Figure 89: Practical structure of an optical fibre.

• Absorption in the glass tube leads to a gradual decrease in light intensity.

• For optical fibres, the glass used for the core has minimum absorption

at a wavelength of 1.3 µm, i.e. in the infra red.

• Hence, this is the laser wavelength used for long-distance signal trans-

mission (∼ 105 m without significant loss).

4.5 Images Formed by the Refraction of Light

A well-known effect in optics is that the apparent depth of an object sub-

merged in a transparent liquid, as seen by an observer in air, appears to be

different to the actual depth in the liquid. For example,

• The apparent depth of a stone on the bottom of a shallow pond appears

to be less than its actual depth.

• A fish in an aquarium (figure 9076) appears to be closer to the glass wall

76Knight, Figure 23.26(b), page 727

147

than it really is.

Figure 90: Refraction of light rays causes the apparent depth of an object to

change.

Consider this second example:

• Let us ignore the presence of the thin glass wall (this produces only a

very small refractive effect).

• The light rays are moving from a liquid (refractive index n1) to air (re-

fractive index n2 < n1).

• The light rays are refracted away from the normal at the water/air

boundary.

• Geometrically, our eye perceives a virtual image of the fish to be formed

at a closer distance to the boundary than that of the object itself.

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This last point is a general rule:

Objects appear to be closer than they really are because of the

refraction of light at the medium/medium boundary.

• We recall from reflection that rays reflected from a plane mirror appear

to diverge from a virtual image.

• Similarly, in refraction, rays from an object point P refract at a bound-

ary between the two media such that the rays appear to diverge from a

point P′ – i.e. a virtual image.

Figure 9177 shows a boundary between two transparent media with refractive

indices n1 and n2.

• Point P, the object, is a source of light.

• Point P′, the virtual image, is the point at which the rays appear to

diverge from.

• Let us define the

– Object distance, s, to be the distance between the object and the

boundary.

– Image distance, s′, to be the distance between the image and the

boundary.

• We want to find s′.

For construction purposes, we set up the optical axis, a line perpendicular

to the boundary that passes through P .

• Consider a ray that leaves the object at an angle θ1 with respect to the

optical axis.

77Knight, Figure 23.27, page 727

149

� �

Figure 91: Finding the position of a virtual image in refraction.

150

• Suppose that the ray meets the boundary at a point which is a distance

l from the optical axis.

• The light ray thus refracts through the boundary into the second medium

such that the angle of refraction is θ2.

• Geometrically, we can trace the refracted ray back to the point P′, where

the angle between this ”imaginary” ray and the optical axis is θ2.

• By simple trigonometry, we see that

tan θ1 =l

s, tan θ2 =

l

s′, (4.9)

i.e.

l = s tan θ1 = s′ tan θ2 , (4.10)

or

s′ = stan θ1

tan θ2

. (4.11)

• But we also recall from Snell’s law that

n2 = n1

sin θ1

sin θ2

. (4.12)

In reality, when we observe the object, the optical axis is a line joining the

object to our eye. In this case,

• The distance l is the radius of your pupil, i.e. we have that

l � s , l � s′ . (4.13)

• The angles θ1 and θ2 are necessarily extremely small (the angles in figure

91 have been grossly exaggerated for clarity).

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• Mathematically, we can employ the small-angle approximation

sin θ ≈ θ , (4.14)

where θ is in radians.

• Thus in (4.12),

sin θ1

sin θ2

=n2

n1

−→ θ1

θ2

≈ n2

n1

. (4.15)

• In a similar manner, we find that the tangent of an angle obeys the

small-angle approximation: i.e. if θ is very small and in radians,

tan θ ≈ θ . (4.16)

• Therefore, in (4.11),

tan θ1

tan θ2

=s′

s−→ θ1

θ2

≈ s′

s. (4.17)

• Equating (4.15) and (4.17), we then find an expression for the image

distance s′, namely

s′ =n2

n1

s . (4.18)

• We note from (4.18) that s′ is independent of θ1, so that all rays parallel

to the axis (so-called paraxial rays) appear to diverge from the same

point P′.

Example:78 a biologist keeps a specimen of his favourite beetle embedded in

a cube of polystyrene plastic. The hapless bug appears to be 3.0 cm within

the plastic. What is the beetle’s actual distance beneath the surface?

