Some Mathematics of Light (Tuklas Vol. 13, No. 8 - Feb. 4, 2012)

SOME MATHEMATICS OF LIGHT

Light, in all intents and purposes, can be classified into two broad forms depending on how 'compact' their waves of propagation are. We have the ordinary household light, when turned on, brightens the room and stuns the would-be thief. On the other hand, there is the LASER, containing different combinations of gasses and emitting different colors after excitation due to electricity.

LASER, even when turned on (and even if they cost over US$1000), doesn't really make reading novels as comfortable as does your cheap reading light. What makes LASERs really expensive is its ability to compress and concentrate light waves into beams and avoid its travel to all directions. Hence a single beam of LASER contains an unbelievable amount of energy, sometimes enough to burn a piece of paper.

Naturally, how bright, how light behaves and what happens before and after passing something will all need to be quantified mathematically.

Matrices are extensively used in optics to compare beams before and after passing through a certain medium, be it a speck of dust, a lens, or mirror. These matrices are called Ray Transfer Matrices (RTM). Different matrices are derived and used in different instances to represent the media where light passes through. For example, if light just travels in free space with a distance $d$, then the RTM associated with this is: $$M = \begin{pmatrix} 1&d\\ 0&1 \end{pmatrix}.$$ If light passes through a thin lens of focal length $f$, then the associated matrix would be: $$M = \begin{pmatrix} 1&0\\ -\frac{1}{f}&1 \end{pmatrix}.$$ And if light passes through free space and then to a thin lens, then the new matrix will just be a product of the individual matrices: $$M = \begin{pmatrix} 1&d\\ 0&1 \end{pmatrix} \begin{pmatrix} 1&0\\ -\frac{1}{f}&1 \end{pmatrix}.$$ We now define a transformation to link these matrices before and after hitting a medium: $$q_2 = \frac{Aq_1+B}{Cq_1+D},$$ where $q_1$ is the beam parameter before, $q_2$ is the beam parameter after, and $ABCD$ are the individual entries of the matrix $$M = \begin{pmatrix} A&B\\ C&D \end{pmatrix}.$$ In general, if light hits medium 1, then 2, then 3, and so on, the effective RTM would be: $$M= \text{[Matrix 1]}\text{[Matrix 2]}\text{[Matrix 3]}\cdots \text{[Matrix n]}.$$ The transformation happens to be the same Möbius Transformation we find in Complex Analysis. This transformation proves to be useful since, together with the $ABCD$ matrices, they are able to derive very fundamental relations such as the focal length of a spherical lens, and a host of other confusing consequences. [1]

As with all scientific relations obtained in various fields, equations are idealized relationships with various assumptions aimed at simplifying matters. The Ray Transfer Matrix is no exception to this rule; where simplifications do not reflect the imperfections of lenses which are almost always present during the lens-making process. Such imperfections, or aberrations, are also accounted for in RTM's and are introduced through an addition we call perturbation. 

A perturbation, to say it simply, is a deviation from the original term. So if aberrations have caused a thin lens presented above to be imperfect, then the new matrix should read $$M = \begin{pmatrix} 1&0\\ -\frac{1}{f} + \Delta f&1 \end{pmatrix}.$$ Achievements have been made on the analysis of matrices and light: by just looking at the matrix, know the standard combinations of some RTM, then we can immediately describe what imperfections exist on the lenses. [2]

ABOUT THE AUTHOR:
Clark Kendrick Go is an Instructor at the Ateneo de Manila University. He is currently taking his M.S. in Physics at the same university.

REFERENCES:

[1] Johanna Mae Indias, Clark Kendrick Go, "Ray Transfer Matrix Model of an Elastomeric Lens", Applied Mechanics and Materials, Mechanical and Aerospace Engineering 110-116 (2011), 4145-4148.
[2] Jerry Barretto and Clark Kendrick Go, "A qualitative approach to aberrations of optical elements," Chin. Opt. Lett. 8 (2010), 1022-1024.

