Chapter 6– Optical Methods - Introduction
Solutions of the Maxwell equations and energy relationships
Sample Problems
6-S1
To define an electromagnetic field it is necessary to define two vectors, the vector that represents the electrical field, and the vector that defines the magnetic field, also called magnetic induction. The propagating electromagnetic field has an effect on material objects. The effect of the electromagnetic field on material objects is defined by a group of vectors, the vector , that represents the electric current density, the vector electric displacement and the magnetic vector .
Write the Gauss equations in the differential form that relate these quantities in Cartesian coordinates.
Hint: Relate the vectors taking into consideration the constitutive equations of vacuum.
Solution to 6-S1
The answer to these questions is provided by,
(Farady’s law of electromagnetism) (6.8)
Expanding the vectorial equations in Cartesian coordinates,
And
(Ampère’s law of electromagnetism) (6.9)
One gets
To relate the vectors ,and it is necessary to consider in the case of vacuum the constitutive laws,
(6.10)
Hence
The above equations relate the vector electric displacement to the vector that defines the magnetic field.
Taking into consideration (6.11) and (6.9)
The above term represents the ordinary electrical current plus the so called displacement current, defined by Maxwell that introduced it to include transient fields. The above set of equations relates the magnetic field to the total current density.
6-S2
The propagation of an electromagnetic field obeys the Maxwell equations. Assume there is a field propagating in the vacuum that is characterized by the fact that the components of the field are contained in one plane and remain in this plane throughout the propagation. It is necessary to utilize a Cartesian coordinate system such that the normal to the plane is the z-coordinate. It is assumed that the field propagates in the z-direction,
Show that the and components are perpendicular to the z-axis.
Solution to 6-S2
It was assumed that the field propagates only in the z-direction hence the field must be independent of x,y, hence it will change only as a function of z, and t, the time. We can write, (z.t) and .
Utilizing
(Gauss Law for the electrical field) (6.6)
In view that the wave propagates in vacuum and there are no charges,
in the expanded form,
The field propagates in the z-direction and hence must be independent of x and y, then the corresponding derivatives must be zero. From the above equation,
. This means that Ez is constant with respect to z and according to our initial assumption that the components are in a plane, we can assume this constant to be zero. From this result it follows that the field can have only transversal components Ex and Ey. Let us assume now that .
The field only propagates in the z-direction hence the field will change only as a function of z, and t, the time. From
(Farady’s law of electromagnetism) (6.8)
Utilizing the expanded form given in sample problem 6-S2, and since and
, , .The above results indicate that Bx and Bz are constant in time, the only variable term is By and hence the magnetic field must be of the form,
, the vector is perpendicular to the direction of propagation. We can therefore conclude that the vectors are perpendicular to the direction of propagation and perpendicular to each other. The plane that contains both vectors is the plane formed by the two vectors and has a normal that is of the direction of . The direction of propagation is constant in space.
6-S3
In section 6.3.1.1. the harmonic solutions of the Maxwell equations were derived. It was shown that the argument of the sinusoidal solution is of the form,
Show that the argument can be written as
Solution to 6-S3
Taking into consideration ( 6.22) and (6.21) multiplying and dividing by T, and recalling that in vacuum v=c gives . Assuming that t>0, that is we take time from an assumed time origin t=0, then the negative sign indicates a wave that propagates in the +x direction and the positive sign indicates a wave that propagates in the – x direction. Hence the solution becomes,
6-S4
Assume that there is a sinusoidal solution of the Maxwell equations (see Section 6.3.1.1), and from the solution of sample problem 6-S1 we have a propagating wave of the form,
.
With the above information determine the B field. Show that Figure 6.6 is correct.
Solution to 6-S4
Since the field is sinusoidal the solution is where the argument is of the form, (6.47), where an initial phase that is of no consequence in this case was removed. Then, , one should remember that c is the speed of light.
It is known that Ez=Ey=0. From the expanded form of (6.8) and the result from sample problem 6-S1,
, then it is possible to write,
Integrating with respect to t,
We can see that the magnitudes of the two vectors are related by the equation E=cB. We know from the previous problem that the two vectors are orthogonal. From the previous derivations the two vector fields are in phase. Consequently Figure 6.6 is correct.The above derivations yield the following result the magnitudes of the magnetic and the electric field in vacuum are related by the equation E=cB.
6-S5
We have shown in sample problems 6-S2 and 6-S3 that two vectors and , are a solution of the Maxwell equations. These two vectors define a plane that we call a plane wave front. In these problems it was assumed that the field propagated normal to one of the coordinate axis in a Cartesian system. Find the equations of propagation in an arbitrary direction. Utilizing (6.22), it is possible to make an expression of the plane as a function of the wave number. Assuming a harmonic solution, provide an equation of the propagating wave front.
