Geometry+ Ellipse Optics
Though the word "optics" sounds very much like a topic of Physics, we will consider strictly mathematical aspect of it.
One of the properties of ellipse is that if its contour is reflective, a ray of light emitted from one of its foci will come to another foci regardless of the direction it was sent.
Let's start with a mathematical meaning of reflection of the contour of ellipse, which is not a straight line.
We do know what a reflection of the straight line is. Simply speaking, angles of incidence and reflection are equal to each other.
Reflection of a curve occurs exactly as if, instead of a curve at the point of incidence, there was a tangential line to a curve, and reflection was of that tangential straight line.
Consider an ellipse with foci F1(−c,0) and F2(c,0), point P(x,y) on this ellipse and tangential to an ellipse at this point P(x,y).
Angle of F1P with X-axis is α.Angle of F2P with X-axis is β.
Angle of a tangent to an ellipse at point P(x,y) with X-axis is γ.
Angle between a tangent and F1P is φ.
Angle between a tangent and F2P is ψ.
To prove that the light emitted from focus F1(−c,0) will reflect at point P(x,y) on an ellipse and will hit point F2(c,0), it is sufficient to prove that angles φ and ψ are equal.
Simple considerations, based on the theorem that the sum of angles of any triangle equals to &pi, lead us to the following equalities
φ = α − γ
ψ = γ − β + π
All the above angles are in the range from 0 to π. In this interval equality of angles follows from equality of their tangents since in this interval tangent is a monotonic function.
So, let's determine tangents of φ and ψ and prove that they are equal.
In both cases we will use the formula for tangent of a difference between angles and, in case of angle ψ, will take into consideration that tangent is periodic with a period π.
| tan(φ) = |
|
| tan(ψ) = |
|
Therefore,
tan(α) = (y−0)/(x+c) = y/(x+c)
tan(β) = (y−0)/(x−c) = y/(x−c)
The issue with tan(γ) is a bit more complicated since we know that this is a slope of a tangential line to an ellipse at point P(x,y), and we don't have a second point on this line to calculate a slope using the same method as with angles α or β.
However, we know that the same slope is a derivative of variable y, as a function of variable x, describing our ellipse.
Using the equation of ellipse in Cartesian coordinates, we can calculate it as follows.
1. Equation of ellipse is
x²/a² + y²/b² = 1
2. Differentiate both sides by x, taking into consideration that y is a function of x
2x/a² + (2y/b²)·y'(x) = 0
(where y'(x) is a derivative of y by x).
3. Find the slope of our tangential to ellipse line at point P(x,y)
y' = −x·b²/(y·a²) = tan(γ)
Now we have all components to calculate tan(φ) and tan(ψ) in terms of calculated tangents of angles α, β, γ.
If we prove that tan(φ)=tan(ψ), we can say that a light ray emitted from focus F1 towards any point P(x,y) on an ellipse will be reflected from point P(x,y) towards focus F2.
So, here are the calculations of tan(φ) and tan(ψ).
We start with calculations of tan(φ) based on values of tan(α) and tan(γ).
We will use the original relation between x and y that reflects the fact that point P(x,y) lies on an ellipse
x²/a²+y²/b²=1,
from which follows
x²·b²+y²·a²=a²·b²
1. tan(α) − tan(γ) =
= y/(x+c) + x·b²/(y·a²) =
= (y²·a²+x²·b²+c·x·b²)/(x·y·a²+c·y·a²) =
= (a²·b²+c·x·b²)/(x·y·a²+c·y·a²)
2. 1 + tan(α)·tan(γ) =
= 1 − x·y·b²/(x·y·a²+c·y·a²) =
= (x·y·a²+c·y·a²−x·y·b²)/(x·y·a²+c·y·a²)
3. tan(φ) = [tan(α)−tan(γ)] / [1+tan(α)·tan(γ)] =
= (a²·b²+c·x·b²)/(x·y·a²+c·y·a²−x·y·b²)
Now let's calculate the value of tan(ψ) based on values of tan(β) and tan(γ).
1. tan(γ) − tan(β) =
= −x·b²/(y·a²) − y/(x−c) =
= (−x²·b²+c·x·b²−y²·a²)/(x·y·a²−c·y·a²) =
= (−a²·b²+c·x·b²)/(x·y·a²−c·y·a²)
2. 1 + tan(γ)·tan(β) =
= 1 + [−x·b²/(y·a²)]·[y/(x−c)] =
= (x·y·a²−c·y·a²−x·y·b²)/(x·y·a²−c·y·a²)
3. tan(ψ) = [tan(α)−tan(γ)] / [1+tan(α)·tan(γ)] =
= (−a²·b²+c·x·b²)/(x·y·a²−c·y·a²−x·y·b²)
What remains is to show that two calculated values, tan(φ) and tan(ψ) are the same.
Both are fractions with numerator and denominator. Equality between two fractions P/Q=R/S is equivalent to equality P·S=Q·R.
Let's check equality of our fractions using this method.
Numerator of tan(φ) multiplied by denominator of tan(ψ) is
(a²·b²+c·x·b²)·(x·y·a²−c·y·a²−x·y·b²) =
= x·y·a4·b² + x²·y·a²·b²·c −
− y·a4·b²·c − x·y·a²b²·c² −
− x·y·a²·b4 − x²·y·b4·c
Notice that
x·y·a4·b² − x·y·a²·b4 =
= x·y·a²·b²·(a²−b²) =
= x·y·a²·b²·c²
which cancels the analogous term with a minus sign in the above expression.
This leaves our expressions equal to
(a²·b²+c·x·b²)·(x·y·a²−c·y·a²−x·y·b²) =
= x²·y·a²·b²·c − y·a4·b²·c − x²·y·b4·c
Next simplification is
x²·y·a²·b²·c − x²·y·b4·c =
= x²·y·b²·c·(a²−b²) = x²·y·b²·c³
That leaves our expression to
x²·y·b²·c³ − y·a4·b²·c
Denominator of tan(φ) multiplied by numerator of tan(ψ) is
(x·y·a²+c·y·a²−x·y·b²)·(−a²·b²+c·x·b²) =
= −x·y·a4·b² − y·a4·b²·c + x·y·a²·b4 +
+ x²·y·a²·b²·c + x·y·a²·b²·c² − x²·y·b4·c
Notice that
−x·y·a4·b² + x·y·a²·b4 =
= −x·y·a²·b²·(a²−b²) =
= −x·y·a²·b²·c²
which cancels the analogous term with a plus sign in the above expression.
This leaves our expressions equal to
(x·y·a²+c·y·a²−x·y·b²)·(−a²·b²+c·x·b²) =
= − y·a4·b²·c + x²·y·a²·b²·c − x²·y·b4·c
Next simplification is
x²·y·a²·b²·c − x²·y·b4·c =
= x²·y·b²·c·(a²−b²) = x²·y·b²·c³
That leaves our expression to
x²·y·b²·c³ − y·a4·b²·c
Final expressions for two cases above are identical, which proves that tan(φ)=tan(ψ).
This, as we mentioned above, is a sufficient condition for a ray of light emitted from one focus point of an ellipse to any direction after reflecting from the ellipse to end up at the other focus point.
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