Q. Explain normal and anamalous dispersion and hence describe Cauchy’s
equation. How this equation is useful in dealing refractive index of different
types of liquid
The variation of refractive index µof the medium with wavelength λ(fre-
quency) constitute the phenomenon of dispersion. The dispersive power
of a medium is repeated by dµ
dλ .
Normal Dispersion
It is generally observed that the refractive index decreases with increase in
wavelength, this is known as normal dispersion. The variation of refractive
index with wavelength for various crystals is shown below.
Dispersion, Optics 2nd Year *****
From the figure, it is seen that
a) the refractive index of the medium decrease with the increase in wavelength
b) the graph is steeper at shorter wavelength.
i.e., dµ
dλ increases more rapidly at shorter wavelength.
c) for a given wavelength, dispersion is more for the medium which is more
transparent.
Dispersion which has all these properties is called normal dispersion. All the
media transparent to light exhibit the phenomenon of normal dispersion.
Normal dispersion more or less obeys the Cauchy’s formula.
µ=A+B
λ2where A, B are called Cauchy’s constants.
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Anamalous Dispersion
In small wavelength ranges there is often increase of refractive index due
to an increased absorption at the incoming radiation through the medium.
It is known as anamalous dispersion. Thus, anomalous dispersion is
present at these wavelengths corresponding to the absorption bands of
the medium. For example, the anamalous dispersion is noticed in the
case of sodium at wavelength 5890 ˚
Aand 5896 ˚
A.
For such disperson, a graph of refractive index µversus wavelength λis a
straight line having positive upward slope.
Dispersion, Optics 2nd Year *****
Lorentz Theory
The theory of dispersion based on electromagnetic theory is based on
following assumption.
a) There is no appreciable interaction between the atoms or between the
molecules.
b) The electric field of the electromagnetic wave induces a dipole moment
in the gas molecules
c) Electrons are bound to the nucleus by linear restoring forces. d) The
electric field E is constant in space over all atom or molecule i.e.,
E=Eoe−iωt (1)
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Now,
The electric force on the electron of mass m and charge e is eE.
The magnetic force can be neglected on account of its motion.
Elastic restoring force on the electron is
F=−kx (2)
x is the displacement from equilibrium
k is constant of proportionality
Damping frictional force against the motion of electron is
F=−αdx
dt (3)
αis constant of proportionality
dx
dt is velocity of electron
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The polarization P is equal to the dipole moment (ex) per unit volume.
If N is no of electrons dispersed in unit volume
Then
P=Nex
P=Ne e
mE
ω2
o−ω2−iγω [∵E=Eoe−iωt]
P=Ne2E
m1
ω2
o−ω2−iγω (6)
Let there are N molecules per unit volume and in each molecule there
are f1oscillators having constants ω1and γ1,f2oscillators having
constants ω2and γ2and so on.
Eqn(6) can be generalized as
P=Ne X
k
fkxk=Ne2E
m X
k
fk
ω2
k−ω2−iγkω!
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The dielctric constant is
r= 1 + χ= 1 + α
o
= 1 + P
oE
∴r= 1 + Ne2
om X
k
fk
ω2
k−ω2−iγkω!
Refractive index is square root of dielectric constant
Hence,
µ2=r= 1 + Ne2
om X
k
fk
ω2
k−ω2−iγkω!(7)
This is dispersive formula. Hence, the refractive index is complex. The
significance of the complex refractive index is that there is an absorption
of energy in the medium.
In the region remote from the natural frequencies of the medium, the
term γkωis very small and neglected in equn(7). Thus,
µ2=r= 1 + Ne2
om X
k
fk
ω2
k−ω2!(8)
Dispersion, Optics 2nd Year *****
This shows that refractive index is real and increases with frequency of
the incident wave. This is normal dispersion.
Since, ω=2πc
λ, equn(8) can be written as
µ2= 1 + Ne2
om X
k
fk
ω2
k−ω2!
or, µ2= 1 + Ne2
om4π2c2 X
k
fkλ2λ2
k
λ2−λ2
k!
or, µ2= 1 + X
k
Akλ2
λ2−λ2
k
(9)
where, Ak=Ne2fkλ2
k
om4π2c2
This is Sellmeier’s equation.
Dispersion, Optics 2nd Year *****
Scattering of Electromagnetic Radiation
Scattering is the deflection of electromagnetic radiation by fine particles
of solid, liquid or gaseous matter from the main direction of a beam. If the
particles are relatively large, reflection and refraction as well as diffraction
play a part. If the particles are small, i.e., smaller than the wavelength,
the effect is diffractive.
The blue colour of the sky is due to scattering by air molecules and the
red sun is due to the removable of the blue colour by scatttering from the
direct beam.
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Scattering is both elastic and inelastic. In elastic scattering, the pho-
tons of the radiation are reflected, i.e., they bounce off the atoms and
molecules without any change of energy. In this type of scattering, known
as Rayleigh’s scattering, there is change of phase but no frequency change.
In Rayleigh’s scattering, the scattered intensity for small particle varies
as 1
λ4so that white light scattered by very small particle is bluish.
In inelastic scattering, there is an interchange of energy between the pho-
tons and the particles. Consequently, the scattered photons have a differ-
ent wavelength as well as different phase. This gives Raman Effect.
Dispersion, Optics 2nd Year *****
Raman Effect
When a monochromatic radiation of very narrow frequency band is scat-
tered by a solid, then the scattered light not only consists of the radiation
of incident frequency but also the radiation of frequencies above or be-
low than that of the incident beam frequency. This form of scattering of
electromagnetic radiation in which light suffers a change in frequency and
a change in phase as it passes through a material medium is the Raman
Effect.
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If f1is the frequency of incident radiation and fsis that of the light
scattering by a given molecular species, than the Raman shift ∆fsis
given by
∆fs=f1−fs
This difference is the characteristics of the substance that can produce
scattering. It doesnot depend on the frequency of the light employed.
Raman shift lies within the range of 1000 cm−1to 3000 cm−1, which falls
in far and near infrared regions of the spectrum.
Dispersion, Optics 2nd Year *****
Q. Calculate the value of Cauchy’s constants A and B for crown glass.
Given,
µR= 1.514, λR= 6563˚
A µV= 1.524 λV= 4862˚
A
According to Cauchy’s formula,
µ=A+B
λ2
∴1.514 = A+B
(6563 ×10−8cm)2...(1)
and 1.524 = A+B
(4862 ×10−8cm)2...(2)
Subtracting (1) from (2),
0.01 = B1
(4862 ×10−8cm)2−1
(6563 ×10−8cm)2
or, B= 5.236 ×10−11cm2
and hence, A= 1.501
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