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The formula for solutions of any quadratic equation is known.
For an equation A·X2+B·X+C=0
the solutions are:
X1,2 =
−B±√B²−4AC
2A
There is a formula for a general cubic equation (Cardano formula), but it's very complex, and we will not list it here.
So, what can someone do to solve a cubic equation?
In some cases there is a possibility to guess one of the solutions. Then, using the polynomial division, we can reduce our cubic equation with one guessed solution to a quadratic one to find two other solutions.
Consider a cubic equation X3 + 4X2 −11X −30 = 0
As we know, any polynomial of the third order, like the one above, has three (generally speaking, complex) roots, that is values of an unknown that cause the value of the polynomial to be equal to zero. Assume these roots are X1 = a X2 = b X3 = c
Then, according to a corollary to a Fundamental Theorem of Algebra, our polynomial can be represented as X3 + 4X2 −11X −30 =
= (X−a)·(X−b)·(X−c)
Then, knowing one root (say, X1 = a) we can construct the result of multiplication (X−b)·(X−c) by dividing our original polynomial by (X−a).
This process is similar to long division of numbers.
Notice that, if we multiply (X−a)·(X−b)·(X−c), the only element without an unknown will be −a·b·c.
Therefore, −30 = −a·b·c.
If the free member of our equation is integer (and it is, it's equal to −30), it's reasonable to attempt, firstly, to find an integer root of our equation among its divisors.
Number 30 has only three divisors: 2, 3 and 5. Let's check if one of them (say, 5) is the root of a given polynomial. 53 + 4·52 −11·5 −30 =
= 125 + 100 −55 − 11 = 40
So, X=5 is not a root.
How about X=−5? (−5)3 + 4·(−5)2 −11·(−5) −30 =
= −125 + 100 + 55 − 30 = 0
Lucky guess! We found a root a=−5 and now we can divide the original polynomial of the third order by X−a = X−(−5) = X+5
obtaining the polynomial of the second order that should have two other roots b and c.
Here is this division, step by step.
Step 1:
Dividend: X3+4X2−11X−30
Divisor: X+5
Quotient: X2
Multiply: (X+5)·X2=X3+5X2
Remainder: (X3+4X2−11X−30)−(X3+5X2)=
= −X2−11X−30
Combine all quotients and get:
Dividend: X3+4X2−11X−30
Divisor: X+5
Quotient: X2−X−6
Multiply: (X+5)·(X2−X−6) =
= X3+4X2−11X−30
Remainder: zero
Now we can either attempt to find a root of the equation X2−X−6 = 0
and perform a polynomial division or, considering this is an equation of the second order, just use the formula for its roots.
The guess and division approach is more educational, so we will choose it.
The free member of the polynomial we deal with now is −6, so we will try its divisors 2 or 3.
Start with a root b=2. 22−2−6 = −4
So, 2 is not a root.
How about b=−2? (−2)2−(−2)−6 = 4+2−6 = 0
Lucky guess! We found a second root b=−2 and now we can divide the original polynomial X2−X−6 of the second order by X−b = X − (−2) = X + 2
Let's do a division to reduce our second order polynomial X2−X−6 as a product of (X+2) and a first order polynomial.
Step 1:
Dividend: X2−X−6
Divisor: X+2
Quotient: X
Multiply: (X+2)·X = X2+2X
Remainder: (X2−X−6)−(X2+2X)=
= −3X−6
Combine all quotients and get:
Dividend: X2−X−6
Divisor: X+2
Quotient: X−3
Multiply: (X+2)·(X−3) =
= X2−X−6
Remainder: zero
By guessing and using the polynomial divisions we have two roots (X=−5 and X=−2) and representation of the original equation as X3+4X2−11X−30 =
= (X+5)·(X+2)·(X−3)
The last multiplier reveals the third root of the original cubic equation X=3
CONCLUSION
Knowing one root X=a of a polynomial of the Nth order P(N)(X), we can represent this polynomial as a product of (X−a) and a polynomial of the (N−1)th order P(N−1)(X), thus simplifying a task of finding all the roots.
In this lecture we will use the material of the previous lecture "Prismatic Lenses" and strongly recommend to be familiar with its context.
Convex lenses serve to change the direction of rays of light (for example, coming from Sun) to produce images for some technical purpose (for example, to generate heat by focusing the Sun rays into one point).
Before discussing convex lenses, consider a case of a double-angled prismatic lens of the following section.
As in the previous lecture, we consider the light going from the medium with higher speed (like air) into a transparent prismatic lens made of a substance with a lower speed of light (like glass or silicon) and then refracted on the way out back into the air.
