Download PHYS20312 Wave Optics -‐ Section 1: Electromagnetism

PHYS20312 Wave Optics -­‐ Section 1: Electromagnetism 1.0 Introductory Remarks Optics – the study of EM waves when: 1. in a vacuum or in a dielectric medium (e.g. air or glass) 2. at very high frequencies ( e.g. ν ~ 1015 Hz for visible light) 2) is very important. It means that only the time-­‐average of the optical field is observed and determines the conditions under which wave-­‐like behaviour is seen. 1.1 A reminder of Maxwell Equations in a Vacuum or in an Isotropic Dielectric and the Wave Equation. c.f. your E&M notes Maxwell’s equations (1.1) : Gauss’ Laws ∇ ∙ 𝑬 = 0 (no free charges) ∇ ∙ 𝑩 = 0 (no magnetic monopoles) Ampere-­‐Maxwell Law ∇×𝑩 = 𝜀𝜇𝑬 Faraday’s Law ∇×𝑬 = −𝑩 (no currents) where 𝜀 = 𝜀! 𝜀! & 𝜇 = 𝜇! 𝜇! , and for a vacuum 𝜀! = 𝜇! = 1 and for a dielectric 𝜀! ≠ 1; 𝜇! ≅ 1 The Wave Equation : ∇! 𝑬 = 𝜀𝜇𝑬 (1.2) We recognise this as the equation for a wave travelling at a velocity, v where v = 1
𝜀𝜇
= 𝑐
𝜀! 𝜇! = 𝑐 𝑛 and c = 1
𝜀! 𝜇!
and refractive index, 𝑛 ≅ 𝜀! 1 | P a g e PHYS20312 Wave Optics -­‐ Section 1: Electromagnetism Properties of EM waves Neatly summarised by: v𝑩 = 𝒌×𝑬 (1.3) where 𝒌 is the unit propagation vector Eqn. (1.3) implies that : 1. EM waves are transverse 2. 𝑬 and 𝑩 are orthogonal and in-­‐phase. Energy carried by EM waves The instantaneous flow of energy per unit area per sec (or flux) is given by the Poynting vector, S: !
𝑺 = 𝑬×𝑩 (1.4) !
the unit vector, 𝑺 , gives the direction of energy flow i.e. the Poynting vector is where a beam of light is pointing Using eqn. (1.3) and v = 1
𝜀𝜇
: !
𝑺 = 𝑆 = 𝐸 𝜂 where the ‘plane-­‐wave impedance’ 𝜂 =
𝜇
𝜀 ; in a vacuum, this becomes 𝜂! =
𝜇!
𝜀! = 377Ω However, since 𝜈 ~1015 Hz for visible light we generally observe the time-­‐averaged flux, which is known as the irradiance or intensity, 𝐼 , i.e. 𝐼= 𝑆
!
1
=
𝑇
!!! !
!!! !
1
𝑆 𝑡′ 𝑑𝑡 ==
𝑇
!
!!! ! !
𝐸
!!! !
𝑡′
𝜂
𝑑𝑡′ where T is the time-­‐averaging interval. In your E&M lectures, you saw that one solution to the wave equation is 𝑬 𝒓, 𝑡 = 𝑬𝟎 cos 𝒌. 𝒓 − 𝜔𝑡 and in the example sheet you will show that for this solution 𝐼 =
𝐸!!
2𝜂 However, it is often mathematically more convenient to use instead 𝑬 𝒓, 𝑡 = 𝑬𝟎 exp 𝑖 𝒌. 𝒓 − 𝜔𝑡 where only ℜ 𝑬 has physical significance. 2 | P a g e PHYS20312 Wave Optics -­‐ Section 1: Electromagnetism But now 𝐸 ! 𝑡 ! = 𝑬𝑬∗ = 𝐸!! i.e. we have lost a factor of one half. Hence for consistency when using the exponential form, we redefine intensity as 𝐼=
𝑬𝑬∗
2𝜂
!
=
𝐸!!
2𝜂
1.2 General Solutions of the Wave Equation. v ! ∇! 𝑬 = 𝑬 •
•
a 2nd order partial differential equation 4 independent variables o 1x time variable o 3x space variables (e.g. x,y,z or r,θ,φ etc) We are now going to use the separation of variables technique to obtain a time-­‐independent version of the wave equation: Let 𝑬 𝒓, 𝑡 = 𝑬! 𝒓 𝐸! 𝑡 and then sub. into the wave equation: v ! ∇! 𝑬! 𝐸! = 𝑬! 𝐸𝒕 v ! 𝐸! ∇! 𝑬! = 𝑬! 𝐸𝒕 divide thru by 𝑬 to get equation 1.5: v ! ∇! 𝑬!
