International GeoGebra Conference for Southeast Europe
January 14-16, Novi Sad, Serbia
Plane Mirror in GeoGebra Laboratory
ABDUL-SAHIB HSANI N.
Islamic Azad University, Abadan Branch, Abadan, Iran
abdulsahiba@yahoo.com
Abstract
The best description that can serve as an abstract and introduction to this article is a quote
from Michelangelo: “A drawing is the source and soul of every image and the root of
every science.” Keep in mind that the drawing in GeoGebra is live and dynamic. We can
easily move the objects, the mirrors, and/or the rays, and change the angles to satisfy or
break the reflection laws and do calculations followed by the desired results. Different
cases of reflection and image formation in plane mirror, for example two intersecting
mirrors, are explored in this paper using a dynamic and interactive worksheet created by
GeoGebra.
Note
All of the discussed problems are adopted from
Halliday & Resnick, Fundametals of Physics, John Wiley & Sons, Inc, New York 1981.
Moving Object and Moving Mirror
What would happen to the image if the mirror or the object move horizontally
and/or vertically?
Move the point object or move the mirror from its lower edge, point B, in the
window titled 'Moving Plane Mirror', and keep an eye on the image. Check the
distances from mirror to the object and image. Move the object downward or the
lower edge of the mirror upward; a checkbox, titled WHY?, will appear if point B
stays higher than point object. Now, the object apparently can see its image
because the mirror can not obscure it! Is it true? Is there any image for such
arrangement? Can you trace the light rays to the image, if any?
Torsion Balance, Transatlantic Cable, Galvanometer!
Do you see any connection between the three names in this strange title? The
following quotes will elucidate the strange title:
"One of the fundamental problems of transmitting a signal through a lengthy
submarine cable was that the electrical current tended to be very low, making
any signals sent difficult to detect. The chief electrician of the company
thought this problem could only be solved by sending larger doses of electric
current down the line. But Thomson had a different approach…
The dynamic worksheet Created with GeoGebra
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…Thanks to Thomson’s invention, the weak currents through the submarine
cable could now be easily read. Thomson’s mirror galvanometer required only
a little battery and enabled operators to read the Morse code coming through
the cable at 10 times the old rate. The invention of the mirror galvanometer led
to Thomson being knighted Lord Kelvin, the first scientist to receive such an
honor.[1]"
In torsion balance the angle of twist is determined by the use of a light pointer. The
Cavendish balance is an example of torsion balance that used for experimental
verification of Newton's law of universal gravitation (or may be used in the grocery
of physicists and physics teachers for selling gravitational law). Click play button
in the left lower corner of the window titled ‘Rotating Plane Mirror’ and watch
what would happen to the reflected ray when the mirror rotates around the incident
point.
You can investigate the reflected ray behavior thoroughly by stopping the mirror
(use stop button) and then moving it manually using slider RotateMirrorBy. How
do the angles beta and rho change? The geometry of the problem shows that:
 = 2 - 2( - ) = 2
A very simple and very important and practical result; rotating the mirror by 
causes the reflected ray to rotate 2 a result that led Thomson to invent the mirror
galvanometer.
Figure 1-Rotating and moving mirrors.
A Full Image in the Plane Mirror
How tall must a vertical mirror be if a person ## cm high is to be able to see her/his
entire length? Assume that her/his eyes are # cm below the top of his head.
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International GeoGebra Conference for Southeast Europe
January 14-16, Novi Sad, Serbia
When I read this problem for the first time, I was a student at high school and I
burst out laughing because I imagined # = 12 cm between the eyes and the crown
as a distorted and caricature figure. I guessed it to be a maximum of four
centimeters at most! But I am a speedy student and I measured the distance
between my eyes and the top of my head more than two decades later! Oh! What a
distorted and caricature figure, it was about 12 cm, ha ha ha!
By the way, I measured the discussed distance for my second daughter, 'Al'a, on
2011-01-10 and it was about 10 cm. Please be kind enough not to think about us as
cartoon. You may do measurement for yourself then you can take a look at the
following lovely picture (taken on 2011-01-10).
Figure 2 - My daughter ALA' the third of the three kids that make me be laughing.
Ok, now we can proceed with experiments:
1) Adjust the two sliders to ensure the law of reflection (use arrow keys for
fine adjustment):
β = γ or their difference becomes zero.
α = δ or their difference becomes zero.
Check the ratio of mirror length to person height. Sometimes there
would be some approximations in the equality or zero being.
2) Move point EYE and repeat the experiment. Different positions of EYE
may represent real configuration or comic one.
The dynamic worksheet Created with GeoGebra
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3) Y-coordinates of the points A, B, C, and EYE show their heights from
the floor. Calculate yB. [Hint: it is very easy just use the Why? checkbox
in the Geogebra construction.]