78Knight, Example 19, page 752

152

Solution: so parallel rays (i.e. paraxial rays) from the beetle refract into the

air and then enter into the observer’s eye. The rays in the air when traced back

into the plastic appear to be coming from the beetle at a shallower location,

a distance s′ from the plastic-air boundary.

If we have an actual object distance s, and an image distance s′ = 3.0 cm,

then using (4.18),

s′ =n2

n1

s =nair

nplastic

s ,

i.e. so if s′ = 3.0 cm, then

3.0 =1.0

1.59s −→ s = 4.77 cm .

4.6 Colour and Dispersion

The phenomenon of colour is associated with the wavelength of light – our

eyes are sensitive to the wavelengths of visible light, and our brains interpret

visible light at these particular wavelengths as colour.

For example, our brains interpret colours with the following wavelengths:

• Deepest red: ∼ 700 nm.

• Red: ∼ 650 nm.

• Green: ∼ 550 nm.

• Blue: ∼ 450 nm.

• Deepest violet: ∼ 400 nm.

Most of the results in optics do not depend upon colour: rays of light corre-

sponding to wavelengths in the red end of the visible spectrum (”red light”)

generally pass through materials in the same way as rays of ”blue” light.

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4.6.1 Light Passing Through a Prism

Newton:

I procured me a triangular glass prism to try there with the celebrated phenom-

ena of colours.

Consider figure 92,79 showing the path of white light refracting through a

prism.

Figure 92: Refraction of white light through a glass prism.

Newton proposed the idea that what we perceive as white light is in fact a

mixture of all colours.

79Pink Floyd, The Dark Side of the Moon

154

• White light can be ”separated out” into its component colours using a

prism (figure 9380).

• Similarly, a prism can be used to mix all of these colours back into white

light.

Figure 93: Dispersion and subsequent recombination of white light through a

prism.

So how does this effect occur?

• The refractive index of a material is slightly different for different wave-

lengths of light – i.e. for different colours – this effect is known as dis-

persion.

– For example, glass has a slightly different refractive index for violet

light than for green light or red light.

– Therefore, since the refractive indices are different, by Snell’s law

we see that different colours refract through different angles and

consequently follow slightly different paths.

80Knight, Figure 23.28(a), page 728

155

Figure 9481 shows a dispersion curve – i.e. a curve showing how the refrac-

tive index of a material varies with wavelength – for glass.

Figure 94: Dispersion curves of two types of glass.

Notice, for example,

• n is larger when λ is shorter.

• n is smaller when λ is larger.

Thus, violet light refracts more than red light.

4.6.2 Rainbow

Figures 95 and 9682 gives a basic description of how we observe a rainbow.

• When the sun is behind you, and a ray of sunlight is incident upon a

water droplet in the atmosphere:

81Knight, Figure 23.29, page 72982Knight, Figure 23.30(a), page 730

156

Figure 95: Production of a rainbow.

Figure 96: Production of a rainbow.

– Refraction occurs at the first boundary – the white light is dis-

persed into its component colours.

– Refraction/reflection occurs at the second boundary – while

most of the rays are refracted out of the droplet and into the air, a

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small number are reflected back into the droplet.

– Refraction occurs at the third boundary – the ”coloured” rays are

dispersed even more.

So white light enters the droplet and is dispersed; while most of the light

is refracted out of the droplet at the second boundary, a small percentage of

light ends up being refracted out of the droplet on the same side from which

it entered.

Figure 95 suggests that violet light should be at the top of the rainbow, while

red light is at the bottom. However, in reality, it is the opposite:

• This is because as shown in figure 96, red light is refracted at 42.5◦, while

violet light is refracted at 40.8◦.

• Consequently, both red and violet light dispersed from the same droplet

do not enter your eye.

• It is a combination of refracted light from different droplets that leads

to a rainbow:

– Since red light is refracted through a greater angle than violet light,

you have to look higher in the sky to see the red light - hence red

is at the top of the rainbow.

4.6.3 Coloured Filters and Coloured Objects

It is well known that coloured filters, or objects (e.g. coloured glass) appear

to refract only light of the same colour – why is this?

A filter/object will absorb all light falling upon it, and re-radiate only the

light corresponding to the particular colour.

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