘wavelength increases’. Above the band draw a frequency scale reading 7.5 × 10^14 Hz at the extreme left and 3.75 × 10^14 Hz at the extreme right, with an arrow labelled ‘frequency decreases’.

Example 6.3 — Finding The Wavelength Of Violet Light From Its Frequency

  1. Given: frequency of violet light \(f = 7.5 \times 10^{14}\ \text{Hz}\); speed of light \(c = 3 \times 10^{8}\ \text{m s}^{-1}\). To find: the wavelength in nm and in Å.

  2. Relation used: \(c = f\lambda\), so \(\lambda = \dfrac{c}{f}\). This relation connects the two quantities named in the question, and it applies because the light is travelling in air.

  3. \(\lambda = \dfrac{3 \times 10^{8}\ \text{m s}^{-1}}{7.5 \times 10^{14}\ \text{Hz}} = 4 \times 10^{-7}\ \text{m}\). Since \(1\ \text{nm} = 10^{-9}\ \text{m}\), this is \(\dfrac{4 \times 10^{-7}}{10^{-9}} = 400\ \text{nm}\). Since \(1\ \text{Å} = 10^{-10}\ \text{m}\), it is also \(4000\ \text{Å}\).

  4. The wavelength of violet light is \(400\ \text{nm}\), or \(4000\ \text{Å}\). This matches the violet end of Table 6.1, so the answer is sensible.

Example 6.4 — Finding The Frequency Of Red Light From Its Wavelength

  1. Given: wavelength of red light \(\lambda = 800\ \text{nm}\); speed of light \(c = 3 \times 10^{8}\ \text{m s}^{-1}\). To find: the frequency.

  2. Relation used: \(c = f\lambda\), so \(f = \dfrac{c}{\lambda}\). The wavelength must first be converted to metres, because the speed is given in m s⁻¹ and the two units must match before they are combined.

  3. \(\lambda = 800\ \text{nm} = 800 \times 10^{-9}\ \text{m} = 8 \times 10^{-7}\ \text{m}\). Then \(f = \dfrac{3 \times 10^{8}\ \text{m s}^{-1}}{8 \times 10^{-7}\ \text{m}} = 3.75 \times 10^{14}\ \text{Hz}\).

  4. The frequency of red light is \(3.75 \times 10^{14}\ \text{Hz}\). It is the lowest frequency in Table 6.1, which is what we expect for the longest visible wavelength.

6.2.3 The Angstrom And The Nanometre

Wavelengths of light are tiny fractions of a metre, so writing them in metres means writing a string of zeros every time. Two smaller units are used instead.

The angstrom, written Å, is defined by \(1\ \text{Å} = 10^{-10}\ \text{m}\), which is the same as \(10^{-8}\ \text{cm}\). It is the older of the two units, and the wavelength figures in Table 6.1 are given in it.

The nanometre, written nm, is defined by \(1\ \text{nm} = 10^{-9}\ \text{m}\). Comparing the two definitions gives a conversion worth remembering:

\[1\ \text{nm} = 10\ \text{Å}\]

Nowadays wavelength is expressed mostly in nanometres, since the nanometre is part of the SI system of prefixes while the angstrom is not. Both units appear in this chapter, so be ready to move between them: divide a figure in angstroms by \(10\) to get nanometres, and multiply by \(10\) to go back.

Wavelength λ. SI unit: the metre (m). Practical units for light: 1 Å = 10⁻¹⁰ m and 1 nm = 10⁻⁹ m, so 1 nm = 10 Å. Always convert to metres before substituting into c = fλ, or the frequency will come out wrong by a large power of ten.

A beam of green light has a wavelength of 5500 Å. Express this wavelength in nanometres and in metres, and calculate the frequency of the light. Take the speed of light as 3 × 10⁸ m s⁻¹.

6.3 Dispersion Of White Light Through A Prism And Formation Of A Spectrum

Section 6.1 established that a prism bends violet light more than red light. Section 6.2 established that white light contains every colour from violet to red at once. Put those two facts together, and something must happen when white light meets a prism: the colours cannot all leave along the same path.

6.3.1 Newton’s Experiment With A Prism

The idea that white light is a mixture was settled by Sir Isaac Newton, and it began with an annoyance rather than a discovery. While studying the image of a heavenly body formed by refraction of white light through a lens, Newton found that the image was coloured at its edges.

His first thought was that the lens was at fault — that the colours were a defect of the glass or of the polishing. So he repeated the experiment with a carefully polished lens. The image was still coloured. Newton then concluded that the fault is not with the lens, but that there is something in the nature of white light itself that makes the image coloured at its edges.

To investigate further, he performed another experiment, this time with a prism, shown in Fig. 6.3-1.

The Arrangement And What Was Seen

Newton allowed white light from the Sun to enter a dark room through a small aperture in a window, and placed a glass prism in the path of the light rays. The light emerging out of the prism was received on a white screen.

On the screen he obtained a coloured patch like that of a rainbow. This band of colours was termed the spectrum.

Starting from the side of the base of the prism, the order of colours in the spectrum on the screen is violet, indigo, blue, green, yellow, orange and red. This order is easily remembered by the word VIBGYOR, made from the first letters of the seven colours.

Diagram

Fig. 6.3-1 — Newton’s Experiment: Dispersion Of White Light By A Prism A dark room shown in section. On the left, a small aperture in a window through which a narrow beam of white light enters, labelled ‘white light’ and ‘aperture in a window’. The beam falls on the face AB of a triangular glass prism ABC drawn with vertex A at the top and base BC at the bottom. Inside the prism draw the beam fanning out slightly, and from the face AC draw seven diverging emergent rays reaching a vertical white screen on the right. On the screen show a continuous band with the colours in order from the top downwards: red, orange, yellow, green, blue, indigo, violet, so that violet lies nearest the base side of the prism. Label the band ‘spectrum’ and label the screen ‘white screen’. Add a note: the colours are drawn widely separated for clarity, and there are no dividing lines between them on a real screen.

Newton’s conclusion followed directly from what the screen showed. White light consists of seven prominent colours, each corresponding to a small range of wavelength. White light is therefore a mixture of a large number of wavelengths, that is, it is polychromatic in nature. Light of a single colour, made of one narrow range of wavelength, is called monochromatic light.

One warning about every diagram of a spectrum, including Fig. 6.3-1. In the spectrum each colour is mixed with the colour next to it, so there is no sharp boundary line separating the colours. In diagrams the colours are shown widely separated just for clarity; the total spread on a real screen is much less than the drawing suggests, and the different colours occupy bands of different width.

The word spectrum was Newton’s own choice. It comes from the Latin word for an appearance or an image, the same root that gives us the word spectre. He also fixed the count at seven colours, adding indigo so that the list would match the seven notes of a musical scale.

6.3.2 Dispersion And Spectrum

The splitting seen in Newton’s experiment has a name of its own, dispersion, and it is worth keeping that name separate from the name of the band that appears on the screen. One is the process; the other is the result.

NoteDefinition

Dispersion The phenomenon of splitting of white light by a prism into its constituent colours is known as dispersion.

NoteDefinition

Spectrum On passing white light through a prism, the band of colours seen on a screen is called the spectrum.

Deviation and dispersion. Deviation is the turning of a ray from its original direction, and it happens even to light of a single colour. Dispersion is the separation of white light into its colours, and it happens only because different colours are deviated by different amounts.

6.3.3 Cause Of Dispersion And Formation Of The Spectrum

The cause of dispersion of white light is that light of different wavelengths travels with different speeds in a medium. Everything else follows from that one fact, in two stages, one at each face of the prism. The stages are set out in FC 6.1 and drawn in Fig. 6.3-2.

Flowchart

FC 6.1 — How A Prism Forms A Spectrum White light falls on the first surface of the prism → different colours travel at different speeds in the glass → each colour is deviated through a different angle towards the base, so dispersion takes place here → the colours follow different paths inside the glass → the separated colours strike the second surface at different angles of incidence → each colour is refracted out through a different angle, so the colours separate further → a band of colours, the spectrum, is formed on the screen

What Happens At The First Surface

When white light enters the first surface of a prism, the lights of different colours, because of their different speeds in glass, get deviated through different angles towards the base of the prism. This is the splitting of white light into its constituent colours, so the dispersion of white light takes place at the first surface of the prism.

Violet colour is deviated the most, while red colour is deviated the least, for the reason established in Section 6.1.3. From this point onwards the light of different colours present in white light follows different paths inside the glass prism.

What Happens At The Second Surface

The separated colours then strike the second surface of the prism. On this surface only refraction takes place, from glass to air — there is no fresh splitting to be done, because the colours are already travelling separately.

Since the colours arrive at this surface along different paths, they arrive at different angles of incidence. Each is therefore deviated through a different angle as it leaves. Here again violet is deviated the most and red the least, so the colours get further separated on refraction at the second surface.

The light emerging out of the prism thus has different colours spreading out at slightly different angles. Caught on a screen, they form the spectrum.

Diagram

Fig. 6.3-2 — Where The Colours Separate An enlarged ray diagram of a prism ABC, vertex A at top, base BC at bottom, showing only two colours for clarity. A single white ray strikes face AB at point Q; from Q draw two rays inside the glass, a violet one bent more towards the normal and a red one bent less, with a label beside Q reading ‘dispersion takes place here — first surface’. The two rays travel to different points on face AC. At each of those points draw the normal dashed and the emergent ray, with a label beside face AC reading ‘only refraction here — colours separate further’. Show the two emergent rays diverging, with the violet ray closer to the base BC than the red ray, and label them violet and red. Mark the angle of incidence inside the glass at face AC for each colour to show that they are different.

Three points about this process are easily lost, and they are exactly the points that are asked about.

  1. Dispersion of white light occurs only at the first surface of a prism.
  2. Deviation of light rays occurs at both the surfaces of a prism.
  3. A prism does not produce colours; it only splits the various colours already present in the light incident on it.

Q. If a prism does not produce colours, where do the colours on the screen come from?

A. From the incoming white light, which already contained all of them mixed together. The prism only sorts them out by bending each one through a different angle.

