Geometry Forum - Problem of the Week
Solutions - Angles of Incidence and Reflection, Dec. 5-9, 1994
Annie says:If nothing else, an excellent exercise in drawing pictures in ASCII! (Hint: never use tabs, always use spaces.)
This was sort of a rerun of a problem from last year - that the shortest path from one point to another while stopping elsewhere along the way involves a reflection. This group of students picked up on that part, and from there, it's not hard to show that i=r (they sure never showed ME why in high school physics! I just never argued because it looked right).
As Pat Daley's calculus class showed, you don't have to have the reflection part to get the right answer, but it sure makes life easier!
Pawel Gobisand Imran Hayat
The angle of incidence (i) is equal to the angle of reflection (r).
P <-Person viewing the object
\ Q <-The Object
\ /
\ /
_______i_\/_r_________________ <-Mirror
We can prove this by drawing the problem a little bit differently:
P <-Person viewing the object
\ Q <-The Object
\ /|
\ / |
_________\/__|________________ <-Mirror
|
|
|
R <-Reflection in the mirror
We can draw the reflection of the object in the mirror as showed
above, the reflection in the mirror will be the same distance
from the mirror as the object and when a line is draw between the
object and its reflection, it will form a 90 degree angle.
P <-Person viewing the object
\ Q <-The Object
\ /|
\ / |
_________\/__|________________ <-Mirror
\ |
\ |
\|
R <-Reflection in the mirror
Because light travels in straight lines, and the person(P) is
really looking at the object's reflection(R) and not the object
itself, we can draw a line between them.
\ / We know that when two lines cross each other the opposite
\ / angles are the same. Since the angle between
__a_\/_c____ the mirror and the reflection(R), and the angle between
\ b the mirror and the object(Q) are the same, and since the
\ opposite angles are the same, then the angle between
\ the person(P) and the object(Q) are the same.
a=b=c
Kristina Almquist
Known : light takes the shortest possible path Prove angle I = angle RIf the mirror was taken away, light ray P would (theoretically) go on forever. But, when the light bounces off of the mirror it acts as if it had gone through & that path was then reflected over the mirror. This is the shortest path & the angle of incidence is always equal to the angle of reflection.
The angle of incidence is measured from the ray to the mirror, so if the ray were to continue through the mirror it will create a vertical angle to the angle of incidence , measured from the ray to the backside of the mirror. Since, this is a vertical angle to the angle of incidence they are equal in measure.
Since, the light ray's path off the mirror acts is as if it'd gone & been reflected over the mirror, the angle measured from the continued ray to the back of the mirror is = to the angle of reflection. (reflections preserve angle measure)
By the Transitive Property of Equality, the angle of incidence is equal to the angle of reflection.
Sean Nichols
If we have the mirror (XY) reflecting the image of point B, defined as B', then the shortest distance between the two points A and B' is the straight line AB', crossing XY at P:
A.
.
. .B
. .
. .
. .
X____________*________Y
P .
.
.
.B'
AB' will have the same length as AB, and since it is the shortest
distance from A to B', then the shortest path from A to B
touching the mirror XY (PB' = PB), is that which goes through the
point P, as shown above. Now, m1) Since AB is the mirror image of AB', these two "lines" should be the same length (assuming the mirror is held vertical, and along the same line as is indicated). All properties of both lines (except the direction) should be the same.
2) You're right - transitivity, not substitution. Sorry about that.
Anita Lau
a
* b
* *
* *
i * * r
--- C ----- ***light beam
+ y
+ +++straight path
+
d i=r=y
bc=cd ac+bc=ac+cb
The shortest path between 2 points is a straight line. now, angle
r and angle y are the same, and the length of bc and dc are
the same. then, bc is bent and is the same length as cd, thus it
is the line itself.
Mrs. Daley's calculus class
With special insight from Kiki David, Jessica Nahabedian, Adam Smith, Christina Smith, and Tom Soper--all grade 12, Fairfield HS.Drop perpendiculars from P and Q so that they intersect l at A and B. The point C is where the light hits l before bouncing back to Q.
