3.18. Defects caused by reflections• We have assumed that the paraxial approximation applies in that

all rays make small angles with respect to the optical axis. We also assume that the rays strike close to th e centre of the mirror.

• In real systems the rays hit the whole mirror. We find that rays that hit near the outside of mirror are focused at a different point when compared with those that are close to the centre.

• This causes a blurring of the image, and is known as spherical aberration.

Note that to show the effect the reflection of the outer rays has been greatly exaggerated. The normals for the outer rays come from the same point as those for the inner rays. This problem is eliminated by use of a parabolic mirror.

3.19. Refraction and refractive index

• The wave speed of light in a vacuum is 3 x108 m/s. • When light propagates through a medium the

interaction between the atoms in the material and the light waves causes a reduction in the speed of the wave .

• This change in speed is quantified by the refractive index n

€

n λ( ) =speed of light in vacuum

speed of light in medium

€

n λ( ) =c

v λ( )

3.20. Dispersion and Refraction

• The refractive index n( ) is wavelength dependent which means that different colours travel with speeds through a material.

€

n λ( ) =c

v λ( )• This is known as dispersion.

3.20. Dispersion and refraction

• When light propagates through a material the speed of light is reduced owing to the response of the medium to the electric and magnetic fields.

• Although the wave speed changes the energy of the wave does not change. As E = hf the frequency remains constant.

• This means that the wavelength must reduce. The effective wavelength in the material is given by λ/n.

• There is a phase shift imposed by the material. If the material of length l and refractive index n the phase shift is 2πnl/λ

3.21. Refraction at an interface

• Consider a perfectly flat interface between two transparent materials.

• The wave speed in material 1 is v1 and the refractive index is n1. The wave speed in material 2 v2 and the refractive index is n2.

• A plane wave propagates from material 1 to material 2.• The wave is incident at an angle i to the interface.• The transmitted wave makes a angle r to the interface.• All the rays and normals lie in the same plane - the

plane of incidence

A

B

C

N N

D

ii

r

r

Medium 1

Medium 2

• This sytem can be understood by use of Huygens’ Principle.

• Consider the incident wave front AB. Point A is in contact with the interface.

• In the time taken for point B to move to D the point A moves to C to form the refracted wave front CD.

• Time, T, to for A to move to C is

• T = AC/v2

• Time, T, to for B to move to D is

• T = BD/v1

• So AC/v2 = BD/v1

• But AC = ADsin r• And BD = ADsin i

• So sin r/v2 = sin i/v1

• Hence

€

sini

sinr=v1

v2

€

sini

sinr=n2

n1

• but n2=c/v2 and n1= c/v1

• Hence

• This is Snell’s law.• When light goes from a less dense to a more dense medium

n2>n1 the rays are refracted towards the normal i>r.

• When light goes from a more dense to a less dense medium n1>n2 the rays are refracted away from the normal r>i.

3.22. Deviation caused by refraction• Consider a parallel plate of glass of thickness t and

refractive index n in air. This plate causes a deviation of the beam from its original path. The angular deviation = i - r

i

r

Medium 1

Medium 1

Medium 2ta

b

e

d

c

i-r

• The spatial deviation d can be found from the figure

• d = ce• Triangle ace gives

– ce = acsin(i-r)

• Triangle abc gives– ac = ab/cosr

• But ab = t• So ac =t/cosr• Hence d = ce = tsin(i-r)/cosr

3.23. A method of measuring the refractive index• A plate of glass has a thickness t and refractive index n.• A mark is placed under the plate and the mark is

viewed from above the plate.• The rays that come from the mark are refracted at the

interface between the plate and air. The rays are refracted away from the normal. Thus the mark appears to be inside the block. The refractive index n is given by

real

depthapparent

depth

c

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n=real depth

apparent depth

3.24. Total Internal Reflection• What happens when light passes from a more dense medium to a

less dense medium?N

i

r

Medium 1, n1

Medium 2, n2

• Here n1 > n2

• From Snell’s law we have

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sini

sinr=n2

n1

• As n1 > n2

• Then r > i• Hence the ray is refracted away from the

normal.• When r = 90˚ the refracted ray moves along

the interface. The angle of incidence at which this happens is the critical angle c.

c is given by

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sinc =n2

n1

3.24. Total Internal Reflection• Here n1 > n2

Medium 1, n1

Medium 1, n1

Medium 1, n1

Medium 2, n2

Medium 2, n2

Medium 2, n2

i

r

i

r

i i

i < c, r < 90˚

i = c, r = 90˚

i > c, tir

•When r = 90˚ the refracted ray moves along the interface. The angle of incidence at which this happens is the critical angle c.

•For angles of incidence greater than the critical angle light cannot exit at the boundary and so is reflected. This is total internal reflection.

3.25. How an optical fibre works• An optical fibre relies on total internal reflections to guide light down the fibre. The fibre consists of a core of

refractive index nco and a cladding layer of refractive index ncl (nco > ncl). There is a specific angle at the input facet which if exceeded means that light is no longer guided down the fibre. This is given by the following

ncl

na

nco

3.26. Refraction and prisms• Consider a prism of apex angle A and refractive index

n. What happens when monochromatic light is incident on the prism?

•The beam is deviated from its original path by the effects of refraction at both surfaces.•The total angular deviation dt is given by•dt = d1 + d2

•With d1 and d2 the angular deviation at both surfaces•So dt = (i1 - r1) + (i2 - r2)

A

dt

d1 d2i1

i2r 1 r 2

undeviated

beam

deviated

beam

3.27. Minimum deviation • There is an angle of incidence i whereby the beam path through

the prism is symmetric. At this angle of incidence the deviation is a minimum

•As the beam path is symmetric dmin = 2i -2 r•Now from the figure we find that 2r = A•So dmin = 2i -A•Hence 2i = A+dmin •Thus the refractive index n is given by•We assume that the prism is in air

A

dmind di

ir r

undeviated

beam

deviated

beam

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n=

sinA +dmin

2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

sinA

2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

3.28. Prisms and Polychromatic Light • So far we have assumed that monochromatic light has been

incident on the prism. When polychromatic light is shone on to the prism we find that the light is dispersed.

red beamwhite

light

blue beam

• The dispersion occurs because the refractive index is wavelength dependent. The refractive index for blue light is greater than that of red light. Hence the blue light is refracted more at the prism interfaces and so experiences a greater deviation when compared to the red light.

• The dispersive power of a prism is given by

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=nblue −nred

nyellow −1