Example 6.5 — Three Colours Meeting A Right-Angled Prism

  1. Given: a beam of red, blue and yellow light falls normally on the face AB of an isosceles right-angled glass prism ABC, with the right angle at B, as drawn in Fig. 6.3-3; the critical angle of the glass–air interface for yellow colour is \(45^\circ\). To find: the path of each colour after it meets the face AC.

  2. Relations used: light falling normally on a surface passes through undeviated, since \(i = 0^\circ\) gives \(r = 0^\circ\); and \(\sin C = \dfrac{1}{\mu}\), which fixes the critical angle \(C\) for light going from glass into air. From Section 6.1.3, \(\mu_{\text{blue}} > \mu_{\text{yellow}} > \mu_{\text{red}}\).

  3. The beam meets AB at \(i = 0^\circ\), so all three colours pass in without deviation and without dispersion. Inside the glass they strike the face AC at an angle of incidence of \(45^\circ\). For yellow, \(C = 45^\circ\), so yellow is refracted along the surface AC itself, at \(r = 90^\circ\). For red, \(\mu\) is smaller, so from \(\sin C = 1/\mu\) its critical angle is greater than \(45^\circ\); red strikes AC at less than its critical angle and is refracted out of the prism, bending away from the normal, so the emergent red ray bends towards the base. For blue, \(\mu\) is larger, so its critical angle is less than \(45^\circ\); blue strikes AC at more than its critical angle and suffers total internal reflection.

  4. Red emerges through the face AC, bent towards the base of the prism. Yellow grazes along the face AC. Blue is totally internally reflected at AC, then falls normally on the base BC and emerges without deviation, at right angles to its original direction. The answer is physically sensible: the colour with the largest refractive index is the one trapped inside, which is what a smaller critical angle must mean.

Diagram

Fig. 6.3-3 — Paths Of Red, Blue And Yellow In A Right-Angled Prism An isosceles right-angled prism ABC with the right angle at B, face AB vertical on the left, base BC horizontal at the bottom, and hypotenuse AC running from the top vertex A down to C on the right. A horizontal beam labelled ‘red + blue + yellow’ arrives from the left and meets AB normally; draw it continuing straight through the glass to strike AC. At that point draw the normal to AC as a dashed line and mark the angle of incidence as 45 degrees. From that point show three separate paths: a yellow ray travelling along the face AC itself, a red ray refracting out of AC and bending away from the normal towards the base, and a blue ray totally internally reflected downwards to meet the base BC at right angles and then emerging vertically downwards out of BC. Label each emergent ray with its colour.

Q1. A triangular prism spreads white light into a band of colours, but the flat window glass you look through every day does not. What must the shape of the glass have to do with it?

A. Everything. In a prism the two refracting faces are inclined, so the bending at the second face adds to the bending at the first and the colours are pushed further apart. In window glass the two faces are parallel, so the second refraction undoes the first, as Section 6.3.4 shows.

6.3.4 Why A Parallel-Sided Glass Slab Gives No Spectrum

A rectangular glass slab is made of the same glass as a prism, and light of different colours travels through it at different speeds just the same. So dispersion does begin inside a slab. What is different is what the second surface does about it.

In a slab, refraction takes place at two parallel surfaces. At the first surface the colours are separated, because each is deviated by a different amount — that is, dispersion takes place, exactly as in a prism.

At the second surface, the faces being parallel, each colour is refracted in such a way that its emergent ray is parallel to the ray that was incident on the first surface. There is no further separation of the colours at this surface, unlike a prism, where the second surface pushes them further apart.

The emergent coloured rays are therefore parallel to one another, and they are so close together that it is difficult to see the separate colours. The light coming out of a slab appears to be almost white. This is compared with the prism in Fig. 6.3-4.

Example 6.6 — A Mixed Beam Through A Prism And Through A Slab

  1. Given: a narrow beam containing blue, red and green light falls on a triangular glass prism, and an identical beam falls on a parallel-sided glass slab, as drawn in Fig. 6.3-4. To find: what emerges in each case, and why the two results differ.

  2. Principle used: at every glass surface each colour is refracted through its own angle, because \(\mu\) is different for each colour; whether the colours end up separated depends on whether the second surface adds to the separation or cancels it.

  3. In the prism, refraction takes place at two inclined surfaces. Both refractions turn the rays towards the base, and both separate the colours further, so the prism causes dispersion together with deviation and a spectrum is seen. In the slab, refraction takes place at two parallel surfaces. The first separates the colours; the second brings each emergent ray back parallel to the original incident ray, adding no further separation.

  4. From the prism, three clearly separated coloured rays emerge, spread out and turned towards the base. From the slab, three coloured rays emerge parallel to one another and parallel to the incident beam, so close together that the emergent light looks almost white. The physics is the same in both; only the geometry of the two surfaces differs.

Diagram

Fig. 6.3-4 — A Prism Disperses, A Parallel-Sided Slab Does Not Two panels side by side, sharing one caption. Panel (a): a triangular glass prism, vertex up, base down, labelled ‘prism’; a single narrow beam labelled ‘blue + red + green’ arrives from the upper left and strikes the first face; inside the glass draw three separating rays; from the second face draw three clearly diverging emergent rays labelled red, green and blue in order, all turned towards the base, with red furthest from the base and blue nearest it. Panel (b): a rectangular glass block labelled ‘glass’, with air outside labelled ‘air’; the same beam labelled ‘blue + red + green’ arrives from the upper left; inside the block draw three slightly separated parallel rays; from the far face draw three emergent rays that are parallel to one another and parallel to the original incident beam, drawn very close together and labelled red, blue, green, with a note reading ‘emergent light appears almost white’.

Stand a small plane mirror slanting inside a shallow tray of water, in a spot where sunlight falls on it, and hold a sheet of white paper above to catch the reflected light. Look for a band of colours on the paper, and check whether the colours appear in the same order as VIBGYOR.

6.4 The Electromagnetic Spectrum

Everything so far in this chapter has concerned the spectrum of white light, the band that lies between the red and the violet regions. Our eye is sensitive only to this narrow range of wavelengths. It is easy to assume, therefore, that this range is all there is.

6.4.1 The Visible And The Invisible Spectrum

The spectrum from violet to red is called the visible spectrum, and it is only a very small part of the entire electromagnetic spectrum. Experiments show that the spectrum of radiation coming from the Sun is not limited to the region between the violet and the red extremes. It extends on either side beyond these extremes, into regions to which our eyes do not respond.

NoteDefinition

Invisible Spectrum The part of the spectrum lying beyond the red extreme and beyond the violet extreme of the visible spectrum, which our eyes cannot see, is called the invisible spectrum.

These radiations are invisible for a definite reason. The portions of electromagnetic waves other than the visible part do not excite our retina, so they produce no sensation of vision. That does not make them harmless — some electromagnetic waves can damage the retina, which is why the eye must never be pointed at a strong source of them.

The waves of wavelength longer than the red part of the visible spectrum are, in increasing order of wavelength, infrared radiations, microwaves and radio waves. The waves of wavelength shorter than the violet part are, in decreasing order of wavelength, ultraviolet rays, X-rays and gamma rays. Together with visible light, these radiations form the electromagnetic spectrum.

6.4.2 Broad Classification Of Electromagnetic Waves

The complete electromagnetic spectrum, written in increasing order of wavelength — which is the same as decreasing order of frequency — is:

  1. Gamma rays
  2. X-rays
  3. Ultraviolet rays
  4. Visible light
  5. Infrared radiations
  6. Microwaves
  7. Radio waves

These electromagnetic waves are of different wavelengths and have different properties, which is why each has its own name and its own uses. At the same time they share many properties with light waves, and Section 6.4.5 collects the properties that all of them have in common.

Increasing wavelength and increasing frequency. The two orders are exact reverses of each other, because c = fλ with c fixed. Gamma rays head the list by wavelength but come last by frequency; radio waves are the other way round. Always check which of the two the question asks for.

6.4.3 Speed, Frequency And Wavelength Of Electromagnetic Waves

However different their wavelengths, all these waves travel with the same speed, equal to \(3 \times 10^{8}\ \text{m s}^{-1}\), in vacuum or in air. For every one of them the speed, the frequency and the wavelength are tied together by

\[c = f\lambda\]

where \(c\) is the speed of the wave in m s⁻¹, \(f\) is its frequency in Hz, and \(\lambda\) is its wavelength in m. Because \(c\) is fixed, fixing either one of \(f\) and \(\lambda\) fixes the other, so a member of the electromagnetic spectrum can be named equally well by either.

The whole spectrum is laid out on this scale in Fig. 6.4-1, with wavelength increasing towards the right and frequency increasing towards the left.

Diagram

Fig. 6.4-1 — The Electromagnetic Spectrum A long horizontal chart. Along the top edge a wavelength axis labelled ‘wavelength in nm’ with an arrow pointing right, marked at 10^-2, 10, 400, 800, 10^6 and 10^9. Along the bottom edge a frequency axis labelled ‘frequency in Hz’ with an arrow pointing left, marked at 3 × 10^19, 3 × 10^16, 7.5 × 10^14, 3.75 × 10^14, 3 × 10^11 and 3 × 10^8, aligned with the matching wavelength marks above. Between the two axes divide the chart into seven bands in this order from left to right: gamma rays, X-rays, ultraviolet, visible, infrared, microwave, radio waves. Draw the visible band as a very narrow strip between 400 nm and 800 nm, shaded from violet at its left edge to red at its right edge and labelled with the letters V I S I B L E written vertically. Draw the boundaries between the bands as dashed lines, with a note under the chart: the boundaries are not sharp and neighbouring regions overlap.

Two cautions go with any such chart. First, although a boundary line is drawn between one region and the next, there is no sharp boundary of wavelength, or of frequency, between two successive electromagnetic waves; in fact they overlap to a good extent. Second, note where the two invisible neighbours of visible light sit: the infrared spectrum is the part of the spectrum just beyond the red end, and the ultraviolet spectrum is the part just before the violet end.