Assume two rays of light A1B1 and A2B2 fall perpendicularly onto the top surface PQ of this lens, continue on the same trajectory down without changing a direction (because they are perpendicular to the top surface) and are refracted by the bottom surfaces OR (for A1B1 refracted to B1C) and RQ (for A2B2 refracted to B2C).
Our goal is to find conditions on the shape of this prismatic lens and location of points A1 and A2 such that refracted rays intersect the Y-axis at the same point C, that we assume to be located at a distance f from the point O at the base of a lens.
Let distances PA1 and PA2 be, correspondingly, x1 and x2.
Let angles of deviation of the light be ∠CB1D1=δ1 (from A1B1 to B1C) and ∠CB2D2=δ2 (from A2B2 to B2C).
Then, according to the results of the previous lecture "Prismatic Lenses",
∠CB1D1 = δ1 =
= arcsin[(ni/nr)·sin(α1)] − α1
Analogously,
∠CB2D2 = δ2 =
= arcsin[(ni/nr)·sin(α2)] − α2
where ni is the index of refraction of the medium of the incident ray of light and nr is the index of refraction of the medium of the refracted ray of light.
All elements used in these formulas (indices of refraction and angles of the prism) are known, therefore angles of deviation are fully defined.
Let's determine the locations of points A1 and A2 to assure that refracted rays of light intersect the Y-axis at the same point C.
The length of segment OC=f.
Then, according to the results of the previous lecture "Prismatic Lenses", on one hand, considering the ray of light A1B1, f = x1·[cot(δ1) − tan(α1)]
and, considering the second ray of light A2B2, f = x2·[cot(δ2) − tan(α2)]
where the angles δ1 and δ2 were represented above in terms of α1 and α2 correspondingly.
As we see, the necessary condition for these two rays to come to the same point C on the Y-axis is the above formulas that combine the angle between the refracting bottom surface of a prism with the horizontal plane (α1 for OR and α2 for RQ) and the distance of the incoming ray of light from the Y-axis (x1 for PA1 and x2 for PA2).
The next step on the way to study convex lenses is to consider a multi-angled prismatic lens with set of rays of light falling perpendicularly to its top surface at points Ai with many refracting surfaces, one per each ray of light, each having its own angle with a horizontal plane αi.
Now for each refracted ray of light we can use the same formulas as above to represent the dependency between
(a) the desired distance f,
(b) the distance xi=PAi of the falling light from the Y-axis,
(c) the angle αi between the corresponding refracting surface with a horizontal plane: f = xi·[cot(δi) − tan(αi)]
where δi = arcsin[(ni/nr)·sin(αi)] − αi
Consider now a convex lens that is a three-dimensional object obtained by a rotation of figure PQO around the Y-axis.
Line PQ is perpendicular to the Y-axis, curve OQ can be represented by some function y=y(x), which would be desirable to find, based on certain properties of a lens, which we need to have, primarily, its ability to concentrate the parallel rays of light falling perpendicularly to its top surface in one point on the Y-axis called the focal point.
Considering the symmetry, we can analyze the two-dimensional section of this lens formed by a plane going through the axis of rotation Y and any particular ray of light AB falling perpendicularly to PQ at a distance PA=x from the Y-axis.
All actions then will be occurring within this section plane.
The direction of a refracted ray of light BC can be obtained by
(a) drawing a tangential RS to curve OQ at the point of refraction B,
(b) drawing a normal to a surface of rotation at this point, which is a perpendicular MN to tangential RS and
(c) calculating the angle of refraction ∠CBN between the refracted light and a normal to a surface of refraction at point B.
Let's assume that curve OQ is a graphical representation of some function y=y(x), where x is a distance of the incident ray of light from the Y-axis.
If tangential RS makes an angle α with X-axis, the incident light makes the same angle α with a normal MN to a refracting surface at point of refraction B, because these two angles, ∠ABM and an angle formed by a tangential RS and X-axis, have correspondingly perpendicular sides.
If function y=y(x) represents curve OQ, the tangent of the angle tangential RS makes with X-axis is the first derivative of this function:
tan(α) = y'(x)
Using the Law of Refraction, we determine the refraction angle ∠CBN: ni·sin(α) = nr·sin(∠CBN)
A refraction angle ∠CBN can be viewed as a sum of two angles:
(a) a deviation angle ∠CBF=δ between the original direction of the light AB and the refracted ray BC and
(b) an angle ∠FBN that is equal to an incidence angle ∠ABM as vertical and, therefore, is equal to α
Putting all this information together, we can construct an equation that defines the refraction of the ray of light AB off the surface defined by a rotation of the curve OQ that we describe as a function y=y(x).