𝐸𝒕
=
𝑬!
𝐸𝒕
now LHS is a function of r only and RHS is a function of t only; this can only be true if both sides are equal to a constant. 1.2.1 Variation with time. Consider RHS of equation 1.5 and set it equal to a constant, −𝜔 ! (chosen because we are expecting an oscillating solution): 𝐸𝒕
= −𝜔 ! 𝐸𝒕
𝐸! = −𝜔 ! 𝐸𝒕 3 | P a g e PHYS20312 Wave Optics -­‐ Section 1: Electromagnetism which has the solution (eqn 1.6): 𝐸𝒕 = 𝐸𝒕,! 𝑒𝑥𝑝 ±𝑖𝜔𝑡 •
from now on we will set 𝐸𝒕,𝟎 = 1 i.e. all the amplitude information is carried with 𝑬! •
•
𝜔 = 2𝜋𝜈 is the frequency of oscillation 𝐸! = 𝐸!,! 𝑐𝑜𝑠 𝜔𝑡 and 𝐸! = 𝐸!,! 𝑠𝑖𝑛 𝜔𝑡 (or a linear combination) are also solutions but we will use eqn 1.6 because it is mathematically more convenient with the real part representing the physical quantity. 1.2.2 Variation with space. Now consider LHS of equation 1.5, also set equal to the constant, −𝜔 ! : ∇! 𝑬! = −
𝜔!
𝑬 v! !
∇! 𝑬! = −𝑘 𝟐 𝑬! (1.7) (1.8) or if the vector direction is constant then a scalar version can be used: ∇! 𝐸! = −𝑘 ! 𝐸! Eqn (1.7) is known as the “Helmholtz equation” •
•
𝑘 is the wave number 𝒌 is the propagation vector or wave vector 𝑘 and 𝒌 are related by: 𝑘= 𝒌 =
𝜔 𝑛𝜔
2𝜋
=
= 𝑛 v
𝑐
𝜆
N.B. sometimes ‘wave number’ is defined as 𝑘=
1 𝜈
= 𝜆 𝑐
and given units of cm-­‐1. The full solution to the wave equation is the product of the time-­‐ and space-­‐varying parts: 𝑬 𝒓, 𝑡 = 𝑬! 𝒓 𝑒𝑥𝑝 ±𝑖𝜔𝑡 However, for much of this course 𝐸! will not be important and so we’ll just consider 𝑬! and the Helmholtz equation, or 𝐸! and equation 1.8. (And from now on I’ll drop the subscript for the sake of brevity i.e. 𝑬 = 𝑬! and 𝐸 = 𝐸! ). 4 | P a g e PHYS20312 Wave Optics -­‐ Section 1: Electromagnetism 1.3 Particular Solutions of the Helmholtz Equation . The Helmholtz equation (1.7) has many different solutions, which depend on the symmetry and the boundary conditions. We’ll now consider three different particular solutions which we will need later in the course: 1. Plane waves (see section 1.3.1) are used for: a. distant sources and parallel rays b. spatial frequencies c. Fourier optics d. Fraunhofer diffraction 2. Spherical waves (see section 1.3.2) are used for: a. Huygens wavelets b. Huygens-­‐Fresnel principle c. Fresnel diffraction d. near sources e. focussed light 3. Cylindrical waves (see example sheet) are used for: a. Young’s slits b. diffraction from a slit 1.3.2 Plane waves. Consider the case when the 𝑥, 𝑦, 𝑧 components are independent and thus separable, i.e.: 𝐸 𝑥, 𝑦, 𝑧 = 𝐸! 𝑥 𝐸! 𝑦 𝐸! 𝑧 Substituting this into the scalar Helmholtz wave equation (eqn 1.8) and dividing by𝐸 𝑥, 𝑦, 𝑧 : 𝐸!!! 𝐸!!! 𝐸!!!
+
+
= −𝑘 ! = − 𝑘!! + 𝑘!! + 𝑘!! 𝐸! 𝐸! 𝐸!