Example 6.7 — Naming A Wave From Its Wavelength

  1. Given: an electromagnetic wave of wavelength \(\lambda = 1\ \text{Å}\); speed of electromagnetic waves in vacuum \(c = 3 \times 10^{8}\ \text{m s}^{-1}\). To find: the frequency of the wave, and the region of the electromagnetic spectrum it belongs to.

  2. Relation used: \(c = f\lambda\), so \(f = \dfrac{c}{\lambda}\). It applies to every electromagnetic wave, since all of them travel at \(c\) in vacuum.

  3. First convert the wavelength: \(\lambda = 1\ \text{Å} = 10^{-10}\ \text{m}\). Then \(f = \dfrac{3 \times 10^{8}\ \text{m s}^{-1}}{10^{-10}\ \text{m}} = 3 \times 10^{18}\ \text{Hz}\).

  4. The frequency is \(3 \times 10^{18}\ \text{Hz}\). A wavelength of \(1\ \text{Å}\) lies between \(0.1\ \text{Å}\) and \(100\ \text{Å}\), so the wave is an X-ray, and the frequency obtained lies inside the X-ray range given in Table 6.2.

Because every electromagnetic wave travels at 3 × 10⁸ m s⁻¹, a radio signal sent to an astronaut on the Moon takes about 1.3 s to reach there, the Moon being roughly 3.8 × 10⁸ m away. No conversation across that distance can ever be quicker than that.

6.4.4 Approximate Ranges Of Wavelength And Frequency

The seven regions are set out with their limits in Table 6.2. Each row’s frequency limits follow from its wavelength limits through \(f = c/\lambda\), so the two halves of the table say the same thing in two ways.

Table 6.2 — Approximate ranges of wavelength and frequency in the electromagnetic spectrum

Radiation Wavelength range The same range in nm, mm or m Frequency range
Gamma rays shorter than 0.1 Å shorter than 0.01 nm above \(10^{19}\ \text{Hz}\)
X-rays 0.1 Å to 100 Å 0.01 nm to 10 nm \(3 \times 10^{19}\) to \(3 \times 10^{16}\ \text{Hz}\)
Ultraviolet rays 100 Å to 4000 Å 10 nm to 400 nm \(3 \times 10^{16}\) to \(7.5 \times 10^{14}\ \text{Hz}\)
Visible light 4000 Å to 8000 Å 400 nm to 800 nm \(7.5 \times 10^{14}\) to \(3.75 \times 10^{14}\ \text{Hz}\)
Infrared radiations 8000 Å to \(10^{7}\) Å 800 nm to 1 mm \(3.75 \times 10^{14}\) to \(3 \times 10^{11}\ \text{Hz}\)
Microwaves \(10^{7}\) Å to \(10^{10}\) Å 1 mm to 1 m \(3 \times 10^{11}\) to \(3 \times 10^{8}\ \text{Hz}\)
Radio waves above \(10^{10}\) Å longer than 1 m below \(3 \times 10^{8}\ \text{Hz}\)

Read the table downwards and the wavelength grows enormously at every step, from a fraction of the size of an atom at the top to more than the height of a person at the bottom. The visible strip that Sections 6.1 to 6.3 were entirely about occupies one narrow row in the middle.

Each region also has its characteristic sources and its own way of being detected, and these are collected in Table 6.3. A radiation you cannot see has to be detected by what it does — by the heat it produces, the chemical change it causes, the fluorescence it triggers, or the current it sets up in an aerial.

Table 6.3 — Sources, discovery and detection of the electromagnetic waves, in order of increasing wavelength

Name of the wave Wavelength in nm Discoverer or pioneer Source Method of detection
Gamma rays below 0.01 Villard In cosmic rays, and from radioactive substances By their large penetrating power
X-rays 0.01 – 10 Roentgen From a heavy metal target of high melting point when highly energetic electrons are stopped by it By the fluorescence produced on a zinc sulphide screen; the photographic film gets affected
Ultraviolet 10 – 400 Ritter Sunlight, arc lamp, spark By their chemical activity on dyes; photographic plates get affected
Visible light 400 – 800 Newton Sunlight, light from an electric bulb, flame, white hot bodies It causes fluorescence; other objects can be seen in its presence
Infrared waves 800 – \(10^{6}\) Herschel Lamp with thoriated filament, heated silicon carbide rod, red hot bodies Heating effect is more; the mercury rises rapidly when a thermometer with a blackened bulb is kept in these radiations
Microwaves \(10^{6}\) – \(10^{9}\) Hertz Electronic devices such as crystal oscillators Oscillatory electrical circuit
Radio waves above \(10^{9}\) Marconi TV and radio transmitters Aerials of radio and TV receivers

Example 6.8 — Wavelength Of A Radio Broadcast

  1. Given: a radio station transmits at a frequency \(f = 100\ \text{MHz}\); speed of electromagnetic waves in air \(c = 3 \times 10^{8}\ \text{m s}^{-1}\). To find: the wavelength of the transmitted wave, and a check that it is indeed a radio wave.

  2. Relation used: \(c = f\lambda\), so \(\lambda = \dfrac{c}{f}\).

  3. Convert the frequency first, since the speed is in m s⁻¹: \(1\ \text{MHz} = 10^{6}\ \text{Hz}\), so \(f = 100 \times 10^{6}\ \text{Hz} = 1 \times 10^{8}\ \text{Hz}\). Then \(\lambda = \dfrac{3 \times 10^{8}\ \text{m s}^{-1}}{1 \times 10^{8}\ \text{Hz}} = 3\ \text{m}\).

  4. The wavelength is \(3\ \text{m}\). It is longer than \(1\ \text{m}\), and the frequency is below \(3 \times 10^{8}\ \text{Hz}\), so by Table 6.2 the wave is a radio wave, exactly as it should be for a broadcast station.

6.4.5 Properties Common To All Electromagnetic Waves

Different as the seven regions are, five properties belong to every member of the electromagnetic spectrum.

  1. Electromagnetic waves do not require any material medium for their propagation. This is why sunlight crosses the empty space between the Sun and the Earth, while sound cannot.
  2. They all travel with the same speed in vacuum and in air, namely \(3 \times 10^{8}\ \text{m s}^{-1}\), and with different speeds in different media.
  3. They exhibit the properties of reflection and refraction. In refraction, when an electromagnetic wave passes from one medium to the other, there is a change in its direction of travel, its speed and its wavelength, but its frequency remains unchanged.
  4. These waves are not deflected by electric and magnetic fields.
  5. These waves are transverse in nature, the vibrations being at right angles to the direction in which the wave travels.

The third property is the hinge of this whole chapter. The frequency of a wave is set by the source that produced it, and no medium alters it. So when light enters glass and slows down, the only quantity that can change to keep \(c = f\lambda\) true is the wavelength, which shortens. That change of speed inside the glass, different for each colour, is what produced the spectrum in Section 6.3.

The fourth property is a useful test in a laboratory. A beam that a magnet or a charged plate cannot bend aside is not a stream of charged particles; it is an electromagnetic wave.

The electromagnetic waves used for cooking in a microwave oven have a frequency of 2450 MHz. Calculate their wavelength in metres and in centimetres, and use Table 6.2 to confirm that this wavelength lies in the microwave region. Take the speed of electromagnetic waves as 3 × 10⁸ m s⁻¹.

6.5 Properties And Uses Of Different Radiations Of The Electromagnetic Spectrum

Table 6.2 gave each region of the electromagnetic spectrum a range of wavelengths. This section gives each one a character: where it comes from, what it does, how it is detected, and what we use it for. Work down the list and you are moving steadily from the shortest wavelengths to the longest.

6.5.1 Gamma Rays

Gamma rays are the most energetic electromagnetic radiations, of wavelength less than \(0.1\ \text{Å}\), that is, less than \(0.01\ \text{nm}\).

They come from deep inside the atom. Gamma rays are obtained in radioactive emissions, when the nuclei of radioactive atoms pass from the excited state to the ground state. They are also present in cosmic radiations reaching the Earth from space.

Properties

Like X-rays, gamma rays cause fluorescence — the glow given out by certain materials, such as zinc sulphide, when radiation strikes them. They can easily penetrate through thick metallic sheets; an iron sheet 30 cm thick is not enough to stop them.

That penetrating power is exactly what makes them dangerous. Gamma radiations pass easily through the human body and cause immense biological damage, so sources of gamma radiation are handled remotely and stored behind thick lead or concrete shielding.

Uses

Gamma rays are used in a controlled manner in medical science to kill cancer cells, a treatment called radiotherapy. In industry they are used to check welding, since a flaw hidden inside a welded joint changes how much radiation gets through.

6.5.2 X-Rays

X-rays are obtained from a heavy metal target of high melting point, when a highly energetic electron beam, that is, cathode rays, is stopped by it. The target must have a high melting point because most of the energy of the electron beam arrives as heat. X-rays have wavelength in the range \(0.1\ \text{Å}\) to \(100\ \text{Å}\), or \(0.01\ \text{nm}\) to \(10\ \text{nm}\).

Properties

X-rays are chemically more active than ultraviolet radiations, and they strongly affect a photographic plate. Like gamma rays, they cause fluorescence in certain materials such as zinc sulphide.

The property that made them famous is selective penetration: X-rays can penetrate through human flesh, but they are stopped by the bones. A photographic film held behind the body is darkened everywhere except in the shadow of a bone, which is why the bone appears light against a dark background in Fig. 6.5-1.

Diagram

Fig. 6.5-1 — Why A Bone Shows Up In An X-Ray Photograph A simple cross-section. On the left an X-ray tube emitting a broad beam of parallel arrows labelled ‘X-rays’ travelling to the right. In the middle a forearm drawn in section, with the soft flesh drawn light and a bone drawn as a solid oval inside it. On the right a vertical photographic film. Show the arrows passing straight through the flesh and reaching the film, and the arrows meeting the bone being stopped, with a label ‘stopped by bone’. On the film, shade the region reached by the rays dark and leave the shadow of the bone light, labelling the two regions ‘film darkened’ and ‘shadow of the bone’.