From triangle ΔOCE we can find OE=f·tan(δ).
From triangle ΔBDE we can find DE=y(x)·tan(δ).
Adding these two segments together, we obtain OE+DE = x = [f+y(x)]·tan(δ)
Nice equation above, easily solvable for y(x), unfortunately, has an unknown angle δ. This angle should be determined based on the shape of curve OQ defined by function y=y(x) and the distance x of the incident light from the Y-axis.
We know that δ+α constitute the refraction angle, which in turn, can be obtained from the incidence angle and the Law of Refraction: nr·sin(δ+α) = ni·sin(α)
This allows to find angle δ in terms of the incidence angle α: sin(δ+α) = (ni/nr)·sin(α) δ+α = arcsin[(ni/nr)·sin(α)] δ = arcsin[(ni/nr)·sin(α)] − α
It's convenient to use relative index of refractionn=ni/nr to simplify the above as δ = arcsin[n·sin(α)] − α
Also, recall that tangent of an angle from the X-axis to a tangential RS is the first derivative of function y(x).
Therefore, tan(α)=y'(x) and α = arctan[y'(x)]
Summarizing all of the above, we have come to a differential equation for function y(x) that describes the shape of a bottom refracting surface of a lens that perfectly focuses all the rays of light falling perpendicularly onto its top flat surface into one focal point at a given focal distance f from the lens x = [f+y(x)]·tan(δ)
where δ = arcsin[n·sin(α)] − α and α = arctan[y'(x)]
We deliberately didn't substitute δ and α to obtain one huge differential equation that contains only x, y(x) and y'(x) to preserve some readability.
The solution to this differential equation is not a circle, nor a parabola, nor hyperbola, nor exponent. It's much more complex. All the above elementary functions, when used in lenses, produce an effect called aberration, they do not concentrate all the rays of light going parallel to the Y-axis in one focal point.
There are numerical methods to approximate the solution to this differential equation.
Also, relatively recently the analytical solution to a more general problem with any top surface of a lens (like spherical) was obtained by young scientists in Mexico. Anyone interested can search for Wasserman-Wolf problem.
The main feature of the refraction of the light is the change of the direction of the ray of light after crossing from one transparent medium into another.
This feature is used in lenses to redirect the parallel rays of light.
This lecture is about right angle triangular prismatic lenses.
Specifically, we go through some calculations of the trajectory of light after it passes this type of a lens.
Consider a side section view of the right angle triangular prismatic lens with a ray of light falling perpendicularly to its side as on the following picture.
The ray of light (solid blue line on a picture above) entering a prism perpendicularly to its surface, goes through it without changing its direction (because it's perpendicular to a surface) and at the exit from a prism changes its direction, according to the Law of Refraction ni·sin(θi) = ni·sin(θi)
where
indices i and r represent, correspondingly, incident and refracted rays of light, ni is a refraction index of the air above and below the prism, nr is a refraction index of the glass the prism is made of,
∠MBA = θi is the angle of incidence between incident ray of light and a normalMN to a surface of refraction,
∠CBN = θr is the angle of refraction between refracted ray of light and a normalMN to a surface of refraction.
Notice that refraction index of a medium is a ratio between the speed of light in vacuum to the speed of light in this medium. Its always greater than 1 because the speed of light in vacuum is its maximum possible speed. Also, the speed of light in glass is less than in the air, that's why the glass refraction index is greater than that of the air.
Assume, the light falls on the upper surface of the prism PQ at a distance PA=x from the Y-axis.
Our task is to determine the angle ∠CBD of the deviation of the light from the original trajectory and the distance OC=y, where point C is the intersection of the refracted ray of light with Y-axis.
Our first step is based on the fact that angle of incidence ∠MBA=θi equals to the angle of a prism ∠PQO=α as angles with mutually perpendicular sides, and, therefore, should be considered as given.
Also, angle ∠DBN equal to angle of incidence ∠MBA=θi as vertical.
That implies that the angle of deviation of the ray of light from its original direction ∠CBD equals to
∠CBD = ∠CBN− ∠DBN =
= θr − θi = θr − α = δ
Since angle of incidence θi is defined by the given prism's angle ∠PQO=α and the angle of refraction θr is related to the angle of incident through the Law of Refraction ni·sin(θi) = nr·sin(θr)
we have a clear path to find the angle of deviation:
1. sin(θr) = (ni/nr)·sin(α)
2. θr = arcsin[(ni/nr)·sin(α)]
3. ∠CBD = δ =
= arcsin[(ni/nr)·sin(α)] − α
The last statement represents a dependency of the angle of light deviation δ from the angle α between top and bottom surfaces of a prism.