Solution to 6-S5
In Figure P6.1there is a plane whose normal is the direction . In a three dimensional coordinate system the equation of planes normal to the direction is given by
Figure P6.1.Plane wave front propagating in space
. The vector is normal to the plane, and we .can adopt for the normal the vector wave number with the modulus . The preceding equation becomes,, this equation implies that , where C is a constant. This equation represents a family of planes that are solutions of the Maxwell equations corresponding to the direction of propagation given by,, that gives the Cartesian equation . If the solution is harmonic then assuming that the propagating vector has amplitude A, , this function defines a family of planes perpendicular to . Moving along , every time that the argument makes a cycle, that is every time that the distance d between two points increases by the vector takes the same amplitude zero. To represent a wave advancing in the positive direction it should be expressed as.
This solution can be represented in the exponential form,
6-S6
Show that the equation is the solution of a 3-D equation representing the propagation of a spherical wave front. In addition indicate the properties of the Poynting vector.
Solution to 6-S6
Either the vector or satisfy the wave equation (6.20a) and (6.20b). Recalling (6.24) the Laplace equation in Cartesian coordinates become,
By utilizing the spherical coordinates shown in the FigureP.6.1, x=r cos ϕ sinθ, y=r cosθ sin ϕ,
z= r cosθ, the wave equation becomes:
The wave front must have spherical symmetry because it propagates equally in all directions. Hence the wave equation must be independent of θ and ϕ. Consequently the expression of the vector field must be. The consequence of the symmetry is that the derivatives with respect to θ and ϕ must be zero. Consequently the wave equation should be of the form:
We can use the identity,
The solution of the equation must be valid for any r independently of t or for any t independently of r. Then
With the substitution
It is important to recall that in the case of plane wave fronts there are two solutions 1) the outgoing wave fronts and 2) the converging wave fronts. Hence the solutions are of the form:
From the solution above it can be concluded that the surfaces of constant phase are spheres. Indeed the arguments of the vectorial function will be the same for any possible direction of r in space. Also the Poynting vector will be radial and the corresponding energy will go down with 1/r2. The solution satisfies the principle of conservation of energy.
6-S7
Give arguments that explain that as an approximation in many cases the vectorial functions that are solutions of the Maxwell equations can be replaced by scalar functions. Thus many problems in optics can be solved ignoring the vectorial nature of light propagation.
Solution to 6-S7
Starting with the Maxwell equations,
(Gauss Law for the electrical field) (6.6)
(Gauss Law for the magnetic field) (6.7)
(Farady’s law of electromagnetism) (6.8)
(Ampère’s law of electromagnetism) (6.9)
Any solution of an optical problem has to satisfy the above equations. Recall that besides the field equations of the continuum fields postulated by the Maxwell theory of light, it is necessary to have constitutive equations that represent the interaction of the field with the medium it is interacting with. One assumed property is that the media with which the light is interacting is linear, this property is represented by (6.12) and (6.13), Section 6.3 The Electromagnetic Theory of Light. The second property is that the media is isotropic, this means that the vectors and are independent of their particular position in space. Another property of the media of interest is that they are homogeneousindicating that the properties are the same in the region of interest. The media is non dispersive indicating that the permittivity ε of the material is independent of the light frequency (wave length) in the region under analysis. Finally the media under analysis is non magnetic, then the permeability of the media is always the permeability of the vacuum μ0. Under all these assumptions it is possible to derive the equations of wave propagation of the electromagnetic field (6.20), Section 6.3 The Electromagnetic Theory of Light.The wave equation is satisfied by both vectors and and also by all the components of these vectors in any system of coordinates. The equation(6.20) is also satisfied by any scalar function of the form, where P indicates a position in space and the propagating field must satisfy the equation of propagation,
Therefore under the assumed conditions the solutions of optical problems can be described by the scalar function. In view of the linearity of the Maxwell equations more complex problems can be solved by adding the corresponding scalar functions that are solutions of the wave equation. This means that the vectorial functions representing solutions of plane wave propagation and the spherical wave fronts could have been obtained by scalar functions as well. In many problems the main quantity of interest is the intensity of light I(P), since the measurements are carried in times that are extremely large compared to the times involved in the light oscillations. Hence (6.60) and (6.61)are replaced by. In one word the final outcome that can be measured, the modulus of the Poynting vector is independent of the type of representation of the electromagnetic field (vectorial or scalar). If some of the postulated conditions are violated, the propagation of the electromagnetic field will no longer be provided by solutions of (6.20).
Note: A principle similar to the Saint Venant principle in Solid Mechanics is valid for the Maxwell equations. At some distance of the region where the prescribed conditions are violated the scalar theory can be applied again. An example is the problem of diffraction caused by an aperture, at some distance from the aperture the scalar theory can be utilized to obtain the diffracted field.
6-S8
Prove that (6.61.a) and (6.61b) are correct.
Solution to 6-S8
The Poynting vector was defined by (6.56). The magnetic field is given by and the magnetic induction is related to the electrical field by (6.45). Hence .If we consider a harmonic solution of the Maxwell equations of the form and a similar equation for the magnetic induction field,, then the Poynting vector is,
(6.60)
Calculating the average value of the modulus of the vector during an interval of time T, .