Uses
  1. For the detection of fracture in bones, teeth and so on — this is radiography.
  2. For diagnostic purposes such as the CAT scan in medical science.
  3. For studying the arrangement of atoms in crystals, as well as in complex molecules.
  4. By detective agencies, to detect concealed precious metals.

Why Everyone Steps Behind A Screen X-rays affect living tissue as well as photographic film, so a hospital keeps the dose as small as the picture allows. The radiographer steps behind a lead-lined screen before making the exposure, and a lead apron may be laid over the parts of the patient’s body that are not being examined, because lead is dense enough to absorb X-rays strongly. A single radiograph is a very small dose; the precautions exist because the staff take such pictures all day, every day.

X-rays and gamma rays. Their wavelength ranges nearly touch, and both are penetrating and both cause fluorescence. The difference is where they are born: X-rays are produced outside the nucleus, when fast electrons are stopped by a metal target, while gamma rays come out of the nucleus itself.

6.5.3 Ultraviolet Radiations

The electromagnetic radiations of wavelength from \(100\ \text{Å}\) to \(4000\ \text{Å}\), that is, from \(10\ \text{nm}\) to \(400\ \text{nm}\), are called ultraviolet radiations. This part of the spectrum was first detected by Prof. J. Ritter in 1801.

How They Were Detected

Ritter’s method used a chemical rather than a thermometer, and it is drawn in Fig. 6.5-2. A silver chloride solution is exposed to the electromagnetic waves of a spectrum, starting from the red end, moving towards the violet end, and then beyond it.

From the red to the violet end, the solution remains almost unaffected. But just beyond the violet end, where the eye sees nothing at all, the solution first turns violet and then finally becomes dark brown, or black.

That colour change shows that there exist certain radiations beyond the violet extreme of the visible part, which are chemically more active than the visible light. These radiations are called ultraviolet radiations, or actinic rays. In general, ultraviolet radiations are detected by their chemical activity on dyes and on photographic plates.

Diagram

Fig. 6.5-2 — Detecting Ultraviolet Radiation Beyond The Violet End A prism at the left casting a horizontal spectrum across the figure, the band shaded red at one end through to violet at the other, with a dark region drawn continuing beyond the violet edge and labelled ‘no visible colour here’. Below the band place four small identical dishes of silver chloride solution: one under the red part, one under the green part, one under the violet part, and one under the dark region beyond violet. Shade the first three dishes pale and unchanged, labelled ‘almost unaffected’, and shade the fourth dish dark brown, labelled ‘turns violet, then dark brown’.

There is one practical difficulty in studying these radiations. The spectrum of ultraviolet radiations is obtained by passing the radiations through a quartz prism in place of a glass prism, because glass absorbs these radiations.

Sources

The electric arc and sparks give ultraviolet radiations. A mercury vapour lamp emits radiations, a part of which is ultraviolet, along with the visible light.

The Sun is also a source of ultraviolet radiations, but a large fraction of it is absorbed by the ozone layer present in the Earth’s upper atmosphere, which protects us from its harmful effects. Because of large-scale pollution there is a depletion of this ozone layer, and that is dangerous for us, since greater exposure to ultraviolet radiations is harmful.

Properties
  1. Ultraviolet radiations can pass through quartz, but they are absorbed by glass. This is why a quartz prism is used instead of a glass prism, and why ultraviolet bulbs have an envelope made of quartz instead of glass.
  2. They travel in a straight line with a speed of \(3 \times 10^{8}\ \text{m s}^{-1}\) in air or vacuum.
  3. They are usually scattered by the dust particles present in the Earth’s atmosphere.
  4. They obey the laws of reflection and refraction.
  5. They strongly affect a photographic plate, as they are chemically more active than visible light.
  6. They produce fluorescence on striking a zinc sulphide screen.
Harmful Effect And Uses

Ultraviolet radiations cause health hazards like skin cancer if the human body is exposed to them for a long period. This is the same radiation that a sunscreen and a wide-brimmed hat are meant to keep off the skin.

Used deliberately and in the right place, they are valuable. Ultraviolet radiations are used for sterilising air and surgical equipment; for detecting the purity of gems, eggs and ghee; and in producing vitamin D in the food of plants and animals.

Q2. Herschel found heating beyond the red edge of the band. If something can also lie beyond the violet edge, how could anyone ever detect it, since the eye cannot see it and it may not warm a thermometer?

A. By its chemical action instead of its heating action. Ritter placed silver chloride solution just beyond the violet edge in 1801 and it darkened, although the same solution under the visible colours barely changed. An invisible radiation can always be found by what it does to something else.

6.5.4 Visible Light

The electromagnetic radiations of wavelength from \(4000\ \text{Å}\) to \(8000\ \text{Å}\) are called visible radiations, or visible light. They are called visible because it is in the presence of these radiations that other objects are seen by us.

The visible part of the spectrum was discovered by Newton, while passing sunlight through a glass prism, as described in Section 6.3.1. Its prominent colours are violet, indigo, blue, green, yellow, orange and red.

The main sources of visible light are the Sun, an electric bulb, a flame and white hot bodies. Visible light is used in photography, in photosynthesis, and to see the objects around us.

6.5.5 Infrared Radiations

Infrared radiations are electromagnetic waves of wavelength in the range \(8000\ \text{Å}\) to \(10^{7}\ \text{Å}\), that is, \(800\ \text{nm}\) to \(1\ \text{mm}\). This part of the spectrum was first detected by William Herschel in about 1800, in the experiment described at the opening of this chapter.

How They Are Detected

The method Herschel used is still the standard one. If a thermometer having its bulb blackened is moved from the violet end towards the red end of the spectrum of visible light, there is a very slow rise in temperature. But when this thermometer is moved beyond the red extreme, a rapid rise in temperature is noticed.

It means that the part of the spectrum beyond the red extreme of the visible light, although not visible to us, has certain radiations which produce a strong heating effect. These radiations are called infrared, or heat, radiations.

A thermopile can also be used to detect these heat radiations. It is a device that produces a small electric current when heat radiation falls on it, and the galvanometer connected with the thermopile shows a deflection when infrared radiations fall on it, as in Fig. 6.5-3.

Diagram

Fig. 6.5-3 — Detecting Infrared Radiation With A Thermopile A prism on the left with a horizontal spectrum spreading to the right, shaded violet along its top edge through to red along its bottom edge, and a dark unshaded region continuing just below the red edge, labelled ‘infrared region’. Place the face of a thermopile, drawn as a small cone-shaped device on a stand, in that dark region. Connect the two terminals of the thermopile by wires to a galvanometer drawn as a circle with a pointer and a central zero scale, with the pointer shown deflected to one side. Label the thermopile, the galvanometer and the deflection.

The spectrum of infrared radiations is obtained by using a rock-salt prism, because a rock-salt prism does not absorb infrared radiations, whereas a glass prism absorbs them. Notice the neat contrast with ultraviolet, where glass was equally useless but quartz was the answer.

Sources

All red hot bodies, such as a heated iron ball, a flame or a fire, are sources of infrared radiations. The Sun is the natural source of infrared radiations.

Properties
  1. They travel in straight lines like light, with the same speed as light, that is \(3 \times 10^{8}\ \text{m s}^{-1}\) in vacuum or air.
  2. They obey the laws of reflection and refraction. If a source of heat, say an infrared lamp, is placed at the focus of a parabolic mirror, a parallel infrared beam is obtained. A burning glass, that is, a convex lens, focuses the more energetic, shorter-wavelength infrared radiations from the Sun on to a paper, and the paper chars or burns.
  3. They do not affect ordinary photographic film; however, a specially treated photographic film is affected by them.
  4. They are absorbed by glass, but they pass through rock-salt.
  5. They are detected by their heating property, using a blackened bulb thermometer or a thermopile.
  6. They are scattered less by the Earth’s atmosphere because of their long wavelength, since the intensity of the scattered radiation is proportional to \(1/\lambda^{4}\). Hence they are able to penetrate deep inside the atmosphere even in fog.
  7. The green house gases, such as the carbon dioxide present in the Earth’s atmosphere, absorb the low energy infrared radiations and keep the Earth’s surface warm.

Property 6 will return in Section 6.7, where the same \(1/\lambda^{4}\) dependence explains the colour of the sky.

Harmful Effect And Uses

A high dose of infrared radiations may cause skin burns, which is why one does not hold a hand close to a glowing heater element for long.

  1. Infrared radiations are used for therapeutic purposes by doctors.
  2. They are used in photography at night, and also in mist and fog, because they are not much scattered and so can penetrate appreciably through it.
  3. Infrared lamps are used in dark rooms for developing photographs, as they provide some visibility without affecting the photographic film.
  4. They are used as signals during war, as they are not visible and are not absorbed much in the medium.
  5. They are used in the remote control of television and other gadgets.

Point a television remote at the camera of a mobile phone, look at the phone’s screen, and press a button on the remote. The screen shows a flash at the tip of the remote that your eye cannot see directly, because the camera responds to infrared while the eye does not.

6.5.6 Microwaves

Microwaves have wavelength in the range \(10^{7}\ \text{Å}\) to \(10^{10}\ \text{Å}\), that is, \(1\ \text{mm}\) to \(1\ \text{m}\), or frequency in the range \(3 \times 10^{11}\ \text{Hz}\) to \(3 \times 10^{8}\ \text{Hz}\).

These waves are produced by electronic devices such as crystal oscillators. They are used for satellite communication, for the analysis of atomic and molecular structure, for cooking in microwave ovens, and in radar communication.

6.5.7 Radio Waves

Radio waves are the waves of longest wavelength amongst all the electromagnetic waves. They have wavelength above \(1\ \text{m}\), that is, above \(10^{10}\ \text{Å}\), or frequency below \(3 \times 10^{8}\ \text{Hz}\). They also show all the properties of electromagnetic waves listed in Section 6.4.5.

These waves are used mainly in radar communication, and also in radio and television transmission.

Example 6.9 — Naming Three Radiations From Their Wavelengths

  1. Given: three electromagnetic radiations of wavelength (i) \(0.005\ \text{nm}\), (ii) \(250\ \text{nm}\) and (iii) \(1500\ \text{nm}\). To find: the name of each radiation and one use of each.

