Sample Solutions

Sample Solution of Exercise 1 (b2)

A sketch of the problem stated in exercise 1 (b2) is shown in the right illustration: A horizontal ray coming from the right side is reflected two times on circular shaped mirrors of radius r and is horizontally oriented again. This is repeated until the end of the circle lattice on the left side is reached. The distance between to circle centers is the lattice spacing and is denoted by g. The angle 'a' between the original ray and the normal (line from circle center to reflection point) and the angle between the normal and the reflected ray are the same due to the law of light reflection. From the horizontal symmetry of the problem it follows that the intersection point of the first reflected ray and the lattice lies on the vertical line connecting the two circle centers. The problem is now to find the right value for y.

The following equations describe the geometry of the problem:
Next we simplify the argument of the tan function:
We plug this into the equation of the second line and replace tan(a) and x by the equations of the first line:
Finally we multiply both sides with the denominator of the right side to obtain a quadratic equation from which the solution for y immediately follows:
Here we omitted the second solution of the quadratic equation as y must be positive. To calculate a height for a 10x10 circle lattice and a radius of 50% of the maximum radius as it is required in the exercise we use the arbitrary precision calculator:
The value of the first line represents the lattice spacing g because the square containing the circle lattice has always a side length of 2. The second line divides the spacing by 2 and multiplies it by 50 % to get the required radius. The expression in the values column (g=$1) means that the result from line 1 should be used for the variable g. Line 3 represents the solution for y and line 4 calculates a possible height h. This notation has the advantage that only a single number (10) has to be changed to recalculate the height for another lattice size.
Next we copy and paste the result for the height into the height field of the variables window (Double-click on the value in the calculator and press ctrl-c. Then double-click on the height field in the variables window and press ctrl-v). The result should look like the above image. Pressing enter on that field gives us the solution of the exercise:
Note: By default the variables window shows only those decimal places that are necessary to reproduce the pattern. That means that only a few of the last digits need to be removed until the pattern starts to change. Via the menu of the variables window this truncation behavior can be switched off. The maximum amount of possible digits (precision) can be controlled with the precision menu entry of the main window.

Sample Solution of Exercise 3

1. On the main page it has been shown that unavoidable initial errors of the light source position quickly blow up with subsequent reflection points calculation. This initial error is about 10^-250 if a precision of 250 has been chosen (available precisions are 125, 250, 500, ...). The error estimation can therefore be done with infinitesimal quantities denoted here by da, db and dx (dphi has been replaced by da).
The right image shows a horizontal beam of length l1 and initial angle uncertainty da1 which is reflected into a beam of length l2 and angle uncertainty of da2. dx denotes the uncertainty that the end point of the reflected beam has. We assume that a uncertainty of the height of the initial beam has been taken into account by making da1 bigger.

To obtain the first equation for r*db only simple trigonometry is needed. r*db is the length of the bold arc. The second equation giving da2 follows from the two sketches at the bottom of the image: The fist sketch shows an angle change of 2 db/2 of the reflected beam if the normal is rotated from n to n' by an angle of db/2. The initial ray is still horizontal here. The second sketch rotates the initial beam by an angle of da1/2 which causes the same angle change on the final beam. These two contributions need to be multiplied by 2 because of the similar situation at normal n''.

The third equation for dx has two contributions: The first follows from the observation that the projected bold arc stays the same if we look from l1 or l2. This yields the first contribution l1*da1. The second comes obviously from rotating l2 by da2. Here we do not have to take into account that the two l2 boundary beams start from different height as this would only make a da2^2 contribution.

The second equation for dx defining da2' is done to construct the same situation as we have with the initial ray: All uncertainty about the position of the reflection point is subsumed in da2' > da2. da2' can be obtained in the image by shifting da2 up to the reflection point.

Inserting all previous equations into the second dx equation yields the last one. Let n1 and n2 denote the normalized direction vectors of the two beams. Then we have n1*n2 = cos(2b). This can be expressed in terms cos(b): cos(2b) = 2*cos(b)^2 - 1. Now we can express the last equation in terms of n1*n2 and the first part of this exercise is solved.

2. The solution follows from the equation for da2. To get the required formula one needs to multiply the following contributions:
dan" / da1 = [dan" / da(n-1)"] [da(n-1)" / da(n-2)"] ... [da3" / da2] [da2 / da1]
As the uncertainty of the position of the reflection point is always neglected we obtain a dai" < dai for i > 2.

3. The hint works out to: li >= 2*(rmax-a*rmax) which is the distance between two circles and sqrt(1-n1*n2) <= 2.

To find out how this formula affects the number of accurate decimal decimal places set
4/a - 3 = 10^y => log(4/a - 3) = y
and plot the graph for y. Example: a = 0.5 => y = 0.7 . This means that for a radius of r = 0.5rmax the number of accurate decimal places is reduced at least by 0.7 per reflection point. So three reflection points make a reduction of about 2 (3*0.7) decimal places.

Sample Solution of Exercise 4

The height numbers are constructed in such a way that the last nonzero digit controls the final beam without changing the complete path. The data shows a nearly linear increment of the precision with little derivation from the average. A number characterizing this experiment is given by the gradient of the linear fit and can be inerpreted as the average error increment per reflection. The open source software gnuplot shows the following result for the fit made with y = mx+b:
  m = 1.199 +/- 0.013 (1.1%)
  b = 3.3   +/- 0.9   (28%)
The error increases 1.2 digits per reflection on average. b is the average precision the last reflection point had in step 6 of the experiments.

Sample Solution of Exercise 5

1/0.001 = ( 4/0.5 - 3 )^(n-1) => log(1000) = (n-1) log (5) =>
n - 1 = 3/log(5) = 4.3
A laser with a given angle uncertainty of dphi1 = 0.001 rad would allow patterns with a maximum of 4 reflection points for r = 0.5 rmax.

Sample Solution of Exercise 6

1. A circle lattice of dimension s x s has s-1 rows and columns. The first and the last columns are occupied by line transitions so the total number n of characters that can be displayed works out to:
n = (s-1)*(s-1)-2*(s-1) = (s-3)*(s-1).

2. The distance between the two circle centers in the image on the right side is given by 2/s where 2 is the side length of the square containing the circle lattice. The solution for the required angle is given by the condition:
sin(phi) = r / (2/s - r) (*)

3. rmax = 2/(2*s) and r = a*rmax. Now we plug this two equations into the equation for sin(phi):
sin(phi) = a / (2 - a)

Sample Solution of Exercise 8

1. A direction vector n of a line with gradient m looks like this: n = (Δ x, Δ y) = (Δ x,m Δ x). This vector is parallel to the vector n' = (1,m). To find the law of reflection in terms of gradients we start with the following equation:
The first vectors on either side are the normalized direction vectors of the lines with gradients m1 and m2. When these normal vectors are scalar multiplied with the direction vector of the line with gradient m we must get the same result. The plus and minus sines account for the two possible solutions for the line with gradient m.
Next we calculate the dot products and square both sides:
This equation is quadratic in m and can be brought into the following form:
Using the quadratic formula yields the the two possible solutions for m and the first part of the exercise is answered.

2. As the two possible solutions for m are perpendicular to each other we get m' * m'' = -1 .

Last edited: 30.06.2014