When the limit of the above expression is . Then
and (6.61a)
In the case that the propagation takes place in vacuum (or as a good approximation in air), , in vacuum
Multiplying and dividing by .By utilizing the same steps applied in the preceding derivations, (6.61b)
6-S9
A solid state laser emits a collimated circular beam of 1 mm diameter from the beam conditioner system, a power meter indicates a 50 mW output. Compute the irradiance of the laser.
Solution to 6-S9
Since the power outputis given, irradiance is given by the power of the laser divided by the cross section of the beam that is a circle.
6-S10
In an optical system the axis of propagation is labeled the z-axis. A plane wavefront is propagating along the optical axis; the amplitude of the electrical field vector is 50 V/m. Calculate the flux density of the plane wave front.
Solution to 6-S10
Applying (6.61), from section 6.3 The Electromagnetic Theory of Light.
Rounding the value of , the value of
Taking into consideration units relationships , the resulting units are:
6-S11
The plane wave front of the preceding problem is converted into a spherical wave front by a converging lens. We assume that the plane wave front is square in shape and has a side of 7.5 cm. The converging lens is a plane convex lens of diameter 7.5 cm and focal distance f=24.5 cm. If the source is a helium-neon laser of λ=632.8 nm ,write the equation representing the spherical wave front.
Solution to 6-S11
From the solution of sample problem 6-S5 and taking into consideration that the wave front is converging towards the focal point of the lens, the lens the equation of the wave front is
Since is of the same direction than r, . The value of . The frequency of the 632.8 wavelength is then
The values of the flux density of the wave front must be computed. The power density in the wave front is 3.32 W/m2. Figure P.6.2 shows the front view and the side view of the optical system. The area of the wave front entering the lens is a circle of radius 3.5 cm. This area is converted in a spherical sector. Assuming very small loses in the lens, the density of energy has to change according to the ratio of the area of the circle of 3.5 cm that enters the lens and the spherical sector that emerges and is shown in Figure P.6.3.The tangent of the angle θ is equal to
. Then h=24.5(1-cos 8.13) = 24.50.010=0.245 cm.
The area of the spherical sector Figure P.6.3, is A=2πrxh and r=(h2+r12)/2h=25.12 cm2. Then A=38.67 cm2. The area of the entering wave front is, A=πx3.52=38.5 cm2.The energy density of the spherical wave front is the: 3.3238.5/38.67=3.305 W/m2. The equation of the spherical wave front is
Figure P6.2. Front and side view of the optical system.
Figure P6.3.Spherical sector.
6-S12
A plane polarized beam of light of =632.8 nm propagating in air enters in the direction of the normal of the surface of separation between air and glass. The index of refraction of the glass is na=1.55. What is the speed of propagation of the light in the glass? Give the wavelength of the light in glass and the state of polarization.
Solution to 6-S12
According to (6.26),
According to (6.24), , we can also write , where is the wavelength in air (vacuum).
Then since the frequency of light is a constant we arrive at. Then
. The state of polarization remains unchanged upon entering the glass.
6-S13
Compute the optical path of a light beam of wavelength λ=540 nm going through a glass plate of thickness 20 mm. The index of refraction of the glass is 1.52. Compute the optical path of a parallel beam going through air. Compute the difference of path of both beams. Convert the difference of optical path into a phase difference.
Solution to 6-S13
It is possible to represent the electrical and magnetic fields in exponential form as shown in (6.38) and (6.39), see section 6.3 Then taking into consideration sample problem 6-S4 they can be written as,
and ,
where and are constant vectors. Make the following consideration; in most of our optical developments of interest measurements are made on periods of times that are extremely long when compared to the frequencies of the propagating light beams. Utilizing this argument it is possible to ignore the time argument in the preceding equations. The spatial argument can be analyzed further. The modulus of the vector was derived analyzing the propagation in vacuum and the wavelength in vacuum is given as λ0 ; this expression is also a good approximation for propagation in air. To generalize the argument for other linear media write λi, calling i=1,2, 3…the wavelength of the different media that the beam of light goes through, and taking into consideration the result of sample problem 6-S4, ,we getthen the scalar product is. Looking at Figure 6.5it is possible to see that in the case of a plane wave front the vector is of the same direction of the normal to the wave front. Since the modulus of the normal vector is unity it is possible to write,
.
The above product is called the optical path of the light beam. In simple terms the above result tell us that when the light enters a linear medium of index of refraction nia the path is increased of the amount of the index of refraction with respect to the same path in vacuum and as a good approximation in air. The above property has been derived for a plane wave front. However it can be generalized for linear media for an arbitrary wave front propagating along a general trajectory in the linear space. Since according (6.61a) this is also the direction of the average Poynting vector, this implies that the energy propagates in the same direction. It follows that the average Poynting vector is normal to the geometrical wave front. Geometrical light rays can be defined as orthogonal trajectories to the wave fronts defined by S=constant. The direction of the rays coincides with the direction of the Poyting vector and the velocity of the vector is v=c/nia. Then the concept of optical path can be extended to a general light ray.