  2. Method used: compare each wavelength with the ranges of Table 6.2, since a radiation of the electromagnetic spectrum is identified by nothing but its wavelength. All three are already in the same unit, so they can be compared directly.

    1. \(0.005\ \text{nm}\) is below \(0.01\ \text{nm}\), so it lies in the gamma ray region. (ii) \(250\ \text{nm}\) lies between \(10\ \text{nm}\) and \(400\ \text{nm}\), so it lies in the ultraviolet region. (iii) \(1500\ \text{nm}\) lies between \(800\ \text{nm}\) and \(10^{6}\ \text{nm}\), so it lies in the infrared region.
  3. The radiations are (i) gamma rays, used in a controlled manner to kill cancer cells, (ii) ultraviolet radiations, used for sterilising surgical equipment, and (iii) infrared radiations, used in the remote control of a television. All three lie outside the visible range, which stretches only from \(400\ \text{nm}\) to \(800\ \text{nm}\), so none of them can be seen.

A gamma ray has a frequency of 3 × 10²⁰ Hz. Calculate its wavelength in metres and in angstroms, and show from Table 6.2 that the value obtained does fall in the gamma ray region. Take the speed of electromagnetic waves as 3 × 10⁸ m s⁻¹.

6.6 Distinction Between Ultraviolet, Visible And Infrared Radiations

These three regions are neighbours in the electromagnetic spectrum. Visible light lies in the middle, ultraviolet radiation sits just beyond its violet edge, and infrared radiation just beyond its red edge. Being neighbours, they are easily confused, and the differences between them are set out in Table 6.3.

Read the table by rows rather than by columns. Each row asks one question — how long is the wavelength, can the eye see it, does it heat, does it darken a photographic plate, does it make a screen glow, does it harm the body, what glass will it pass through — and each row gives the answer for all three at once.

Table 6.3 — Ultraviolet, Visible And Infrared Radiations Compared | Property | Ultraviolet radiations | Visible radiations | Infrared radiations | |—|—|—|—| | Wavelength range | 100 Å to 4000 Å | 4000 Å to 8000 Å | 8000 Å to 10⁷ Å | | Effect on the eye | They are invisible | They are visible | They are invisible | | Heating effect | They produce no heating effect | They produce slight heating effect | They produce a strong heating effect | | Effect on a photographic plate | They affect photographic plate | They affect photographic plate | They do not affect photographic plate | | Fluorescence | They cause fluorescence on a zinc sulphide screen | They do not cause fluorescence | They do not cause fluorescence | | Effect on the human body | They cause health hazards like skin cancer | They do not affect the body | They do not affect the body, but a high dose may cause skin burns | | Passage through materials | They can pass through quartz, but they do not pass through glass; they are absorbed by glass | They can pass through glass; they are not absorbed by glass | They can pass through rock-salt, but they do not pass through glass; they are absorbed by glass |

The last row is the one that decides what a piece of apparatus must be made of. A glass prism will give you the visible spectrum but will swallow the radiations on either side of it, so the ultraviolet spectrum needs a quartz prism and the infrared spectrum needs a rock-salt prism.

For all their differences, these three regions remain members of one family. All three travel at \(3 \times 10^{8}\ \text{m s}^{-1}\) in air, all three need no material medium, all three obey the laws of reflection and refraction, and all three are transverse waves undeflected by electric and magnetic fields.

Q. Why does a car left standing in the sun with its windows shut become so hot inside?

A. Visible light and short infrared pass in through the glass and warm the seats. The warm seats give out longer infrared, which glass absorbs rather than lets out, so the energy is trapped inside.

A radiation passes easily through a rock-salt prism but is stopped by a glass plate, produces no fluorescence on a zinc sulphide screen, and leaves an ordinary photographic film unaffected. Name the radiation, state the range of wavelength it belongs to, and give one method by which its presence can be detected.

6.7 Scattering Of Light

The last part of this chapter needs no prism at all. The atmosphere itself sorts the colours of sunlight, and it does so by a different mechanism from the one a prism uses.

6.7.1 What Scattering Of Light Means

When light from the Sun enters the earth’s atmosphere, it gets scattered by the dust particles and air molecules present in the atmosphere, that is, the light spreads out in all directions instead of continuing in one straight beam.

You have seen the evidence indoors. If sunlight enters a darkened room through a fine hole, you can see the path of the beam stretching across the room, as in Fig. 6.7-1. You are not looking at the beam itself; you are looking at light that dust particles lying in its path have thrown sideways towards your eye. Where the air is free of dust, the beam is invisible from the side.

Diagram

Fig. 6.7-1 — Seeing A Beam Of Sunlight By Its Scattered Light A darkened room drawn in section. On the left wall a small hole in a closed shutter, with sunlight entering through it as a narrow straight beam that crosses the room and makes a bright patch on the floor. Along the beam show many tiny dots labelled ‘dust particles’, and from several of these dots draw short arrows going outwards in many directions, labelled ‘scattered light’. On the right draw an observer standing to one side of the beam, with a dashed line from one dust particle to the observer’s eye. Label the beam ‘straight beam of sunlight’ and add a note: the observer sees the beam only because light is scattered sideways out of it.

The scattering of light was first studied by the scientist Rayleigh.

NoteDefinition

Scattering Of Light Scattering is the process of absorption and then re-emission of light energy by the dust particles and air molecules present in the atmosphere.

The definition says absorption first, then re-emission, and the order matters. The air molecules of size smaller than the wavelength of the incident light absorb the energy of the incident light and then re-emit it without any change in its wavelength. The colour of the light is not altered by being scattered; only its direction is.

In 1871, Lord Rayleigh in England took up a question that had no accepted answer: why is the sky blue and not simply bright white like the Sun? He worked out mathematically what happens when light falls on particles much smaller than its own wavelength, and found that the amount thrown sideways depends very sharply on wavelength — as the inverse fourth power of it. The sharpness of that dependence is the whole explanation. A modest difference in wavelength between violet and red becomes a large difference in how much of each colour the sky sends to our eyes.

Dispersion and scattering. Dispersion separates the colours of white light by refraction through a prism, and every colour continues forward. Scattering separates them by throwing some colours sideways more than others, and needs no prism at all.

6.7.2 Dependence Of Scattering On Wavelength

The scattering of light is not the same for all wavelengths of incident light. The intensity of the scattered light is found to be inversely proportional to the fourth power of the wavelength of light:

\[I \propto \frac{1}{\lambda^{4}}\]

where \(I\) is the intensity of the scattered light and \(\lambda\) is the wavelength of the light in m. This relation holds only when the scattering particles are smaller in size than the wavelength of the incident light.

A fourth power is a steep dependence. Halve the wavelength and the scattered intensity does not merely double — it becomes sixteen times as large. That steepness is drawn in Fig. 6.7-2.

Diagram

Fig. 6.7-2 — How Scattered Intensity Falls With Wavelength A graph. The x-axis is labelled ‘wavelength λ in Å’ and runs from 4000 to 8000, marked at 4000, 5000, 6000, 7000 and 8000, with the colours of the spectrum shown as a thin coloured strip along the axis from violet at 4000 to red at 8000. The y-axis is labelled ‘intensity of scattered light I in arbitrary units’ and runs from 0 to 16, marked at 1, 4, 8, 12 and 16. Plot a smooth curve of I against λ that starts at the value 16 when λ is 4000 Å and falls steeply, passing through about 4.9 at 6000 Å and reaching the value 1 at 8000 Å, flattening as it goes. Label the curve ‘I is proportional to 1/λ⁴’. Mark the point at 4000 Å ‘violet, scattered most’ and the point at 8000 Å ‘red, scattered least’.

Fig. 6.7-2 — How Scattered Intensity Falls With Wavelength The horizontal axis carries the wavelength of the light, in angstroms, and it covers exactly the visible range, 4000 Å at the left to 8000 Å at the right. The vertical axis carries the intensity of the light scattered by the air, in arbitrary units, chosen here so that red light comes out as 1. Start at the right-hand end of the curve. At \(\lambda = 8000\ \text{Å}\), which is red, the height of the curve is 1 unit. Now move left along the axis to \(\lambda = 4000\ \text{Å}\), which is violet, and read the height there: 16 units. So violet light is scattered nearly 16 times more than red light, and the graph is only restating \(I \propto 1/\lambda^{4}\), since halving the wavelength multiplies the intensity by \(2^{4} = 16\). Next, look at the shape rather than the two end values. The curve is steepest at the left, near the violet end, and flattens towards the right. That means a small change of wavelength among the blues and violets changes the scattering a great deal, while the same change among the oranges and reds hardly matters. There is no meaning to the area under this curve, and none is asked for. What the graph is for is comparing heights at two wavelengths, and the height at any point is simply how much of that colour the sky throws sideways.

In the visible spectrum, the wavelength of violet light is the least, about \(4000\ \text{Å}\), and that of red light is the most, about \(8000\ \text{Å}\). Therefore, from the incident white light, violet light is scattered the most and red light is scattered the least — violet light being scattered nearly 16 times more than the red light.

As a result, light from the Sun when it reaches the earth’s surface has less intensity of light of the violet end, such as blue, indigo and violet, and more intensity of light of the red end, such as orange and red. Whatever is missing from the direct beam has not vanished; it has been thrown sideways into the sky.

Example 6.10 — Comparing The Scattering Of Blue And Red Light

  1. Given: wavelength of blue light \(\lambda_{\text{blue}} = 4800\ \text{Å}\); wavelength of red light \(\lambda_{\text{red}} = 7200\ \text{Å}\); the scattering particles in the air are smaller than these wavelengths. To find: how many times more strongly blue light is scattered than red light.

  2. Relation used: \(I \propto \dfrac{1}{\lambda^{4}}\), which applies because the scattering particles are smaller than the wavelength of the light.

  3. Taking the ratio for the two colours, \(\dfrac{I_{\text{blue}}}{I_{\text{red}}} = \left(\dfrac{\lambda_{\text{red}}}{\lambda_{\text{blue}}}\right)^{4} = \left(\dfrac{7200\ \text{Å}}{4800\ \text{Å}}\right)^{4} = (1.5)^{4} = 5.06\). The unit Å cancels inside the bracket, as it must, since the answer is a ratio.