To determine the length of a segment y=OC, we see that OC = BD − BE
From the right triangle ΔBDC we can express the length of segment BD: BD = CD·cot(δ) = x·cot(δ)
From the right triangle ΔOBE we can express the length of segment BE: BE = x·tan(α)
Therefore, OC = x·cot(δ) − x·tan(α) =
= x·[cot(δ) − tan(α)]
We came to the results of our calculations.
1. The angle of deviation δ, as demonstrated above, depends on the given prism angle α and refraction indices of media the light goes through.
2. The length y of a segment between the point O (the edge of a prism) and an intersection of a ray of light, falling on the top of prism at distance x and, subsequently, refracted by another prism surface, from the Y-axis is y = x·[cot(δ) − tan(α)]
where α is a prism angle between the plane the original ray of light is perpendicularly falling on and the plane refracting this light, δ is an angle of deviation of the refracted ray of light from its original trajectory.
As a conclusion, we observe that
(a) the angle of deviation δ of the refracted ray of light from its original trajectory depends only on the physical characteristics of a prism and surrounding environment (prism angle α and indices of refraction inside and outside of a prism),
(b) the distance y=OC from the edge of a prism to a point of intersection of the refracted light with Y-axis is proportional to a distance x of the incident ray of light from the Y-axis. The further from the Y-axis the trajectory of the incident ray of light goes - the further from the point O the refracted ray of light intersects this axis.
Prism does not gather the parallel rays of light falling perpendicularly to its surface into one point, but maintains their parallelism on a different trajectory.
Refraction is an effect of changing the direction of light propagation after the light hits a border between two different media and penetrates into a new medium in, generally speaking, different direction and with different speed, as compared to the original direction and speed.
In this lecture we will quantitatively evaluate the effect of refraction from the viewpoint of the Fermat's Principle of the Least Time presented in the lecture about reflection of light.
Many experiments have shown that the direction of the ray of light after it crosses the border between two different media is determined by its initial direction before it hits the border and the properties of both media. In particular, it's the speed of light propagation before and after the border, as the main property of the medium, is taken into consideration.
The effect of refraction of light going from the air into the water is pictured below. The choice of air and water was not very important. Instead of them, any other two media could be mentioned, as long as the speed of light is greater in the top medium.
It can be shown that the plane that goes through points P, R and Q also contains the normal to a border surface at point R. That's the reason to present the refraction using only two dimensions of this plane.
The frame of reference has X-axis along the surface of water and Y-axis is normal to it.
The ray of light goes from the source P(a,b), which is in the air, into the water to point Q(c,d) by going along one straight line before it hits the water at point R(x,0) and then along the other straight line within the water to a destination point Q.
Our task is to determine the point R on the surface of the water, that is to determine the X-coordinate of this point, where the ray of light changes the direction.
The main criteria we will choose to determine the trajectory of light is the same Fermat's Principle of the Least Time we used to analyze the reflection of light in the corresponding lecture.
Let's determine the time needed for the ray of light to go along the trajectory from point P(a,b) in the air to point Q(c,d) in the water through point R(x,0) on the surface of water, where b is positive, d is negative and a ≤ x ≤ c.
The first segment of this trajectory PR has the length, as a function of x s1(x) = √(x−a)²+b²
The second segment RQ has the length s2(x) = √(x−c)²+d²
Let's assume that the speed of light in the air is V1 and its speed in the water is V2.
Length and speed values determine the time needed for light to go through an entire trajectory.
The first segment requires the time T1(x) = s1/V1 = √(x−a)²+b²/V1
The second segment requires the time T2(x) = s2/V2 = √(x−c)²+d²/V2
According to Fermat's Principle of the Least Time, sum of these two times needs to be minimized to find the value of x - the position of the point R of refraction. T(x) = T1(x) + T2(x) =
= √(x−a)²+b²/V1 +
+ √(x−c)²+d²/V2
The necessary condition for this function T(x) to have its minimum at x=x0 would be for its derivative by x to be equal to zero at this point.