4) Do you think that a suitable arrangement of the mirror depends on the
distance between the person and the mirror? Move points C and D to the
left or right and watch the light rays and their corresponding angles. You
can change the person's height but be aware to set equal x-coordinates
for points C and D - a healthy person always stands vertically.
Figure 3 -A Full Image in the Plane Mirror
Area of the Mirror Used In Observing the Point Object
A point object is at a certain distance from a mirror while the eye of an observer,
with a known pupil diameter, placed at a distance twice that of the object. Find the
area of the mirror used in observing the reflection of the point.
To solve this problem do the following experiments:
Set the sliders alpha and beta to the value 90 degrees. The sliders titled
ObjectPosition must show the value 0.5 and sliders l and m adjust the incident
points, E and F, to ensure the reflection law. The segment D'D'' is the diameter of
the pupil!
Note: for fine adjustment of the sliders use the arrow keys.
1) To extend the upper incident ray type the command ray[E,O] into input
bar (O stands for Object).
2) select perpendicular line from the toolbar followed by selecting the
point O then segment AB (a line perpendicular to AB at point O)
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International GeoGebra Conference for Southeast Europe
January 14-16, Novi Sad, Serbia
3) select intersect two objects from the toolbar followed by selecting the
drawn perpendicular line then the segment ED' (point D appeared)
4) select the drawn ray then the segment D'D'' (point I appeared)
Take a look at triangles IED' and OED, they are similar therefore:
OB/AB = OD/ID' and the first side of this equality is equal to the value of slider
ObjectPosition, say s, and the segment OD is equal to the segment FE which is
the radius of the vision area, say x. Also D'I = D'D'' - x and finally we get
D'D'' / x = (1+s) / s
This equation holds for any object distance and represented as dynamic text in the
following GeoGebra applet.
The goal was to calculate the area of the mirror used in observing the reflection of
the point object but I calculated the ratio of pupil diameter to the diameter of the
desired area. Never mind:
Once upon a time there was a physics and mathematics teacher that wrote on
the blackboard (or anycolorboard) 'a' saying "the value b" and when a student
reminded h(er/im) about the value 'a' s(he) said: "oh, excuse me, the desired
value is 'c' not b"!
We assumed that both the eye pupil and the object were on the same line
perpendicular to the mirror - recall the values of sliders alpha and beta. Moving the
point PupilDiameter changes the pupil diameter and break the reflection law, to
ensure the reflection law adjust the sliders l and m. Moving point A horizontally
also changes the pupil diameter but the reflection laws hold for any arrangement
that satisfies them because the point O defined as (1 - s)*B + s*A where s is the
value of slider ObjectPosition that ranges from zero to one.
What will happen if the eye pupil and the mirror are not perpendicular to the
segment AB? Change the values of sliders alpha and beta degree by degree and
repeat the above experiments.
The dynamic worksheet Created with GeoGebra
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Figure 4-Area of the Mirror Used In Observing the Point Object
Two Intersecting Mirrors
A Step Toward Corner Reflector
The following construction shows a light source emitting a 'ray' that strikes a plane
mirror M1 at an angle of incidence i. find the angle between incident and emerging
rays. To solve this problem input the following commands into the Input Bar (you
can use copy/paste method):
line[LightSource,I_1]
line[I_2,C]
intersect[a,b]
angle[I_2,D,I_1]
(draw line a)
(draw line b)
(point D, the intersection of lines a and b)
(marks the angle  between lines a and b)
To see the value of angle , right click on the angle, click Properties, on tab Basic
select the option Name & Value from the drop down menu next to the checkbox
Show Label. Try different values of angles i and m; do you see any relationship
between angles , i, and m? In triangle I1I2D we have:
I1 + I2 = 180o - 
and due to reflection laws the mirrors bisect angles I1 and I2 of triangle I1I2D so in
triangle I1I2B we can write:
m + (I1 + I2) / 2 = 180o
thus
 = 2 m - 180o
The angle  is independent of incident angle. It governed only by the angle
between the mirrors and gets the value zero when the two mirrors are at right angle
- the incident and emerging rays are parallel. The angle  is an interior angle of a
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International GeoGebra Conference for Southeast Europe
January 14-16, Novi Sad, Serbia
triangle; sometimes you may need to use angle[I_1,D,I_2] instead of
angle[I_2,D,I_1] to get the proper angle. Delete the wrong angle before using the
new command.
If you still want to proceed with this experiment then you can think about a
construction with an added mirror perpendicular to the common axis of the two
mirrors. This form a corner reflector which is not easy to create with two
dimensional GeoGebra.