(N.B 𝑘 ! is a constant and so we are free to say it is the sum of three others). This yields the following solutions: 𝐸! 𝑥 = 𝐸!,! exp 𝑖𝑘! 𝑥 ; 𝐸! 𝑦 = 𝐸!,! exp 𝑖𝑘! 𝑦 ; 𝐸! 𝑧 = 𝐸!,! exp 𝑖𝑘! 𝑧 ; Hence, 𝐸 𝑥, 𝑦, 𝑧 = 𝐸! exp ±𝑖 𝑘! 𝑥 + 𝑘! 𝑦 + 𝑘! 𝑧
= 𝐸! exp ±𝑖𝒌. 𝒓 and if we also include the time-­‐varying part 𝐸 𝑥, 𝑦, 𝑧, 𝑡 = 𝐸! exp 𝑖 𝒌. 𝒓 ± 𝜔𝑡
(1.9) 5 | P a g e PHYS20312 Wave Optics -­‐ Section 1: Electromagnetism 1.3.2 Wavefronts. A wavefront is a surface of constant phase and its shape can be determined from the wave equation at a particular instant of time. For example, the wavefront corresponding to equation 1.9 is given by: 𝒌. 𝒓 ± 𝜔𝑡 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 At a particular instant (i.e. 𝑡 fixed) this yields 𝒌. 𝒓 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, which is the equation for a plane (hence the name of this solution to the wave equation). 𝒌 is normal to the wavefront in isotropic media but, as we will see in the next section, this is not always the case in anisotropic media. The idea of a ‘ray’ used in geometric optics corresponds to the normal to the wavefront even in anisotropic media, and hence is sometimes called the wave normal. Wavefront velocity. If the phase is then its rate of change with time is 𝜙 = 𝒌. 𝒓 ± 𝜔𝑡 𝑑𝜙
𝑑𝒓
= 𝒌. ± 𝜔 𝑑𝑡
𝑑𝑡
but for a wavefront 𝜙 = 𝑐𝑜𝑛𝑠𝑡, hence and 𝑑𝜙
= 0 𝑑𝑡
𝑑𝒓
𝜔
= ∓ 𝑑𝑡
𝒌
The velocity in the 𝒌-­‐direction is thus 𝑑𝑟
𝜔
= ∓ = ∓v! 𝑑𝑡
𝑘
i.e. the phase velocity. Hence, 𝐸 𝑥, 𝑦, 𝑧, 𝑡 = 𝐸! exp 𝑖 𝒌. 𝒓 − 𝜔𝑡 travels in the +𝒌-­‐direction at v! 𝐸 𝑥, 𝑦, 𝑧, 𝑡 = 𝐸! exp 𝑖 𝒌. 𝒓 + 𝜔𝑡 travels in the -­‐𝒌-­‐direction at v! 6 | P a g e PHYS20312 Wave Optics -­‐ Section 1: Electromagnetism 1.3.2 Spherical Waves Consider a point source from which light spreads out (or converges to) isotropically so that 𝐸 𝒓 =
𝐸 𝑟 For spherical symmetry, the scalar wave equation becomes ∇! 𝐸 =
1 𝜕 ! 𝑟𝐸
= −𝑘 ! 𝐸 𝑟 𝜕𝑟 !
Regarding the product 𝑟𝐸 as the dependent variable: 𝜕 ! 𝑟𝐸
= −𝑘 ! 𝑟𝐸 𝜕𝑟 !
𝑟𝐸 = 𝐴𝑒𝑥𝑝 ±𝑖𝑘𝑟 𝐸 𝑟 =𝐴
𝑒𝑥𝑝 ±𝑖𝑘𝑟
𝑟
(1.10) where 𝐴 is a constant. N.B. 𝐼 𝑟 ∝ 𝐸𝐸 ∗ =
𝐴
𝑟!
Hence, this is a spherical wave as expected. Full solution is 𝐸 𝑟, 𝑡 = 𝐴
exp 𝑖 𝑘𝑟 ± 𝜔𝑡
𝑟
As for the plane wave, the wavefront velocity can be found by considering how a surface of constant phase travels and yields: v! =
Hence, 𝜔
𝑑𝑟
= ∓ 𝑘
𝑑𝑡
𝐸!
exp 𝑖 𝑘𝑟 − 𝜔𝑡 𝑟
corresponds to a spherical wavefront expanding or diverging at v! , and 𝐸 𝑟, 𝑡 =
𝐸 𝑟, 𝑡 =
𝐸!
exp 𝑖 𝑘𝑟 + 𝜔𝑡 𝑟
corresponds to a spherical wavefront contracting or converging at v! . 7 | P a g e PHYS20312 Wave Optics -­‐ Section 1: Electromagnetism 1.4 Optical Spectra. 1.4.1 Frequency spectrum. If 𝐸𝒕 = 𝐸𝒕,! 𝑒𝑥𝑝 −𝑖𝜔𝑡 is a solution of the wave equation then so is any linear combination, i.e.: !
𝐴! 𝑒𝑥𝑝 −𝑖𝜔! 𝑡 𝐸! 𝑡 =
!!!
or if 𝜔 varies continuously, we can define !