Let's differentiate T(x) and see the implications of the condition T'(x0)=0. T'(x) =
= 2(x−a)/(2√(x−a)²+b²·V1) +
+ 2(x−c)/(2√(x−c)²+d²·V2) =
= (x−a)/(√(x−a)²+b²·V1) +
+ (x−c)/(√(x−c)²+d²·V2) =
At this point it's convenient to express ordinates of points P and Q in terms of their abscissas and angles of incidence θi and refraction θr as follows b = (x−a)/tan(θi) d = (x−c)/tan(θr)
Using these expression, notice that √(x−a)²+b² =
= √(x−a)²+(x−a)²/tan²(θi) =
= √(x−a)²·[1+1/tan²(θi)] =
= |x−a| /sin(θi) √(x−c)²+d² =
= √(x−c)²+(x−c)²/tan²(θr) =
= √(x−c)²·[1+1/tan²(θr)] =
= |x−c| /sin(θr)
Substituting these expressions into T'(x), we obtain T'(x) =
= (x−a)/[|x−a|·V1/sin(θi)] +
+ (x−c)/[|x−c|·V2/sin(θr)] =
= sin(θi)/V1 − sin(θr)/V2
Notice:
since a ≤ x ≤ c, (x−a)/|x−a| = 1 (x−c)/|x−c| = −1
that's why the first item in the final expression for T'(x) is positive, the second one is negative.
Equality of T'(x) to zero implies: sin(θi)/V1 − sin(θr)/V2 = 0
Therefore, the necessary condition for minimizing the time for light to travel from P to Qis not a straight line, but a set of two linked segments PR and RQ, inclined to the normal to a surface of refraction at angles that satisfy the equality sin(θi)/V1 = sin(θr)/V2
Speed of light in any medium is a pretty large number, and it might be convenient to use a different form of the above formula. Let's introduce a concept of refraction index of any medium n=c/V, where c is the speed of light in the vacuum and V is the speed of light in the medium in question. Using n1=c/V1 and n2=c/V2 as the refraction indices of media involved in the refraction, the above condition on incidence and refraction angles looks like sin(θi)·n1 = sin(θr)·n2
When the light falls perpendicularly to a surface that separates two media, an incidence angle is zero, θi=0. Under this condition the value of a refraction angle θr must also be equal to zero to satisfy the refraction equation above. It means that if the incident ray of light is perpendicular to a surface that separates two media, the refracted ray is also perpendicular to this surface, light does not change its direction, only the speed.
When the light goes from the medium of higher speed of light to the medium of a slower speed at any incidence angle greater than zero, the refraction angle will be smaller than incidence. As an incidence angle grows from 0 to π/2, a refraction angle grows from 0 to some maximum smaller than π/2.
A more interesting case can be observed, when the light goes from a medium with a slower speed into a medium, where its speed is higher. For example, from glass to air.
The following picture illustrates this case.
As an incidence angle grows from 0 to π/2, a refraction angle also grows from 0 up, but always greater than an incidence angle. That will result in some critical value of an incidence angle, when the refraction angle reaches π/2, that is will go parallel to a surface border between two media. Any further increase in incidence angle value will cause the light to stay within the area of a lower speed of light, it will be reflected by the surface border and will not go through it. This is called a total internal reflection.
This is the property of light used in fiber optics and jewelry.
Consider now a more complicated case of a curved reflection surface.
Any smooth surface can be considered as an infinite set of infinitesimally small flat pieces with each piece reflecting light in a direction that can be determined by a plane tangential to a surface at that point.
So, to determine the reflected light at some point of a surface we can just replace a surface with a tangential plane at that point and use the Laws of Reflection presented in the previous lecture.
Let's illustrate this on a concrete example of a paraboloid as a surface reflecting the light. Paraboloid is a surface obtained by rotating a parabola z=k·x² in the XZ-plane around the Z-axis.
As a result of this rotation, the three-dimensional formula for a paraboloid is z = k·(x² + y²)
We will examine how vertically going down rays of light are reflected by this surface.
Assume, a light ray falls down parallel to the Z-axis of a paraboloid within XZ-plane at distance a from this axis and hits paraboloid at point B (vertical blue line on the picture below).
After the reflection off the surface of paraboloid, which we will analyze as if reflected off the tangential plane to paraboloid at point B, the reflected ray of light crosses the Z-axis of this paraboloid at point C (black line BC on the picture below).
The reflected ray of light should cross the Z-axis at some point C from the considerations of rotational symmetry of the paraboloid.
We will analyze this using a two-dimensional cut along the plane going through a point B on the surface of paraboloid, where the light ray hits its surface and the vertical axis of this paraboloid with Z-axis coinciding with the axis of paraboloid
The light blue colored line represents the light going down at a distance a=OA from the Z-axis. It hits a point B on a parabola z=k·x² (red curved line) and the solid black line represents the reflected ray of light that hits the Z-axis at point C.
The green line is tangential to a parabola at point B and should be used to determine the direction of the reflected light by establishing a normal to a parabola line (a purple line perpendicular to a green tangential line) and using the law of reflection about equality between an incidence angle θi and the reflection angle θr.