  4. Blue light is scattered about \(5.06\) times as strongly as red light. The answer is sensible: it is less than the factor of 16 that separates violet from red, because these two wavelengths are closer together than \(4000\ \text{Å}\) and \(8000\ \text{Å}\).

When The Particles Are Large

Everything above assumed particles smaller than the wavelength of light. If they are not, the outcome is completely different.

The air molecules and particles of size bigger than the wavelength of the incident light scatter the light of all wavelengths of white light to the same extent. No colour is favoured, so the scattered light remains a mixture of all colours — and a mixture of all colours is white. This single sentence explains several of the everyday sights in the next section.

I ∝ 1/λ⁴ The shorter the wavelength, the more strongly that colour is thrown sideways out of a beam — and it grows not gently but as the fourth power. Light of half the wavelength is scattered sixteen times as much. It holds only for particles smaller than the wavelength of the light.

Stir two or three drops of milk into a glass of water and shine a torch through it in a dim room. Look at the glass from the side, then look straight through it at the torch, and note which of the two views looks bluish and which looks orange.

6.8 Some Applications Of Scattering

One relation, \(I \propto 1/\lambda^{4}\), together with one exception for large particles, is enough to explain a whole set of everyday sights. Some effects of scattering of sunlight by the earth’s atmosphere are:

  1. the red colour of the Sun at sunrise and sunset,
  2. the white colour of the sky at noon,
  3. the blue colour of the sky,
  4. the black colour of the sky in the absence of atmosphere,
  5. the white colour of clouds, and
  6. the use of red light for the danger signal.

6.8.1 Red Colour Of The Sun At Sunrise And Sunset

At the time of sunrise and sunset, light from the Sun has to travel the longest distance of atmosphere to reach the observer. The Sun is then near the horizon, so its light comes in almost along the ground rather than straight down, and the length of air it must cross is at its greatest. Compare the two paths drawn in Fig. 6.8-1.

Over that long path, the scattering has time to take its toll. Since the blue light of short wavelength is scattered more, much of it is lost from the beam, while the red light of long wavelength is scattered a little, so it is not much lost.

Thus blue light is almost absent in the sunlight reaching the observer, and mainly the light of the red end reaches us. As a result, the Sun and the region nearby it are seen red. The chain of reasoning is set out in FC 6.2.

Flowchart

FC 6.2 — Why The Sun Looks Red At Sunrise And Sunset Sun near the horizon → its light travels the longest path through the atmosphere → blue and violet, having the shortest wavelengths, are scattered strongly out of the beam and lost → the beam that survives the journey is left rich in orange and red → the Sun and the sky near it are seen red

Diagram

Fig. 6.8-1 — Red Colour Of The Sun At Sunrise And Sunset The Sun drawn at the left with rays travelling to the right towards the Earth, which is drawn as a circle surrounded by a shaded ring labelled ‘earth’s atmosphere’. Draw one ray reaching an observer standing at the top of the Earth, where the Sun is overhead, and mark this short path through the atmosphere ‘shortest path — noon’. Draw a second ray reaching an observer standing at the edge of the Earth’s disc, where the Sun is on the horizon, and mark this long slanting path through the atmosphere ‘longest path — sunrise and sunset’. Along the long path show many short arrows leaving the beam sideways, labelled ‘scattered blue light’, and label the beam that continues to the observer ‘red light reaches the observer’. Label both observers.

6.8.2 White Colour Of The Sky At Noon

At noon, the Sun is directly above our head, so we get light rays directly from the Sun after travelling the shortest distance, without much scattering of any particular colour.

Since no one colour has been preferentially removed, the light arriving from the direction of the Sun is still the full mixture. Hence the sky in that direction is seen white.

6.8.3 Blue Colour Of The Sky

Light from the Sun has to travel a long distance of the earth’s atmosphere before reaching us. As light travels through the atmosphere, it gets scattered in different directions by the air molecules present in its path.

The blue, or violet, light due to its short wavelength is scattered more as compared to the red light of long wavelength. Thus, the light reaching our eye directly from the Sun is rich in red colour, while the light reaching our eye from all other directions is the scattered blue light, as Fig. 6.8-2 shows.

Therefore the sky in a direction other than the direction of the Sun is seen blue. Look away from the Sun and every bit of sky you see is sending you second-hand light, thrown sideways by molecules of air.

Further, for some time after sunset and before sunrise, there is no sunlight reaching directly to us. Thus, at these hours the blue scattered light reaching our eyes makes the entire sky appear blue.

Diagram

Fig. 6.8-2 — Why The Sky Is Blue A person standing on the ground, looking upwards, with the Sun drawn low in the sky to one side. Show a broad band of sunlight entering the atmosphere and passing across the sky above the observer, with many small dots along it labelled ‘air molecules’. From several of these dots draw short blue arrows going down towards the observer’s eye, labelled ‘scattered blue light from every direction’. Draw one straight line from the Sun directly to the observer, labelled ‘direct light, rich in red’. Add labels ‘looking towards the Sun — light is reddened’ and ‘looking anywhere else — light is blue’.

Q4. The sky overhead is blue in the afternoon, yet the Sun near the horizon at sunset looks red. Could one single reason explain both of these at once?

A. Yes — both come from blue light being scattered far more strongly than red. The blue that is scattered out of the beam has to go somewhere, and it arrives at our eyes from every part of the sky, making the sky blue. The beam it was scattered out of is left short of blue, and so the low Sun looks red.

6.8.4 Black Colour Of The Sky In The Absence Of Atmosphere

If there would have been no atmosphere, there would be no scattering of light, and no scattered light would reach our eyes. Thus the sky would appear black instead of blue.

Since there is no atmosphere on the Moon, no scattered sunlight reaches the Moon’s surface. Hence, to an observer on the surface of the Moon, no light reaches the eyes of the observer except the light reaching directly from the Sun. Thus, the sky in a direction other than that of the Sun will appear black to an observer on the Moon’s surface.

Similarly, when an astronaut goes above the atmosphere of the earth in a rocket, he or she sees the sky black. But to the astronaut the earth appears blue, due to the blue colour of sunlight scattered by the earth’s atmosphere reaching him or her. The stars and other heavenly bodies are seen as usual, but without twinkling.

Diagram

Fig. 6.8-3 — The Sky Seen From Above The Atmosphere Two panels side by side. Panel (a): an observer standing on the ground on Earth, with the shaded band of atmosphere above, the Sun in one corner, and the sky drawn light blue everywhere except the Sun; label it ‘from the earth’s surface — the sky is blue’. Panel (b): an astronaut floating above the shaded band of atmosphere, with the sky drawn completely black, the Sun shown as a bright disc, stars drawn as steady points, and the curved earth below drawn blue; label it ‘above the atmosphere — the sky is black, the earth looks blue’. Add a small note in panel (b): stars do not twinkle here.

6.8.5 White Colour Of Clouds

The clouds are nearer to the earth’s surface, and they contain dust particles and aggregates of water molecules of size bigger than the wavelength of visible light. This is the exception met at the end of Section 6.7.2, and it changes the answer completely.

Therefore the dust particles and tiny ice particles present in clouds scatter all colours of incident white light from the Sun to the same extent. Hence, when this scattered light reaches our eyes, the clouds are seen white.

On a sea voyage in 1921, the Indian physicist C. V. Raman refused to accept that the deep blue of the Mediterranean was merely the sky reflected in it. His later work showed that the water molecules themselves scatter light, and led on to the discovery of the Raman effect, for which he received the Nobel Prize in 1930.

6.8.6 Use Of Red Light For The Danger Signal

In visible light, the wavelength of red light is the longest. Therefore the light of red colour is scattered the least by the air molecules of the atmosphere.

Hence the light of red colour, as compared to the light of other colours, can penetrate to a longer distance without becoming weak. Thus red light can be seen from the farthest distance in comparison to the light of other colours having the same intensity.

Hence red light is used for the danger signal, so that the signal may be visible from a far distance even in fog and similar conditions. Similarly, the remote control of electrical gadgets such as a television also makes use of infrared radiations, which lie beyond red and are scattered even less.

Q5. Danger signals, stop lights and railway signals are red almost everywhere in the world. Is there a reason from physics for that choice, or is it only a habit?

A. There is a reason. Red has the longest wavelength in the visible spectrum, so the air scatters it the least. A red lamp therefore keeps its brightness over a longer distance, and stays visible through mist and fog, where a lamp of any other colour of the same power would already have faded.

On a clear day, compare the colour of the sky straight overhead with the colour near the horizon, and look again about half an hour before sunset. Note where the blue is deepest and where it turns pale. Never look directly at the Sun, at any time of day.

A ship’s lookout can see a red lamp from a greater distance than a green lamp of the same power, and the difference grows larger on a hazy night. Give a reason for both observations, naming the property of light on which your answer depends.

