Approximation: UNIZOR.COM - Math4Teens - Algebra - Real Numbers

Notes to a video lecture on http://www.unizor.com

Real Numbers - Approximation

From Exact to Approximation

As we know, we can use decimal notation to represent real numbers.
Integer numbers are represented in a decimal form without decimal point.
Rational non-integer numbers use finite or infinite periodical sequence of decimal digits after the decimal point to represent their fractional part.
Irrational numbers in decimal form have infinite sequence of non-periodical decimal digits after the decimal point.

In this lecture we will talk about how to use approximation of those real numbers, whose exact decimal representation is impossible (because they have an infinite sequence of decimal digits after the decimal point) or inconvenient (for lengthy sequences of digits) to show.

As a simple example, consider a task of measuring 1/7^th of 1 meter of a rope.
The metric ruler has centimeters and millimeters on it. It means, it's impossible to measure exactly 1/7^th of 1 meter because
1/7 = 0.(142857) - an infinite periodical fraction.
The best we can do is to measure 14 cm and either 2 mm or 3 mm because
0.142 < 1/7 < 0.143.
It's better to choose
14 cm and 3 mm
because the exact value 1/7 = 0.(142857) is closer to it.

The need for approximation in the above example is dictated by our practical ability and limitations to deal with exact number. These limitations determined the level of precision required from approximation - 1 millimeter.

Consider another example - a population of the United States of America.
Obviously, we cannot know the exact number of people living in the country because people get born, die, move to another country or come from another country all the time.
So, for practical reasons we approximate this number to, say, millions. If we say that the population of the USA is, approximately, 330 millions, it's sufficient to address most issues related to the population, like how much water is consumed by all people in a country or average density of the population.
Again, practical considerations dictate certain level of approximation, its precision - 1 million of people.

Now we approach the approximation more formally.
The purpose of the approximation is to represent the real number in question by another number, according to certain rules:

1. Out of all real numbers we choose a countable subset of numbers equally spaced from each other, we will call them base numbers.
For example, we can have a set of all integer numbers or a set of numbers, starting at zero, with a distance of 1000 from each other, or a set of numbers, starting at zero, with a distance of 0.001 from each other.
In the example above, when we measured the length with a precision of 1 millimeter = 0.001 meter, our base numbers are 0, 0.001, 0.002, 0.003 etc.
In the example above, when we counted people in millions, our base numbers are 0, 1000000, 2000000, 3000000 etc.
This set of base numbers determines the possible approximate values for numbers we would like to represent.

2. For any real number we would like to represent approximately we use the closest to it number from a set of base numbers chosen above.

3. A special case, when the exact number is equidistant from both base numbers, the one less and the one greater than it, needs special consideration that we discuss below.

Example 1
We need to find a length of a side of a square whose area is 2 m².
From geometry we know that this side should be equal to a square root of 2 meters, but it's an irrational number, so we cannot represent it exactly in decimal notation needed for practical purposes.
We choose the precision sufficient for our practical purposes and practically achievable using the tools at hand (say, a measuring tape with meters and centimeters), to be 1 centimeter, that is 0.01 of a meter.
This determines the base numbers for our approximation 0, 0.01, 0.02, 0.03 etc.
Then we approximate square root of 2 beyond the second digit after the decimal point to determine which base number it's closest to:
√2 ≅ 1.414
which is closer to a base number 1.41 than to 1.42.
So, with a precision of 0.01 (that is, in centimeters) the size of a side of a square with area of 2 square meters equals to
1.41 m


Example 2
The distance between two cities is represented in kilometers. The exact distance is usually measured between the main post offices in these cities, and it's never equals to an exact integer number of kilometers.
For example, the distance from Hanoi to Shanghai is listed as 1925 km.
Obviously, it's an approximation to an integer number of kilometers.
The actual exact distance from the entrance door to Hanoi's main post office to the entrance door to Shanghai's main post office along the shortest route might be something like 1924.532 km (that is, 1924 km and 532 m), but we replaced the exact distance with its approximation as a more practical value.


Special case

It's logical and reasonable to approximate any exact number with a base number that is closest to it.

Consider now a case when an exact number is equidistant from two base numbers, one below and another above it.
For example, we would like to approximate to a precision of 1000 (so, base numbers are 0, 1000, 2000, 3000 etc.) and a number we want to approximate is 56500, which is equidistant from base numbers 56000 and 57000.

Or we would like to approximate to a precision of 0.01 (so, base numbers are 0, 0.01, 0.02, 0.03 etc.) and a number we want to approximate is 8.565, which is equidistant from base numbers 8.56 and 8.57.

This is a dilemma that must be resolved by some rule imposed on the process of approximation.
Unfortunately, there are more than one rule, and these rules might contradict each other. Fortunately, one rule does play a dominant role and used in most cases. This is the one we would like to specify as the one to follow, unless a specific other rule is mentioned.
In case an exact number is equidistant from two base numbers, one below and one above it, choose the one that has higher absolute value.