Recall that I1 and I2 are the interior angels of I1I2D. The previous calculations are
valid for m no less than /2, but what about m no more than this value? Conduct
some experiments for the following calculations:
Experiment 1: In I1I2D we have
I1 + I 2 =  - 
And in I1I2B
m + ( / 2 – I1 / 2) + ( – I2) / 2 = 
m - (I1 + I2) / 2 +  = 
2 m = I1 + I2 =  - 
=-2m
Hint: adjust the two sliders so that angle m remains less than 90 degrees and point
D stays in front of the two mirrors.
Experiment 2: In I1I2D we have again
I1 + I 2 =  - 
And in I1I2B
m + I1 / 2 + {I2 + ( – I2) / 2} = 
m - (I1 + I2) / 2 =  / 2
2 m + (I1 + I2) = 
2m+ -= 
=2m
Hint: adjust the two sliders so that angle m remains less than 90 degrees and point
D stays in back of one of the mirrors. Some rearrangements like zooming and
resizing may be needed. To resize and displace the mirrors move points A and B.
The worksheet can be moved by holding the control key and dragging the
construction by mouse.
The dynamic worksheet Created with GeoGebra
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Figure 5-A Step Toward Corner Reflector.
What Is the Number of Images That Can Be Seen?
Two plane mirrors makes an angle  with each other. What is the largest number of
images of an object placed between them that can be seen by a properly placed
eye? The object need not lie on the mirrors' bisector.
Solution:
Suppose that the object always being placed on the bisector of the mirrors and you
need not ‘place’ your eye properly in the GeoGebra worksheet. Each mirror
produces an image that acts as a virtual object for other mirror which in turn leads
to another image. If the real object lied on the mirrors' bisector and the mirrors'
angle is a factor of 360, then the sequence of images will terminate at two
overlaying images. The following constructions started with mirrors' angle of six
degrees which produced fifty-nine images (each mirror produced thirty). Starting
with image C' then C" the green points are images produced by green mirror M1
and the red points are images produced by red mirror M2. The black images started
with C'1 and produced by two mirrors in share. Tick the checkbox Show Bisector
to see two overlaying images on it labeled R and J". To trace image formation step
by step, delete points C' and C'1 then choose the tool Reflect Object about Line,
click on point C, and click a mirror; an image will appear on other side of the
mirror. In the next step choose the image and click the second mirror and so on.
Instead of deleting points C'1 and C', you can delete the object point, C, choose the
New Point tool and create a point object between two mirrors. This point may be a
free object that you can drag it anywhere or a dependent object lying on the mirrors
bisector and constrained to move along it. (A break time: count the word 'object' in
this paragraph, the first two are from physics and the last two are from GeoGebra
and the latter in parenthesis refers to the four words preceding it).
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Conduct experiments by setting the slider to  = 6, 8, 9, 10, 12, 15, 18, 20, 24, 30,
36, 40, 45, 60, 72, 90, 120, 180 degrees and check the number of images. What
about angles more than pi like 240 or 300?
For some values of mirrors' angle, with the object placed on the bisector, there will
be two images that lying on the mirrors' extension. Check the values 8, 24, 40, 45,
72, and 120 degrees. Why these values? Delete the object point C and copy paste
the following commands in the Input Bar then choose the proper value say 24.
rotate[A,β,B]
circularsector[B,A,A'_1]
sequence[rotate[e,k β,B],k,1,2*pi/β]
produces a point say A'1.
produces a sector say e.
see its product after executing it.
tick the show mirrors extension box and investigate the figure carefully. You may
need to move the point A and B to obtain a good scene of the figure. Start from the
sector you created and count sectors until you reach the extension of one of the
mirrors. If you chose 24 degrees then you would count seven sectors and a half in a
semicircle, for such arrangement the extension of each radius bisects the angle of
the opposite sector, hence:
n + /2 = 180o  2n + 1 = 360o / 
therefore, considering Images No. = 360o/ - 1, the number of images will be even,
2n. For odd number of images there will be an image on the extension of the mirror
bisector, prove it.
It would be a good idea to think about the situation of images before we leave the
GeoGebra  laboratory. Click the reset button on the upper right corner then
choose the Circle with Center through Point tool and click on points B and C
successively.
Repeat the same experiments ( = 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60,
72, 90, 120, 180) with object that is not placed on the bisector of the mirrors. Move
the object (physics idiom!) back and forth between the mirrors and go close to
point B and far away from it. What did you see? Nothing for the cases in which the
overlapping images placed on the extension of the bisector except that the images
grouped in pairs and the remaining one grouped with the object. Remaining one?
Why one image remained singular? [Hint: keep an eye on the dynamic texts in the
construction.]
For the cases in which no images placed on the extension of the bisector the
images apparently doubled and grouped in pairs but there is a reminding sign
between the mirrors. Do you notice it?
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Figure 6-What Is the Number of Images That Can Be Seen?
[1] Herman Erlichson, "Kelvin and the TransAtlantic Cable," Phys. Teach.44, 426–427 (Oct. 2006).
Available on: http://scitation.aip.org
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