6.9 Summary

  1. A prism is a transparent block bounded by two plane refracting surfaces inclined to each other. The angle between these two surfaces is the angle of prism \(A\), measured in degrees, and the surface opposite it is the base.
  2. A ray passing through a prism is deviated at both surfaces, and both deviations are towards the base, so they add: \[\delta = \delta_1 + \delta_2\] where \(\delta_1\) is the deviation at the first surface, \(\delta_2\) that at the second surface, and \(\delta\) the total angle of deviation, all in degrees.
  3. The angle of deviation is the angle between the direction of the incident ray and the direction of the emergent ray.
  4. The total deviation \(\delta\) depends on three factors: the angle of incidence \(i\) at the first surface, the angle of prism \(A\), and the refractive index \(\mu\) of the material of the prism.
  5. Light of all colours travels with the same speed in air, but with different speeds in a medium. In a medium, the speed decreases as the wavelength decreases.
  6. Refractive index is defined by \[\mu_{\text{glass}} = \frac{\text{speed of light in air}}{\text{speed of light in glass}}\] with both speeds in m s⁻¹; \(\mu\) has no unit. The refractive index of glass increases as the wavelength of light decreases, so \(\mu_{\text{violet}} > \mu_{\text{red}}\).
  7. Hence the deviation produced by a prism increases as the wavelength decreases. In visible light, violet (\(\lambda \simeq 4000\ \text{Å}\)) is deviated the most and red (\(\lambda \simeq 8000\ \text{Å}\)) is deviated the least.
  8. White light is a mixture of a large number of wavelengths, that is, it is polychromatic. Light of a single colour is monochromatic.
  9. Wavelength is the characteristic of colour, irrespective of its origin. Colour is the subjective property of light related to its wavelength.
  10. For light travelling in air, \[c = f\lambda\] where \(c = 3 \times 10^{8}\ \text{m s}^{-1}\), \(f\) is the frequency in Hz and \(\lambda\) the wavelength in m.
  11. Red light has the longest visible wavelength, \(8000\ \text{Å}\) or \(800\ \text{nm}\), and the lowest frequency, \(3.75 \times 10^{14}\ \text{Hz}\). Violet light has the shortest wavelength, \(4000\ \text{Å}\) or \(400\ \text{nm}\), and the highest frequency, \(7.5 \times 10^{14}\ \text{Hz}\). From the violet end to the red end, the wavelength increases and the frequency decreases.
  12. Units of wavelength: \(1\ \text{Å} = 10^{-10}\ \text{m}\) and \(1\ \text{nm} = 10^{-9}\ \text{m}\), so \(1\ \text{nm} = 10\ \text{Å}\).
  13. Newton’s prism experiment showed that white light consists of seven prominent colours, in the order violet, indigo, blue, green, yellow, orange, red, remembered as VIBGYOR, with violet nearest the base of the prism. There is no sharp boundary between the colours.
  14. Dispersion is the phenomenon of splitting of white light by a prism into its constituent colours. The spectrum is the band of colours seen on a screen when white light is passed through a prism.
  15. The cause of dispersion is that light of different wavelengths travels with different speeds in a medium.
  16. Dispersion of white light occurs only at the first surface of a prism; deviation occurs at both surfaces; and a prism does not produce colours, it only splits the colours already present in the incident light.
  17. In a parallel-sided glass slab the colours separate at the first surface, but the second surface adds no further separation, so the emergent coloured rays are parallel and very close together, and the emergent light appears almost white.
  18. The visible spectrum, from \(4000\ \text{Å}\) to \(8000\ \text{Å}\), is a very small part of the electromagnetic spectrum. The part beyond the red extreme and beyond the violet extreme, to which the eye does not respond, is the invisible spectrum.
  19. The complete electromagnetic spectrum in increasing order of wavelength, that is, in decreasing order of frequency, is: gamma rays, X-rays, ultraviolet rays, visible light, infrared radiations, microwaves, radio waves.
  20. Approximate ranges: gamma rays below \(0.01\ \text{nm}\); X-rays \(0.01\ \text{nm}\) to \(10\ \text{nm}\); ultraviolet \(10\ \text{nm}\) to \(400\ \text{nm}\); visible \(400\ \text{nm}\) to \(800\ \text{nm}\); infrared \(800\ \text{nm}\) to \(1\ \text{mm}\); microwaves \(1\ \text{mm}\) to \(1\ \text{m}\); radio waves above \(1\ \text{m}\). The boundaries are not sharp, and neighbouring regions overlap.
  21. Properties common to all electromagnetic waves: they need no material medium; they all travel at \(3 \times 10^{8}\ \text{m s}^{-1}\) in vacuum or air, and at different speeds in different media; they obey the laws of reflection and refraction, in which the direction, speed and wavelength change while the frequency remains unchanged; they are not deflected by electric and magnetic fields; and they are transverse in nature.
  22. Gamma rays come from the nuclei of radioactive atoms and from cosmic radiation. They penetrate thick metal sheets, cause fluorescence, and cause immense biological damage. They are used in a controlled manner in radiotherapy and in industry to check welding.
  23. X-rays are produced when a highly energetic electron beam is stopped by a heavy metal target of high melting point. They pass through flesh but are stopped by bone, strongly affect a photographic plate, and cause fluorescence. They are used in radiography, in CAT scans, in the study of crystals and complex molecules, and in detecting concealed metals.
  24. Ultraviolet radiations, first detected by Ritter in 1801, pass through quartz but are absorbed by glass, so a quartz prism is used to obtain their spectrum. They are chemically more active than visible light, affect photographic plates, cause fluorescence, and can cause skin cancer on long exposure. They are used for sterilising, for detecting purity, and in producing vitamin D.
  25. Infrared radiations, first detected by Herschel in about 1800, are detected by their heating effect, using a blackened bulb thermometer or a thermopile. They are absorbed by glass but pass through rock-salt, so a rock-salt prism is used for their spectrum. Being scattered little, they penetrate fog. They are used in therapy, in photography in fog and in the dark, as war signals, and in remote controls.
  26. Microwaves are produced by electronic devices such as crystal oscillators, and are used in satellite communication, in the analysis of atomic and molecular structure, in microwave ovens and in radar. Radio waves have the longest wavelength of all, and are used in radar communication and in radio and television transmission.
  27. Scattering is the process of absorption and then re-emission of light energy by the dust particles and air molecules present in the atmosphere. It was first studied by Rayleigh.
  28. The intensity of scattered light depends on wavelength as \[I \propto \frac{1}{\lambda^{4}}\] where \(I\) is the intensity of the scattered light and \(\lambda\) the wavelength in m. This holds only when the scattering particles are smaller than the wavelength of the incident light. Violet light is therefore scattered nearly 16 times more than red light.
  29. Particles bigger than the wavelength of the incident light scatter all wavelengths to the same extent, so the scattered light remains white.
  30. Applications of scattering: the Sun looks red at sunrise and sunset because its light travels the longest path through the atmosphere and loses its blue; the sky is white at noon because the path is shortest and no colour is preferentially removed; the sky in other directions is blue because scattered blue light reaches us from everywhere; the sky is black where there is no atmosphere, as on the Moon; clouds are white because their large particles scatter all colours equally; and red is used for danger signals because, being scattered the least, it can be seen from the farthest distance.

Flowchart

FC 6.3 — The Spine Of This Chapter White light, a mixture of many wavelengths → each colour travels at its own speed in glass, so the glass has its own refractive index for each → each colour is deviated through its own angle by a prism → dispersion, and the spectrum VIBGYOR on a screen → radiations found beyond both ends of that band → the complete electromagnetic spectrum, from gamma rays to radio waves → in the atmosphere the same wavelengths are scattered by amounts varying as 1/λ⁴ → the blue of the sky, the red of the setting Sun and the white of the clouds

6.10 Key Terms

  1. Actinic rays — another name for ultraviolet radiations, given because of their strong chemical action on substances such as silver chloride.
  2. Angle of deviation — the angle between the direction of the incident ray and the direction of the emergent ray after light has passed through a prism; measured in degrees.
  3. Angle of prism — the angle between the two refracting surfaces of a prism; measured in degrees.
  4. Angstrom (Å) — a unit of length used for wavelengths of light, where 1 Å = 10⁻¹⁰ m.
  5. Dispersion — the splitting of white light by a prism into its constituent colours.
  6. Electromagnetic spectrum — the complete family of electromagnetic waves, arranged in order of wavelength, from gamma rays to radio waves.
  7. Electromagnetic waves — travelling disturbances of electric and magnetic fields, which need no material medium and travel at 3 × 10⁸ m s⁻¹ in vacuum or air.
  8. Fluorescence — the glow given out by certain materials, such as zinc sulphide, when radiation falls on them.
  9. Frequency — the number of waves passing a point in one second; SI unit hertz (Hz).
  10. Gamma rays — the most energetic electromagnetic radiations, of wavelength below 0.01 nm, emitted by the nuclei of radioactive atoms.
  11. Infrared radiations — invisible electromagnetic radiations of wavelength from 800 nm to 1 mm, lying just beyond the red end of the visible spectrum and producing a strong heating effect.
  12. Invisible spectrum — the parts of the spectrum lying beyond the red extreme and beyond the violet extreme, to which the eye does not respond.
  13. Microwaves — electromagnetic waves of wavelength from 1 mm to 1 m, produced by electronic devices such as crystal oscillators.
  14. Monochromatic light — light of a single colour, made up of one narrow range of wavelength.
  15. Nanometre (nm) — a unit of length used for wavelengths of light, where 1 nm = 10⁻⁹ m, so that 1 nm = 10 Å.
  16. Polychromatic light — light that is a mixture of a large number of wavelengths, as white light is.
  17. Prism — a transparent block bounded by two plane refracting surfaces inclined to each other.
  18. Radio waves — electromagnetic waves of wavelength above 1 m, the longest in the electromagnetic spectrum.
  19. Scattering — the process of absorption and then re-emission of light energy by the dust particles and air molecules present in the atmosphere.
  20. Spectrum — the band of colours obtained on a screen when white light is passed through a prism.
  21. Thermopile — a device that detects infrared radiations by their heating effect, showing a deflection on a galvanometer connected to it.
  22. Ultraviolet radiations — invisible electromagnetic radiations of wavelength from 10 nm to 400 nm, lying just beyond the violet end of the visible spectrum and chemically more active than visible light.
  23. Visible spectrum — the band of light the eye can see, from 4000 Å to 8000 Å, containing the seven colours from violet to red.
  24. Wavelength — the distance between two neighbouring crests of a wave; SI unit metre (m).
  25. White light — light such as sunlight, which is a mixture of light of all the visible wavelengths.
  26. X-rays — electromagnetic radiations of wavelength from 0.01 nm to 10 nm, produced when a highly energetic electron beam is stopped by a heavy metal target.

6.11 Points To Ponder

  1. A prism separates colours because glass slows each colour by a different amount. Suppose a transparent material could be made in which every colour travelled at exactly the same speed. Would a prism cut from it still deviate a ray of light? Would it still produce a spectrum?

  2. The spectrum on a screen has no dividing lines anywhere in it, yet Newton fixed the count of colours at seven. If the boundaries are not real, what is it that decides how many colours a spectrum is said to contain?