Examples:
Precision is 0.001,
exact value is 25.6575,
approximate value is 25.658.
Precision is 0.001,
exact value is −2.7595,
approximate value is −2.760.
Precision is 10,
exact value is 25,
approximate value is 30.
Precision is 10,
exact value is −95,
approximate value is −100.


From Approximation to Exact

Let's reverse the logic and try to evaluate the exact number, if it's approximation is given.
Again, a lot depends on the precision of approximation. The more precise approximation was applied to exact number - the closer the approximation is to it and, therefore, more precise evaluation of the exact number can be performed, if an approximate number is known.

Let's assume, we have an exact number X and its approximate value with precision δ is R. It means that among base values R−δ, R and R+δ the number X is closer to R than to R−δ or R+δ.

If so, the following inequality must hold
R−δ/2 ≤ X < R+δ/2
as illustrated below with the blue area representing the possible values of X, if R is its approximation with precision δ.
So, if approximate value of the number of people living in New York in 2019 was 8.419 million (implying the precision δ=0.001 of a million, that is δ=1000), the exact number was greater or equal to 8.4185 million (that is, 8,418,500), but less than 8.4195 million (that is, 8,419,500).

If approximate value for number π is 3.14 (implying the precision δ=0.01), that exact ratio of a circumference of any circle to its diameter is greater or equal to 3.135, but less than 3.145.

As you see, the range of values for an unknown exact number X around its known approximation R is equal to a precision δ of approximation:
(R+δ/2) − (R−δ/2) = δ


Approximation Error

What happens, when we make some arithmetic operation with approximate values?
Unfortunately, the error of approximation is accumulated and growing.
Let unknown exact number X be approximated by a known number R with precision δ. Let unknown exact number Y - by a known number S with the same precision δ.
Let's determine the range of values for X+Y.

According to the laws of approximation,
R−δ/2 ≤ X < R+δ/2
S−δ/2 ≤ Y < S+δ/2
Then, adding these inequalities, we obtain
R+S−δ ≤ X+Y < R+S+δ
The range of values for X+Y is
(R+S+δ) − (R+S−δ) = 2δ
As you see, the precision of evaluating the sum of two exact values, using the sum of their approximations, is now 2δ. So, the quality of approximation is worsening, when we do arithmetic operations with approximate values to evaluate the result of these operations with unknown exact values.

Obviously, similar worsening of the quality of approximation is observed with any other operation (like subtraction, multiplication, division etc.), when we use approximate values in order to evaluate the result of this operation on unknown exact values.

For example, we need to evaluate the rise of the sea level in 10 years from now.
According to EPA, if sea level in 1880 is taken as level zero, by 1994 this level was 6.307" = 160.192 mm above zero.

From 1994 to 2019 the sea level was rising by an average factor of μ = 1.016 a year with standard deviation σ = 0.012, reaching the level of 240.775 mm
Let's disregard extreme values and limit ourselves to 68% probability of the exact value of the average factor of the rise of a sea level to be within [μ−σ;μ+σ] range, that is from 1.004 to 1.028.

Then in 10 years from 2019 the minimum sea level will be
W_min = 240.775·1.004¹⁰ ≅ 251 (millimeters above 1880 level)
while its maximum will be
W_min = 240.775·1.028¹⁰ ≅ 317 (millimeters above 1880 level)
Which numbers to believe?

What is the estimate of the sea level in 100 years?
W_min = 240.775·1.004¹⁰⁰ ≅
≅ 359 mm = 0.359 meters above 1880 level
W_min = 240.775·1.028¹⁰⁰ ≅
≅ 3810 mm = 3.81 meters above 1880 level
As you see, the difference is dramatic! This proves how skeptical we have to be when hearing different predictions about future.

Division of Polynomials: UNIZOR.COM - Math4Teens - Algebra - Fundamental...

Notes to a video lecture on http://www.unizor.com

Division of Polynomials

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

Step 2:
Dividend: −X²−11X−30
Divisor: X+5
Quotient: −X
Multiply: (X+5)·(−X)=−X²−5X
Remainder:
(−X²−11X−30) − (−X²−5X) =
= −6X−30

Step 3:
Dividend: −6X−30
Divisor: X+5
Quotient: −6
Multiply: (X+5)·(−6)=−6X−30
Remainder:
(−6X−30) − (−6X−30) = 0

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

Step 2:
Dividend: −3X−6
Divisor: X+2
Quotient: −3
Multiply: (X+2)·(−3)=−3X−6
Remainder: (−3X−6)−(−3X−6) = 0

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.

Convex Lenses: UNIZOR.COM - Physics4Teens - Waves - Properties of Light

Notes to a video lecture on http://www.unizor.com

Convex Lenses

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[(n_i/n_r)·sin(α₁)] − α₁
Analogously,
∠CB₂D₂ = δ₂ =
= arcsin[(n_i/n_r)·sin(α₂)] − α₂
where
n_i is the index of refraction of the medium of the incident ray of light and
n_r 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 A_i with many refracting surfaces, one per each ray of light, each having its own angle with a horizontal plane α_i.

