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.
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denoted here by da, db and dx (dphi has been replaced by da).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.
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.
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)
2. As the two possible solutions for m are perpendicular to each other we get m' * m'' = -1 .
Last edited: 30.06.2014