  3. Our eyes respond to only a thin slice of the electromagnetic spectrum, and that slice happens to be the region in which sunlight reaching the ground is strongest. Is that likely to be a coincidence, or is there a reason for it?

  4. Imagine the earth’s atmosphere were twice as deep as it is, with the same air throughout. What would change about the colour of the midday sky, about the colour of the setting Sun, and about how long after sunset the sky stayed lit?

  5. The radiations on either side of the visible band were found within about a year of each other — one by a thermometer, the other by a chemical. What does that suggest about how a new kind of radiation gets discovered, and about what might still be waiting for the right detector?

  6. A cloud and the clear sky are both matter scattering sunlight, yet one looks white and the other blue. If you were handed the scattered light from an unknown object, what single measurement would tell you which of the two kinds of scattering was taking place?

6.12 Objective Questions

  1. (MCQ) The angle of deviation produced by a triangular prism does NOT depend on: (1 mark)
    1. the angle of incidence at the first surface
    2. the angle of the prism
    3. the intensity of the incident light
    4. the refractive index of the material of the prism
  2. (MCQ) Inside a glass prism, the colour that travels the fastest is: (1 mark)
    1. violet
    2. green
    3. yellow
    4. red
  3. (MCQ) The correct arrangement of the following radiations in increasing order of their frequency is: (1 mark)
    1. radio waves, microwaves, infrared, visible light
    2. visible light, infrared, microwaves, radio waves
    3. infrared, visible light, radio waves, microwaves
    4. microwaves, radio waves, visible light, infrared
  4. (MCQ) A radiation has a wavelength of \(5 \times 10^{-11}\ \text{m}\). It belongs to the region of: (1 mark)
    1. ultraviolet rays
    2. X-rays
    3. infrared radiations
    4. microwaves
  5. (MCQ) Light of wavelength \(4000\ \text{Å}\) is scattered by the air molecules of the atmosphere nearly \(n\) times as much as light of wavelength \(8000\ \text{Å}\), where \(n\) is: (1 mark)
    1. 2
    2. 4
    3. 8
    4. 16
  6. (MCQ) To obtain the spectrum of infrared radiations, the prism used must be made of: (1 mark)
    1. glass
    2. quartz
    3. rock salt
    4. diamond
  7. (Assertion–Reason) Assertion (A): To an astronaut travelling above the earth’s atmosphere, the sky appears black. Reason (R): Scattering of light needs dust particles or air molecules lying in the path of the light. (1 mark)
    1. both A and R are true and R is the correct explanation of A
    2. both A and R are true but R is not the correct explanation of A
    3. assertion is false but reason is true
    4. assertion is true but reason is false
  8. (Assertion–Reason) Assertion (A): A prism does not produce any colour of its own. Reason (R): The dispersion of white light takes place at both the surfaces of a prism. (1 mark)
    1. both A and R are true and R is the correct explanation of A
    2. both A and R are true but R is not the correct explanation of A
    3. assertion is false but reason is true
    4. assertion is true but reason is false
  9. (Match The Following) Match each radiation in Column X with the way it is detected in Column Y. (1 mark)
Column X Column Y
a. infrared radiations i. darkening of silver chloride solution
b. ultraviolet radiations ii. aerial of a television receiver
c. X-rays iii. thermopile connected to a galvanometer
d. radio waves iv. fluorescence on a zinc sulphide screen together with the shadow of bone on a film
  1. (Fill In The Blanks) Complete the following. (1 mark)

    1. The dispersion of white light by a prism takes place only at the …………… surface of the prism.
    2. \(1\ \text{nm} = ...............\ \text{Å}\).
  2. (True Or False) State whether each statement is true or false. (1 mark)

    1. All colours of light travel with the same speed inside a glass prism.
    2. The frequency of an electromagnetic wave remains unchanged when it passes from air into glass.
  3. (Name The Following) Name the following. (1 mark)

    1. The electromagnetic radiation used in the remote control of a television.
    2. The electromagnetic radiation emitted when the nucleus of a radioactive atom passes from the excited state to the ground state.
  4. (Give One Word Or Phrase) Give one word or phrase for the band of colours obtained on a screen when white light is passed through a prism. (1 mark)

  5. (State The SI Unit) State the SI unit of (i) wavelength and (ii) frequency. (1 mark)

  6. (Odd One Out) Pick the odd one out and give a reason for your choice: X-rays, ultraviolet rays, gamma rays, infrared radiations. (1 mark)

6.13 Check Your Concepts

  1. (Short Answer) Define dispersion of light, and name the colour of white light that a prism deviates the most and the colour it deviates the least. (2 marks)

  2. (Short Answer) Give a reason why a rock-salt prism, and not a glass prism, is used to obtain the spectrum of infrared radiations. (2 marks)

  3. (Short Answer) Differentiate between monochromatic and polychromatic light, stating clearly the basis of the difference. (2 marks)

  4. (Short Answer) State the three factors on which the angle of deviation produced by a triangular prism depends, and state how the deviation depends on each of them. (3 marks)

  5. (Short Answer) Dispersion takes place at the first surface of a parallel-sided glass slab, and yet no spectrum is obtained from it. Explain why. (3 marks)

  6. (Short Answer) Differentiate between ultraviolet and infrared radiations on the basis of (i) their range of wavelength, (ii) their effect on an ordinary photographic plate, and (iii) their heating effect. (3 marks)

  7. (Long Answer) Explain the cause of dispersion of white light by a triangular prism. State at which surface of the prism dispersion takes place, and at which surface there is only refraction of the already separated colours. (4 marks)

  8. (Long Answer) Why does the Sun appear red at sunrise and sunset, while the sky in a direction away from the Sun appears blue at the same time? Give reasons for both observations. (4 marks)

  9. (Long Answer) State any four properties that are common to all electromagnetic waves. (4 marks)

  10. (Long Answer) (a) State the range of wavelength of ultraviolet radiations. (b) State any two of their properties. (c) State one harmful effect and two uses of these radiations. (5 marks)

  11. (Long Answer) What would you see if the earth suddenly lost its atmosphere but kept everything else unchanged? Describe the colour of the sky at noon, the appearance of the Sun at sunset, and the appearance of the stars at night, giving a reason in each case. (5 marks)

6.14 Diagrams, Graphs & Constructions

  1. (Diagram) Draw a labelled ray diagram to show a narrow beam of white light falling on one face of a triangular glass prism and forming a spectrum on a screen. Mark the positions of the violet and the red ends on the screen, and mark the angle of deviation of the red ray. (4 marks)

  2. (Diagram) Fig. 6.3-4 shows a beam containing blue, red and green light falling on a parallel-sided glass slab. Redraw the path of the three colours through the slab, and state the appearance of the emergent light, giving a reason. (3 marks)

  3. (Diagram) Fig. 6.3-3 shows a beam of red, blue and yellow light falling normally on the face AB of an isosceles right-angled prism, the critical angle of the glass–air interface for yellow being \(45^\circ\). Redraw the diagram, showing the complete path of each of the three colours, and name the colour that finally emerges at right angles to its original direction. (4 marks)

  4. (Graph) Sketch a graph showing how the intensity of light scattered by the atmosphere varies with the wavelength of the light over the visible range. Label both axes with the quantity and its unit, and mark on the curve the points corresponding to violet light and to red light. (3 marks)

  5. (Graph) From the graph in Fig. 6.7-2, read the intensity of scattered light at \(4000\ \text{Å}\) and at \(8000\ \text{Å}\), and hence find the ratio of the two. State what the changing steepness of the curve tells you about light at the violet end of the spectrum. (3 marks)

  6. (Diagram) Draw a labelled chart of the complete electromagnetic spectrum in order of increasing wavelength. Mark on it the position of visible light, and write the two wavelengths that fix the limits of the visible band. (4 marks)

  7. (Diagram) Draw a labelled diagram showing a source of heat placed at the focus of a parabolic mirror, and the beam of radiation obtained from it. Name the radiation concerned and state the property of that radiation which the diagram illustrates. (2 marks)

6.15 Practice With Numericals

  1. (Numerical) A beam of yellow light has a wavelength of \(5.9 \times 10^{-7}\ \text{m}\). Calculate its frequency. Take the speed of light as \(3 \times 10^{8}\ \text{m s}^{-1}\). (2 marks)

  2. (Numerical) Express a wavelength of \(4500\ \text{Å}\) in nanometres and in metres, and calculate the frequency of this light. Take the speed of light as \(3 \times 10^{8}\ \text{m s}^{-1}\). (3 marks)

  3. (Numerical) An electromagnetic wave has a frequency of \(6 \times 10^{10}\ \text{Hz}\). Calculate its wavelength, and name the region of the electromagnetic spectrum to which it belongs. Take the speed of the wave as \(3 \times 10^{8}\ \text{m s}^{-1}\). (3 marks)

  4. (Numerical) In a certain glass, red light travels at \(2.00 \times 10^{8}\ \text{m s}^{-1}\) and violet light at \(1.95 \times 10^{8}\ \text{m s}^{-1}\). Taking the speed of light in air as \(3 \times 10^{8}\ \text{m s}^{-1}\), calculate the refractive index of the glass for each colour, and state which of the two a prism of this glass will deviate more. (3 marks)

  5. (Numerical) Light of wavelength \(4000\ \text{Å}\) and light of wavelength \(6000\ \text{Å}\) pass through the same dust-free air. Calculate how many times more strongly the light of shorter wavelength is scattered. (3 marks)

  6. (Numerical) A radio station transmits waves of wavelength \(250\ \text{m}\). Calculate the frequency of the waves and express your answer in MHz. Take the speed of the waves as \(3 \times 10^{8}\ \text{m s}^{-1}\). (3 marks)

  7. (Numerical) Light of frequency \(5 \times 10^{14}\ \text{Hz}\) travelling in air enters a block of glass of refractive index \(1.5\). Calculate (i) the wavelength of the light in air, (ii) its speed in the glass, and (iii) its wavelength in the glass. Take the speed of light in air as \(3 \times 10^{8}\ \text{m s}^{-1}\). (4 marks)