Our task is to determine a distance OC from the origin of coordinates to point C, where the reflected ray of light intersects the Z-axis of this parabola.
The analysis of this task, going from what is to be found back to what's known, is:
1. Find OC as the difference between AB (known to be the value of z=k·x² at x=a, that is k·a²) and an unknown BD.
2. To find BD, we will use the formula BD = CD·cot(∠CBD),
where CD=OA=a
3. Angle ∠CBD is the difference between π and angle ∠θi+∠θr, that is
(since θi=θr=θ) ∠CBD = π−2θ
4. Since ∠BEA=∠θ and BE is a tangential to our parabola z=k·x², tangent of ∠BEA equals to a derivative of z=k·x² at point x=a, from which follows: tan(∠θ) = 2k·a
So, as we see, the reflected ray of light will intersect the Z-axis at point C at a distance OC=1/(4k) from the bottom of a paraboloid.
What's remarkable about this result is that the location of point C does not depend on the value of parameter a - the distance of the incident light from the Z-axis.
So, any vertically directed ray of light will be reflected by a paraboloid towards the same point on its axis - its focal point - located at distance f=1/(4k) from the bottom, where parameter k defines the "steepness" of a paraboloid.
Using the parabolic mirror, we can "gather" the sun rays into a focal point and boil the water positioned there to use the steam to generate electricity.
If the source of light is positioned at the focal point of a parabolic mirror, all its emitted light will be directed in one direction parallel to the axis of a paraboloid. That's the principle of work of a projector.
The dish-like parabolic antenna, directed towards a stationary satellite broadcasting some radio signals, collects all the radio waves falling into it, reflecting all these signals towards its focal point, where a radio receiver is located. This allows to catch even a relatively weak radio signal.
When we don't hear a distant sound, we make a sort of a "dish" with our hand, directing the reflected sound towards the ear to hear better.
All the above examples and many others are the usages of a principle of focusing the waves by parabolic (or almost parabolic) reflectors.
Reflection and refraction are effects of changing the direction of light propagation after the light hits some surface or, more precisely, when light reaches the border between two different media, "old", where it's coming from, and "new", which the light hits on its path.
Reflection happens when light returns back to the "old" medium after hitting its border with a "new" medium and continues to propagate there in a different direction, while refraction is the penetration of the light inside the "new" medium, where it continues to propagate in, generally speaking, different direction and different speed, as compared to the original direction and speed.
In this lecture we will address the effect of reflection.
Before addressing the Laws of Reflection, let's accept as an intuitively understood axiom, the Fermat's Principle of the Least Time of light propagation. This principle, proposed by French mathematician Pierre Fermat in 1662, states that the light travels from its source to some point along such a trajectory that the travel time is the least among all possible trajectories.
In particular, if the environment the light travels through is uniform (like vacuum or glass of a uniform consistency), the light travels along a straight line, because a straight line is the shortest distance between any two points, which results in the least travel time for light that travels with a constant speed.
It means that, if the source of light S emits light in all directions in a uniform environment, at some observing point A we see only the ray that travels along a straight line SA.
Reflection is easily understood from the viewpoint of the corpuscular theory of light, which might be a factor in dominance of this theory, when scientists first attempted to understand the nature of light.
Indeed, reflected light behaves exactly like billiard balls hitting the border of a billiard table.
Many experiments have shown that the direction of the reflected ray of light is determined by its initial direction before it hits the reflecting border between two media and the geometry of this border.
Consider the simplest case of a border between two media being an ideal plane that reflects all the light coming on it, like a mirror.
Let's examine how the light is reflected by this mirror from the viewpoint of the Fermat's Principle of the Least Time.
Let point S be a source of light. Choose one particular ray emitted by it at a certain angle to a plane of a mirror (this is an angle between a line of a ray and a plane of a mirror, which is measured as an angle between this line and its projection on the plane).
This ray is reflected by a mirror. Let point A be any point on the reflected ray.
Before hitting a mirror the ray travels within a uniform environment along a straight line. After the reflection light also travels to point A in a uniform environment along another straight line.
Our task is to determine a point R, where the light hits a reflecting plane before traveling to point A.
Since both segments the light travels (SR before hitting a mirror and RA from a reflecting mirror to point A) are in the same environment, where the speed of light is the same, the Principle of the Least Time will be satisfied if the whole distance from the source S to a reflection point R and to point A is minimal among all possible trajectories.
Consider now a purely geometric problem. Given two points in space S and A on the same side of a plane α, find a point R on plane α such that the sum of the lengths of two segments SR and RA is minimal.