𝐸! 𝑡 =
!!
𝐴! 𝜔 𝑒𝑥𝑝 −𝑖𝜔𝑡 𝑑𝜔 (1.11) (1.12) which we recognise as an inverse Fourier transform, where 𝐴! 𝜔
=
!
1
2𝜋
!!
𝐸! 𝑡 𝑒𝑥𝑝 +𝑖𝜔𝑡 𝑑𝑡 is the corresponding Fourier transform and is a definition of ‘the spectrum’ of frequencies, 𝜔 . (For more on Fourier transforms see you Waves & Fields notes or Hecht Section 11.1 and 11.2) 1.4.2. Spatial Frequencies. Similarly, for the space-­‐varying part we have: !
𝐸 𝑥 =
!!
𝐴! 𝑘 𝑒𝑥𝑝 𝑖𝑘𝑥 𝑑𝑘 𝐴! 𝑘
=
1
2𝜋
(1.13) (1.14) !
!!
𝐸 𝑥 𝑒𝑥𝑝 −𝑖𝑘𝑥 𝑑𝑥 where 𝐴! 𝑘 is the “spatial spectrum” comprised of “spatial frequencies”, 𝑘 . Note: 8 | P a g e PHYS20312 Wave Optics -­‐ Section 1: Electromagnetism •
•
Eqn 1.13 and 1.14 imply that any spatial distribution of light can be constructed from a superposition of plane waves. In this example, we have restricted propagation to being along the 𝑥 -­‐axis to ensure it yields the 1D form the Fourier transform that you’ve seen before. In general, however, 𝒌 is a vector and so the spatial spectrum can consist of plane waves with different wavelengths and orientated in different directions. 1.5 Principles of Propagation. Huygen’s Wavelets. •
Every point on a wavefront acts as a source of secondary wavelets (with the same v! and 𝜐 •
•
etc) These wavelets destructively interfere in all but one direction to form a new wavefront The process repeats and thus the wave propagates N.B. That portion of a wavelet that crosses a distance in the minimum time (i.e. by a direct path) will be in phase with portions of the other wavelets that crossed the same distance in the same minimum time. This leads on to… Fermat’s Principle – The Principle of Least Time. Light takes the minimum time to travel between two points In other words, to find the path taken by a light ray we need to minimise the ‘optical path length’, 𝐿 , where !
𝑛! 𝑙! 𝐿=
!!!
where 𝑛! is the refractive index of the ith region through which the light passes and 𝑙! is the length of the light path in that region. 𝐿 takes into account the phase retardation (i.e. speed reduction) caused by higher values of 𝑛. We will now see how this works in practise . 9 | P a g e PHYS20312 Wave Optics -­‐ Section 1: Electromagnetism 1.5.1. Example 1 – Snell’s Law Snell’s law can be derived from Fermat’s principle i.e. by minimising 𝐿: Figure 1 Snell's Law (Image courtesy of A. Pedlar). Optical path SP, 𝐿 = 𝑛!"# 𝑥 ! + ℎ! + 𝑛!"#$$
𝑎−𝑥
Differentiate with respect to 𝑥 𝑑𝐿
𝑥
= 𝑛!"#
+ 𝑛!"#$$
𝑑𝑥
𝑥 ! + ℎ!
!
+ 𝑏 ! 𝑥−𝑎
𝑎−𝑥
!
+ 𝑏!
The optical path is minimised when 𝑑𝐿
= 0 𝑑𝑥
i.e. when 𝑛!"#
𝑥
𝑥 ! + ℎ!
= 𝑛!"#$$
𝑎−𝑥
𝑎−𝑥
!
+ 𝑏!
𝑛!"# sin 𝜃! = 𝑛!"#$$ sin 𝜃! QED 10 | P a g e PHYS20312 Wave Optics -­‐ Section 1: Electromagnetism 1.5.2. Example 2 – The Law of Reflection. Deriving the law of reflection using Fermat’s principle is set as a tutorial question. Use the following the diagram to help: Figure 2 The Law of Reflection. and follow this procedure: 1. Write the optical path, 𝐿 , between S and P as a function of 𝑥 !"
2. Find !"
3. Setting !"
!"
