Transcript Document

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 
speed of light in vacuum
speed of light in medium
c
n 
v 
 


3.20. Dispersion and Refraction
• The refractive index n() is wavelength dependent which
means that different colours travel with speeds through a
material.
c
v 
• This is known as dispersion.
 
n 

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 /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/
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
N
N
Medium 1
B
i
i
A
r
r
D
Medium 2
C
• 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
sin i v 1

sin r v 2
• but n2=c/v2 and n1= c/v1
• Hence
sin i n2

sin r n1
• 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 d = 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
i
Medium 1
a
t
i-r
e
r
b
c
d
Medium 2
Medium 1
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
n
real depth
apparent depth
real
depth
apparent
depth
c
3.24. Total Internal Reflection
• What happens when light passes from a more dense medium to a less
dense medium?
• Here n1 > n2
• From Snell’s law we have sin i 
sin r
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 qc.
 qc is given by
sin qc 
n2
n1
N
r
Medium 2, n
i
Medium 1, n 1
2
3.24. Total Internal Reflection
• Here n1 > n2
r
Medium 2, n
i < qc, r < 90˚
Medium 1, n
2
1
i
r
i = qc, r = 90˚
Medium 2, n
Medium 1, n
i
Medium 2, n
i > qc, tir
i
i
Medium 1, n
2
1
2
1
•When r = 90˚ the
refracted ray moves
along the interface. The
angle of incidence at
which this happens is the
critical angle qc.
•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 q at the input
facet which if exceeded means that light is no longer
guided down the fibre. This is given by the following
na
nco
n cl
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?
A
i1
d1
r1
undeviated
beam
dt
d2
r2
i2
deviated
beam
•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)
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
A
i
d d
r
undeviated
beam
d min
r
i
•As the beam path is symmetric dmin = 2i -2 r
•Now from the figure we find that 2r = A
•So dmin = 2i -A
A  d 
•Hence 2i = A+dmin
min


sin


•Thus the refractive index n is given by
2


•We assume that the prism is in air
n
A 
sin
 

2
 
deviated
beam
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.
white
light
red beam
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 wof a prism is given by
w
n blue  nred
nyellow  1