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 x_i=PA_i of the falling light from the Y-axis,
(c) the angle α_i between the corresponding refracting surface with a horizontal plane:
f = x_i·[cot(δ_i) − tan(α_i)]
where
δ_i = arcsin[(n_i/n_r)·sin(α_i)] − α_i

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:
n_i·sin(α) = n_r·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:
n_r·sin(δ+α) = n_i·sin(α)
This allows to find angle δ in terms of the incidence angle α:
sin(δ+α) = (n_i/n_r)·sin(α)
δ+α = arcsin[(n_i/n_r)·sin(α)]
δ = arcsin[(n_i/n_r)·sin(α)] − α

It's convenient to use relative index of refraction n=n_i/n_r 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.

Prismatic Lenses: UNIZOR.COM - Physics4Teens - Waves - Properties of Light

Notes to a video lecture on http://www.unizor.com

Prismatic Lenses

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
n_i·sin(θ_i) = n_i·sin(θ_i)
where
indices i and r represent, correspondingly, incident and refracted rays of light,
n_i is a refraction index of the air above and below the prism,
n_r is a refraction index of the glass the prism is made of,
∠MBA = θ_i is the angle of incidence between incident ray of light and a normal MN to a surface of refraction,
∠CBN = θ_r is the angle of refraction between refracted ray of light and a normal MN 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=θ_i 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=θ_i as vertical.

That implies that the angle of deviation of the ray of light from its original direction ∠CBD equals to
∠CBD = ∠CBN − ∠DBN =
= θ_r − θ_i = θ_r − α = δ

Since angle of incidence θ_i is defined by the given prism's angle ∠PQO=α and the angle of refraction θ_r is related to the angle of incident through the Law of Refraction
n_i·sin(θ_i) = n_r·sin(θ_r)
we have a clear path to find the angle of deviation:
1. sin(θ_r) = (n_i/n_r)·sin(α)
2. θ_r = arcsin[(n_i/n_r)·sin(α)]
3. ∠CBD = δ =
= arcsin[(n_i/n_r)·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: UNIZOR.COM - Physics4Teens - Waves - Properties of Light

Notes to a video lecture on http://www.unizor.com

Refraction

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 θ_i and refraction θ_r as follows
b = (x−a)/tan(θ_i)
d = (x−c)/tan(θ_r)

Using these expression, notice that
√(x−a)²+b² =
= √(x−a)²+(x−a)²/tan²(θ_i) =
= √(x−a)²·[1+1/tan²(θ_i)] =
= |x−a| /sin(θ_i)
√(x−c)²+d² =
= √(x−c)²+(x−c)²/tan²(θ_r) =
= √(x−c)²·[1+1/tan²(θ_r)] =
= |x−c| /sin(θ_r)

Substituting these expressions into T'(x), we obtain
T'(x) =
= (x−a)/[|x−a|·V₁/sin(θ_i)] +
+ (x−c)/[|x−c|·V₂/sin(θ_r)] =
= sin(θ_i)/V₁ − sin(θ_r)/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(θ_i)/V₁ − sin(θ_r)/V₂ = 0
Therefore, the necessary condition for minimizing the time for light to travel from P to Q is 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(θ_i)/V₁ = sin(θ_r)/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(θ_i)·n₁ = sin(θ_r)·n₂

When the light falls perpendicularly to a surface that separates two media, an incidence angle is zero, θ_i=0. Under this condition the value of a refraction angle θ_r 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.

Parabolic Reflector: UNIZOR.COM - Physics4Teens - Waves - Properties of ...

Notes to a video lecture on http://www.unizor.com

Parabolic Reflector

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 θ_i and the reflection angle θ_r.

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 ∠θ_i+∠θ_r, that is
(since θ_i=θ_r=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

Based on this analysis, we derive the following:
(a) tan(∠BEA) =
= tan(∠θ) = 2k·a
(b) tan(∠CBD) =
= tan(π−2θ) = −tan(2θ) =
= −2tan(θ)/(1−tan²(θ)) =
= 4k·a/(4k²·a²−1)
(c) cot(∠CBD) =
= 1/tan(∠CBD) =
= (4k²·a²−1)/(4k·a) =
= k·a − 1/(4k·a)
(d) BD = a·cot(∠CBD) =
= k·a² −1/(4k)
(e) OC = AB − BD =
= k·a² − (k·a² −1/(4k)) =
= 1/(4k)

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: UNIZOR.COM - Physics4Teens - Waves - Properties of Light

Notes to a video lecture on http://www.unizor.com

Reflection

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 θ_i 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
V_i(t)={c·sin(θ_i);0;−c·cos(θ_i)}.

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
V_r(t)={c·sin(θ_i);0;c·cos(θ_i)}.

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 θ_r, the vector of velocity is
V_r(t)={c·sin(θ_r);0;c·cos(θ_r)}.
Therefore, we have two expressions for the same vector of velocity after the reflection, and they must be equal to each other:
{c·sin(θ_i);0;c·cos(θ_i)} =
= {c·sin(θ_r);0;c·cos(θ_r)}
Obviously, if
sin(θ_i) = sin(θ_r) and
cos(θ_i) = cos(θ_r),
angles θ_i and θ_r 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.