Determination of Shapes by Reflection
Introduction
Direct visual observation, is a method where human beings used their eyes to identify an object. However, not all things in life can be observed directly. For example, how can you tell the position of a broken bone? Is it possible to look at a baby inside a pregnant woman? How about identifying cancer cells inside a brain? All of these require a special technique involving indirect observation.
In this experiment, you are to determine the shape of an object using indirect observation. You will be given two closed cylindrical boxes and in each box, there will be an object with unknown shapes. Your challenge is to reveal the object without opening the box. The physics concepts for this experiment are simple, but creativity and some skills are needed to solve it.
Experiment Apparatus
For this experiment, you will be given two sets of cylindrical boxes consisting of:
(1) An object with unknown shape to be determined (it is a simple geometrical object with either plane or cylindrical sides).
(2) Closed cylindrical box with an angular scale on the top side (2a) and around its circumference (2b).
(3) A knob which you can rotate.
(4) A laser pointer.
(5) Spare batteries for the laser pointer.
(6) A ruler.
Experimental Method
The students are to determine the shape of the object inside a closed cylindrical box. The diameter of the cylinder can be measured by a ruler. Students are not allowed to open the cylindrical box or break the seal to determine the shape of the object. The object is an 8 mm thick metal with its sides polished so that it can reflect light likes a mirror. You can rotate the object using the knob on the top part of the cylinder. This will rotate the object in the same axis as the cylinders axis.
The laser pointer can be switched on by rotating its position. You can adjust the position of the light beam by rotating the laser pointer in either clockwise or anti-clockwise direction. The reflection of the laser beam from the laser pointer can be observed along the circumference of the closed cylinder. Measurement using the angular scale can be used. By rotating the knob on the upper part while the laser pointer is switched on, you will notice that as you rotate the object, the position of the reflected light from the object will change. If the light from the laser pointer dim or the laser pointer fail to work, ask the committee for replacement. By observing the correlation of the angular position of the object and the reflection of the laser beam, you should be able to determine the shape ofthe objec
For every object (the two objects are of different shapes) :
(A) Draw a graph of: reflection angle of the laser beam against the angular position of the object.
(B) Determine the number of edges (sides) in each object.
(C) Use data from the graphic to sketch the shape of the object and find the inside angles positions.
(D) Draw rotating axis of the object and determine the distance to every sides.
(E) Determine the length of sides without error analysis; determine also the angles between neighboring sides.
You must present your result on graph papers and try to deduce the mathematical equations to determine the shape of the object.
Remarks:
(1) One of the objects has only plane sides and the second object has one curved side.
(2) Sometimes you may get two reflections of the beam from the object.
(3) In case of a curved side the determination of the radius of curvature is not required but determination whether it is convex or concave with respect to the axis of rotation is necessary.
Table–1–
There are 3 jumps on the graph. This is observed at α =-10°, 140° and 220°. The jump in the reflection angles are caused by the change of sides, therefore the object has 3 sides and if all the sides are straight sides, we can approximate the lines on the graph using linear regression, i. e.
\beta=m\alpha+c,where α= position angle of the object (in °)
and β = reflected ray angle (in °)
Segment 1 (- 10 to 130): β = 1. 98α + c_{1} , (A1)
Segment 2 (140 to 210): β = 1. 73α + c_{2}, (A2)
Segment 3 (220 to 340) : β = 1. 78α +c_{3}. (A3)
To find the gradient, m, as function of side distance from the rotation
axis,
r, we can simulate it and get a graph and for “small” r:
m =-0. 02 r+2 or r = 100-50 m. (A4)
From (Al) to (A3) and using (A4) we can determine r from the 3 sides:
r_{1} = 100 – 50(1.98) = 1. 5 mm,
r_{2} = 100 – 50(1. 73) = 13.5 mm,
r_{3} = 100-50(1. 78) = 11.0 mm.
For each side, we can use the object position when the reflection angle is 0° to draw with a higher precision. The angle for each segment is:
\alpha_{1} = 66°,
\alpha_{2} = 189°,
\alpha_{3} = 282°.
From data obtain the shape of the object can be determined as:
Table–2–
There are 5 jumps in the graphics. This can be observed at α = 10° , 120° , 190° , 300° and 345° • The jumps in reflection angle are caused by the change of sides, therefore there are 5 sides in the object and if all the sides are straight, we can approximate the lines using linear regression, i. e.