= 0 should then yield 𝜃! = 𝜃! Reflection from a Curved Mirror. This is also set as a tutorial question and yields the following relationship: 1 1 2
+ = so si R
where 𝑅 is the radius of curvature for the mirror, and 𝑠! and 𝑠! are the distances from the mirror to an object and to its reflected image, respectively. 11 | P a g e PHYS20312 Wave Optics -­‐ Section 1: Electromagnetism 1.5.3 Refraction at a Spherical Surface. One or both of the surfaces of a lens are spherical and so understanding refraction at such as surface is key to understanding how light propagates through lenses. We will use Fermat’s principle to find out how light is refracted at a single spherical surface and then combine the effects of two surfaces to derive the lens equation. Figure 3 Refraction at a spherical surface. The optical path: 𝐿 = 𝑛! 𝑙! + 𝑛! 𝑙! where 𝑛! is the refractive index for the object ray (in the medium before the surface i.e. to the left in Fig 3) and 𝑛! is the refractive index for the image ray (in the medium after the surface i.e. to the right in Fig 3). Using the cosine rule for triangle SAC: 𝑙!! = 𝑠! + 𝑅
!
+ 𝑅 ! − 2𝑅 𝑠! + 𝑅 cos 𝜃 Similarly, for triangle CAP: 𝑙!! = 𝑠! − 𝑅
!
+ 𝑅 ! − 2𝑅 𝑠! − 𝑅 cos 180° − 𝜃 𝑙!! = 𝑠! − 𝑅
!
+ 𝑅 ! + 2𝑅 𝑠! − 𝑅 cos 𝜃 Differentiating 𝐿 with respect to 𝜃 𝑑𝐿 𝑛! 𝑅 𝑠! + 𝑅 sin 𝜃 𝑛! 𝑅 𝑠! − 𝑅 sin 𝜃
=
−
𝑑𝜃
𝑙!
𝑙!
12 | P a g e PHYS20312 Wave Optics -­‐ Section 1: Electromagnetism -­‐when 𝑑𝐿
= 0 𝑑𝜃
𝑛! 𝑠! + 𝑅
𝑛! 𝑠! − 𝑅
=
𝑙!
𝑙!
for small 𝜃 , 𝑙! ≈ 𝑠! and 𝑙! ≈ 𝑠! which yields 𝑛! 1 +
𝑅
𝑅
= 𝑛! 1 +
𝑠!
𝑠!
[This is known as the “paraxial approximation” or sometimes the “First order approximation” or “Gaussian optics”]. Re-­‐arranging: 𝑛! 𝑛! 𝑛! − 𝑛!
+ =
𝑠! 𝑠!
𝑅
(1.15). Note the eqn. (1.15) has no dependence on 𝜃 and so all rays from S will converge at P. Since the surface is spherically curved then there is nothing special about the line SP; we can draw other lines through the centre of the lens and the same analysis would be valid. We can thus generalise this result to say that, for small angles, all the rays from any object point will converge on a corresponding image point. In other words, there is a unique ‘mapping’ of points from object to image i.e. the optical field in the object plane is reproduced in the image plane (although it may have been shrunk or expanded). This is what we mean when we say an image has been formed. Now let’s consider two spherical surfaces combined together to form a lens. The image formed by the first will act as the object for the second: Figure 4 Propagation through a lens. 13 | P a g e PHYS20312 Wave Optics -­‐ Section 1: Electromagnetism Writing down eqn. (1.15) for each surface. For the first surface: 𝑛!
𝑛!
𝑛! − 𝑛!
+
=
𝑠!,! 𝑠!,!
𝑅!
and for the second: 𝑛!
𝑛!
𝑛! − 𝑛!
+
=
𝑠!,! 𝑠!,!
𝑅!
The object distance for the second surface, 𝑠!,! , is related to the image distance for the first surface, 𝑠!,! , by 𝑠!,! = − 𝑠!,! − 𝑑 and is negative because it is to the right of the lens. Combining these we have 1
𝑠!,!
+
1
𝑛! − 𝑛! 1
1
𝑛!
𝑑
=
−
+
𝑠!,!
𝑛!
𝑅! 𝑅!
𝑛! 𝑠!,! 𝑠!,! − 𝑑
For a ‘thin lens’, 𝑑 is negligible and we obtain “The Lensmaker’s Formula”: 1 1 𝑛! − 𝑛! 1
1
+ =
−
𝑢 𝑣
𝑛!
𝑅! 𝑅!
(1.16) where 𝑢 = 𝑠!,! is the object distance and 𝑣 =𝑠!,! is the image distance. The RHS of the Lensmaker’s equations is determined solely by the properties of the lens and the refractive index of the surrounding medium, and characterises how it refracts light. Replacing the RHS with a single parameter, we get the “The Lens Equation”: 1 1 1
+ = 𝑢 𝑣 𝑓
(1.17) where 𝑓 is the focal length of the lens. 14 | P a g e