OLYMPIAD CORNER

from the Romanian National Contest, 2000

Problem: Show that there exist infinitely many 4-tuples of positive integers $(x,y,z,t)$ such that the four numbers' greatest common divisor is $1$ and such that $$x^{3}+y^{3}+z^{2}=t^{4}.$$

Solution: 

Setting $a=k^{3}$ for any even $k>0$, we have the identity $$(a+1)^{4}-(a-1)^{4}=8a^{3}+8a$$ which yields $$(2k^{3})^{3}+(2k)^{3}+\left[ \left( k^{3}-1\right) ^{2}\right]^{2}=(k^{3}+1)^{4}.$$ Because $k^{3}+1$ is odd, $\gcd (2k^{3},k^{3}+1)=\gcd (k^{3},k^{3}+1)=1$. Hence, there are infinitely many quadruples of the form $(x,y,z,t)=(2k^{3},2k,(k^{3}-1)^{2},k^{3}+1)$, for $k>0$ even, satisfying the required conditions.

SOLUTIONS

(for January 28, 2012)

19. 
    If $a_{n+1} = \frac{1}{1+\frac{1}{a_n}}, n = 1, 2, \ldots, 2011$ and $a_1 = 1$, find the value of $$a_1 a_2 + a_2 a_3 + a_3 a_4 + \ldots + a_{2011} a_{2012}.$$ (Taken from Lecture Notes on Mathematical Olympiad Courses by Xu Jiagu)
    
    
    
    SOLUTION: 
    Note that for $n = 1, 2, \ldots, 2011$, $$a_{n+1} = \frac{1}{1+\frac{1}{a_n}} = \frac{a_n}{a_n + 1} \Rightarrow a_{n+1} a_n + a_{n+1} = a_n$$ giving us $$a_n a_{n+1} = a_n - a_{n+1}$$ thus $$\begin{align*} a_1 a_2 + a_2 a_3 + a_3 a_4 + \ldots + a_{2011} a_{2012} \\ = (a_1 - a_2) + (a_2 - a_3) + \ldots + (a_{2011} - a_{2012}) \\ = a_1 - a_{2012}. \end{align*}$$ Also, note that the relation gives us the pattern $$a_2 = \frac{1}{2}, a_3 = \frac{1}{3}, a_4 = \frac{1}{4}, \ldots , a_{n-1} = \frac{1}{n-1},$$ from which $a_n = \frac{\frac{1}{n-1}}{\frac{1}{n-1}+1} = \frac{1}{n}$, thus $a_{2012} = \frac{1}{2012}$ and $$a_1 a_2 + a_2 a_3 + a_3 a_4 + \ldots + a_{2011} a_{2012} = 1 - \frac{1}{2012} = \frac{2011}{2012}.$$
20. One commercially available ten-button lock may be opened by depressing - in any order - the correct five buttons. Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many additional combinations should this allow? (American Invitational Mathematics Examination, 1988)
    
    
    SOLUTION: 
    The total number of combinations would be equal to $\binom{10}{1} + \binom{10}{2} + \ldots + \binom{10}{9} = 2^{10} - 2 = 1022$, which implies that the total number of additional combinations is simply $1022 - \binom{10}{5} = 770$.
21. Let $a$, $b$ and $c$ be three distinct positive integers. Show that among the numbers $$a^5 b - ab^5, b^5 c - bc^5, c^5 a - ca^5,$$ there must be one that is divisible by 8. (Taken from Lecture Notes on Mathematical Olympiad Courses by Xu Jiagu)
    
    
    SOLUTION: 
    Factoring, we have $$\begin{align*} a^5 b - a b^5 = ab(a-b)(a+b)(a^2+b^2), \\ b^5 c - b c^5 = bc(b-c)(b+c)(b^2+c^2), \\ c^5 a - c a^5 = ca(c-a)(c+a)(c^2+a^2). \end{align*}$$ If $a$ and $b$ have the same parity, then $a-b$, $a+b$ and $a^2 + b^2$ are all even, hence $a^5 b - ab^5$ is divisible by 8. If $a$ and $b$ are of different parity, then $c$ must be of the same parity as either of the two, and the same result follows (using the other two equations).