\beta=m\alpha+c,where α = angular position of the object (in °)
and β = reflected ray angle (in °)
Segment 1 (10 to 110): β = 1. 56α+ c_{1}, (B1)
Segment 2 (120 to 1 80) : β = 2. 12α + c_{2} , (B2)
Segment 3 (190 to 290) : β = 1. 64α + c_{3} , (B3)
Segment 4 (190 to 290) : β = 0.15α + c_{4} , (B4)
Segment 5 (190 to 290) : β = 0.l0α + c_{5} . (B5)
From (B1) to (B3) and (A4) we can determine r from the 5 sides:
r_{1} = 100 – 50(1. 56) = 22.0 mm,
r_{2} = 100 – 50(2. 12) =-6.0 mm,
r_{3} = 100 – 50(1. 64) = 18.0 mm,
r_{4} = 100 – 50(0.15) = 92.5 mm,
r_{5} = 100 – 50(0.10) = 95.0 mm.
There are weird data for r_{2} , r_{4} and r_{5} . It is impossible to have r with either negative or very large value but not so small angle of reflection. So we can guess that it is either a curve sides or double reflection. For double reflection we need to have two adjacent sides with concave angle, so only r_{4} and r_{5} are possible. So r_{2} can only be a curve side. From segment 2 of the graph we can see that the graph looks like a reverse “S” shape, so it is only possible when the sides is concave.
Considering error in the experiment, we can guess that the shape has reflection symmetry.
For each side, we can use the object position when the reflection angle is 0° to draw with a higher precision. The angle for segment 1 to 3 is
\alpha_{1} = 58°,
\alpha_{2}= 149°,
\alpha_{3} = 237°.
From the data obtained, the shape of the object can be determined as
Table–1–
Object Position (α°) | Object Position (β°) | Cal. Reflected Ray (β°) |
-10 | 211 | -149 |
-5 | 223 | -137 |
0 | 230 | -130 |
10 | 251 | -109 |
20 | 270 | -90 |
30 | 290 | -70 |
40 | 310 | -50 |
50 | 330 | -30 |
60 | 349 | -11 |
Cont.
Object Position (α°) | Object Position (β°) | Cal. Reflected Ray (β°) |
75 | 17 | 17 |
80 | 28 | 28 |
90 | 46 | 46 |
100 | 65 | 65 |
110 | 87 | 87 |
120 | 117 | 117 |
130 | 128 | 128 |
140 | 273 | -87 |
150 | 291 | -69 |
160 | 310 | -50 |
170 | 326 | -34 |
180 | 343 | -17 |
195 | 10 | 10 |
200 | 12 | 12 |
210 | 34 | 34 |
220 | 249 | -111 |
230 | 267 | -93 |
240 | 285 | -75 |
250 | 301 | -59 |
260 | 319 | -41 |
270 | 337 | -23 |
275 | 345 | -15 |
290 | 13 | 13 |
300 | 29 | 29 |
310 | 46 | 46 |
320 | 64 | 64 |
330 | 83 | 83 |
340 | 102 | 102 |
Table–2–
Object Position (α°) | Object Position (β°) | Cal. Reflected Ray (β°) |
10 | 278 | -82 |
20 | 296 | -64 |
30 | 313 | -47 |
40 | 330 | -30 |
50 | 347 | -13 |
65 | 12 | 12 |
70 | 20 | 20 |
80 | 36 | 36 |
90 | 56 | 56 |
100 | 73 | 73 |
110 | 91 | 91 |
120 | 300 | -60 |
130 | 320 | -40 |
Cont.
Object Position (α°) | Object Position (β°) | Cal. Reflected Ray (β°) |
140 | 342 | -18 |
145 | 351 | -9 |
155 | 13 | 13 |
160 | 23 | 23 |
170 | 43 | 46 |
180 | 67 | 67 |
190 | 277 | -83 |
200 | 297 | -63 |
210 | 313 | -47 |
220 | 330 | -30 |
230 | 347 | -13 |
245 | 13 | 13 |
250 | 22 | 22 |
260 | 39 | 39 |
270 | 55 | 55 |
280 | 74 | 74 |
290 | 91 | 91 |
300 | 335 | -25 |
305 | 335 | -25 |
310 | 336 | -24 |
315 | 337 | -23 |
320 | 338 | -22 |
345 | 21 | 21 |
350 | 22 | 22 |
355 | 22 | 22 |
360 | 23 | 23 |
365 | 23 | 23 |