The following picture represents a solution:
Find a point A' symmetrical to point A relatively to a given reflecting plane α by dropping a perpendicular to plane α from point A and choose on this perpendicular point A' on the opposite side of a plane such that AB=BA', where B is intersection point of this perpendicular with plane α.
Next, connect points S and A' by a straight line. Point R is an intersection of line SA' with plane α. From equality of right triangles ΔARB and ΔA'RB, that follows from the equality of their catheti, follows equality of hypotenuses RA and RA'.
The point R is the point where reflection occurs and the sum of distances SR and RA is the least among all other reflection points on plane α.
Indeed, consider any other point R' as the reflection point. It's obvious that R'A=R'A' (analogously to why RA=RA', as proved above) and, therefore, SR'+R'A=SR'+R'A'
is greater than SR+RA=SR+RA'=SA'
because SR+RA' is a straight line, while SR'+R'A' is not.
So, any other ray, not coinciding with AR, will not hit point A because the trajectory from point S to a different reflection point R' and then to point A will be longer than straight line SA'.
The following easily provable statements are direct consequences of the method of construction of the reflection point R.
(a) Plane of light rays β that contains initial ray of light SR and reflected ray of light RA is perpendicular to a reflecting plane α because it contains the point A' that lies on a continuation of line SR and it goes through a perpendicular to α line AA'.
(b) Projection S' of the source of light S onto reflection plane α also lies in the plane β because line SS' is parallel to AA' that belongs to plane β and point S is on that plane as well.
(c) Perpendicular RR' from a reflection point R to reflecting plane α (normal to plane α at the point of reflection) also lies in the plane β because line RR' is parallel to AA' that belongs to plane β and point R is on that plane as well.
(d) Points S', R and B lie on the same straight line - the line of intersection of two planes α and β; from this follows that ∠SRS' equals to ∠A'RB as vertical within plane β.
(e) ∠A'RB equals to ∠ARB from the equality of triangles ΔARB and ΔA'RB within plane β.
(f) ∠SRS' equals to ARB, as follows from the two previous statements.
(g) Complementary to the two equal angles of the previous statement, incidence angle ∠SRR' (between an incident ray and a normal to a reflecting plane at the reflection point) and reflection angle ∠ARR' (between a reflected ray and a normal to a reflecting plane at the reflection point) also are equal to each other.
The last statement about equality of an incidence angle and a reflection angle is very important.
Now, using the properties described above, we can formulate the Laws of Reflection as consequences of the Fermat's Principle of the Least Time.
1. Three lines, an incident ray, a normal to a reflection plane at a point of reflection and a reflected ray, lie in the same plane.
2. An incidence angle equals to a reflection angle.
3. Incident and reflected rays lie on different sides relatively to a normal at a point of reflection.
Let's support our derivation of the above Laws of Reflection, based on the Principle of the Least Time, with more physical considerations from the viewpoint of the corpuscular theory that states that the ray of light is a set of particles flying in the same direction with certain constant speed along a straight line.
Consider a frame of reference with XY-plane being the reflecting plane and a light particle flying with constant linear speed from some point in the second quadrant of the XZ-plane towards the origin of coordinates along a straight line, so its Y-coordinate and Y-component of its speed are always zero.
Then the above picture represents the trajectory in the XZ-plane.
Assume that a ray of light originated at time t=0 at a distance D from the incidence point (from the origin of coordinates) and flies toward it along a straight line at an angle of incidence θi with constant speed c.
The ray will reach a point of incidence at the time moment T=D/c, at which point its coordinates will be {x(T)=0;y(T)=0;z(T)=0}.
At the incidence point the velocity vector of a light particle will be Vi(t)={c·sin(θi);0;−c·cos(θi)}.
Assuming the ideally elastic reflection, the X-component of the particle's velocity will be unchanged because it's parallel to the reflective XY-plane, Y-component will remain at zero, while Z-component after the contact with reflecting XY-plane will be inversed by an ideal reflection.
Therefore, the velocity vector of a light particle after the reflection will be Vr(t)={c·sin(θi);0;c·cos(θi)}.
After the reflection the light will go along the trajectory that coincides with its velocity vector.
Since Y-component of the velocity vector was, is and will always be zero, the reflected ray from the reflection point (the origin of coordinates) will continue its motion within the same XZ-plane it came from. So, the incident ray, normal to a reflecting XY-plane (that is, Z-axis) and reflected ray lie within XZ-plane, which supports the above mentioned first law of reflection.
If the angle of reflection is θr, the vector of velocity is Vr(t)={c·sin(θr);0;c·cos(θr)}.