P
*
| * Q
| * *
| * * |
| * * |
| i * * r |
______|___________*__________|_______l
A x C B
Plan: find the lengths of PA, AC, QB, and CB and show that
triangle PAC is similar to triangle QBC. Therefore, i = r.
Let AC=x, PA=a, QB=b, AB=c. Then BC=c-x.
PC=(a^2+x^2)^(1/2) and QC=(b^2+(c-x)^2)^(1/2) by the
Pythagorean Theorem.
Find x so that PC+QC is least.
L=PC+QC=(a^2+x^2)^(1/2) + (b^2+(c-x)^2)^(1/2)
L'=x(a^2+x^2)^(-1/2) - (c-x)(b^2+(c-x)^2)^(-1/2)
L'=0 ==> x = c - x
--------------- -------------------
(a^2+x^2)^(1/2) (b^2+(c-x)^2)^(1/2)
Square both sides and solve for x.
x = ac = AC c-x = bc = BC
----- -----
a+b a+b
PA QB a b a+b a+b
-- = -- ==> -------- = -------- ==> --- = ---
AC BC ac/(a+b) bc/(a+b) c c
Therefore, triangle PAC is similar to triangle QBC and i = r.
So if light takes the shortest path from P to l to Q, then the angle of incidence equals the angle of reflection.
Leigh Jeitner
If you stuck two four-foot high sticks in the ground 6 feet away from each other, and shone a flashlight (at the top of the 1st stick) at a 53 degree angle so that it hit a mirror on the ground (placed 3 feet from each stick), the light would bounce back up at the same angle as it hit the mirror, and the light would hit the piece of cardboard (at the top of the second stick). The distances from the flashlight to the mirror and from the mirror to the cardboard are both 5 feet, so the two triangles formed are congruent by SSS (the sticks were the same height and the mirror was placed on the ground at a point equidistant from each stick). Thus, the angle of incidence and the angle of reflection are corresponding parts of congruent triangles and are themselves congruent.Susanna Puntel
The 2 triangles formed are congruent by the HL Theorem, since if you drop a perpendicular from P and Q, two right triangles are formed. The angle of incidence and the angle of reflection are then corresponding parts of congruent triangles. As long as the hypotenuses are congruent, this will hold true. It is true that the length of the hypotenuses will differ, but if you drop the perpendicular down closer to the vertex of the angle, you should still be able to prove it this way. No matter what angle the light hits the surface, it will always reflect back at the same angle.Sarah Egner
To prove that i=r, you first get a person to stand behind a sheet that is parallel to a wall. The person holds a flashlight 3 feet off the ground. Measure the sheet so that the light is hitting 6 ft high on a mirror on the opposite wall. Measure the height at which the reflected light shines back on the sheet. Assuming that it is 9 feet high, triangle ABC is isosceles, which proves the base angles, 1 and 2, are congruent. Since the lines are parallel, angle 2 = angle 4 and angle 1 = angle 3 because they are corresponding angles. Therefore, angle 3 is congruent to angle 4 by substitution property. Angle 4 is congruent to angle 8 & angle 3 is congruent to angle 7 because they are vertical angles. Angle 7 is congruent to angle 8 by substitution, which means i=r. (See Sketchpad attachment for diagram)Kim Biedermann
I believe the way to prove the angles are congruent is to prove that they are corresponding parts of congruent triangles.First I made 2 parallel lines. I then made 2 rays (the path of the light). I set the angles i and r equal to each other. Then I made a line fron=m where the ray intersects the top parallel line perpendicular to the bottom parallel line. I know the line connecting the parallels are congruent because if two lines are parallel, all points on the one line are equidistant from all points on the other line. I also know the angles formed by the perpendiculars are congruent. The hypotenuses are congruent because I cut them off in the same place by the top parallel line. Because light travels the shortest path in a straight line, not curved, this also proves they are congruent.
Therefore the 2 triangles are congruent by Hypotenuse/Leg theorem and the angles i and r are corresponding parts of congruent triangles.
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