Therefore, we have two expressions for the same vector of velocity after the reflection, and they must be equal to each other:
{c·sin(θi);0;c·cos(θi)} =
= {c·sin(θr);0;c·cos(θr)}
Obviously, if sin(θi) = sin(θr) and cos(θi) = cos(θr),
angles θi and θr are equal to each other.
This supports the second law of reflection about equality of the incidence and reflection angles.
Since before the reflection X-coordinate of a light particle is negative and it becomes positive after the reflection, while Y-coordinate is always zero and Z-coordinate is always non-negative, incident ray lies in the second quadrant of the XZ-plane, while reflected ray lies in the first quadrant.
This supports the third law of reflection.
When we talk about light, we mean electromagnetic waves (oscillations of the electromagnetic field) that our eyes can detect.
Not all the oscillations of electromagnetic field are sensed by our eyes, but only within a visible spectrum of frequencies. This spectrum of frequencies of visible light varies for different people, but, in general, it's usually defined as from flow = 4·1014Hz to fhigh = 8·1014Hz.
With the speed of light in vacuum approximately c = 300,000,000 m/sec, using the formula for the wavelengthλ=c/f, we can approximate the low and high wavelengths for visible light: λlow = c/fhigh = 750·10−9(m) =
= 750(nm) λhigh = c/flow = 375·10−9(m) =
= 375(nm)
Colors
Traditionally, we divide the visible spectrum of light based on the difference in how we sense it in terms of different colors.
Though different people see colors slightly differently, here is the division by colors, as is traditionally defined, as a function of the wavelength in nanometers:
As you see from the picture above, the color becomes almost black, when we approach high and low boundaries of visible spectrum, that is the light becomes almost invisible for the eyes, though younger people usually have more sensitive eyes and see a slightly broader spectrum of light.
Invisible for an eye light with wavelength of less than 375 nm is called ultraviolet.
Invisible for an eye light with wavelength of greater than 750 nm is called infrared.
Speed
The speed of light mentioned above as 300,000,000 m/sec is an approximation. The exact speed depends on the substance where the light propagates. In vacuum it's the fastest.
In vacuum it's exactly299,792,458 m/sec. We emphasize the exactness of this speed because in SI system of units meter is defined through a speed of light, as the length traveled by light in vacuum during the time T = 1 /299,792,458 sec.
Speed of light in water is slower than in vacuum by, approximately, 1.33 times and equals to 2.25·108m/sec. Obviously, it depends on the chemical composition of water.
Analogously, speed of light is different in all translucent substances, but always slower than in vacuum.
According to the Theory of Relativity by Albert Einstein, the speed of light in vacuum is the fastest speed possible to achieve.
Source
There are many different sources of light.
Chemical reaction can produce a visible light. For example, coal or wood burning is a chemical reaction between carbon in the coal or wood and oxygen in the air, producing carbon dioxide and energy in a form of heat and visible light.
Electric current can be a source of light, when a sufficiently strong flow of electrons passes through a conductor, producing heat and light.
This can be observed in the incandescent lamps.
Nuclear reactions of fission and fusion, occurring within stars, including our Sun, produces visible light.
Luminescence is a general term that encompasses close in their nature but slightly different sources of light: fluorescence, phosphorescence, chemiluminescence.
They all involve absorption of light energy in some form and its emission as a visible light of different wavelengths immediately after absorption or at a later time.
Recently new way of producing light is light emitting diodes (LED).
Theories
There have been many theories of light, each one explaining this or that property of light. Discovery of each new property of light was the cause to re-evaluate the concept of light and, in most cases, developing a new theory.
Particle or corpuscular theory of light was developed, primarily, by Pierre Gassendi, Isaac Newton and other scientists. According to this theory, light consists of particles (corpuscles) emitted by the source and flying in all directions.
The corpuscular theory explained many properties of light, but had problems explaining certain observable phenomenons, like interference. Eventually, this theory was rejected by scientists.
Wave theory explained quite well such property as interference, but required a medium for wave propagation - aether. Many scientists contributed to this theory, including Hooke, Huygens and others. Numerous experiments, however, contradicted the concept of aether and, eventually, this theory was rejected as well.
Electromagnetic theory of light became the dominant because of work by Faraday, Maxwell and Hertz. According to this theory, light is the oscillations of electromagnetic field with variable electric and magnetic components causing each other. This theory is the foundation of contemporary usage of radio waves, including TV, cell phones, remote controls etc.
Quantum theory complemented the electromagnetic theory and provided a better explanation of certain corpuscular properties of light, like photoelectric effect. Works of Planck, Einstein and other physicists were essential to developing the quantum theory of light, which is now considered as the current model.
