## Saturday, July 30, 2022

### Rope & Springs: UNIZOR.COM - Physics4Teens - Waves - Light Energy

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

Rope & Springs

A wave on a rope, initiated by some forceful oscillations of its end, propagates along a rope, thereby transferring the initial boost of energy to distant locations.

Our task is to determine how much energy is carried by a single wave on a rope - a part of a rope from crest to crest or from trough to trough, the length of which is the wavelength λ of rope's oscillations.

Let's define the X-axis along the direction the rope is stretched (we will call it "horizontal"). The forced oscillations of the rope's end will be along the Y-axis ("vertical") that is perpendicular to X-axis.

Previous lecture was dedicated to analysis of the vertical displacement of each piece of a rope as a function y(x,t) of its distance from the endpoint of a rope x and time t, if the rope's end (x=0) is forced to perform harmonic oscillations of the form y(0,t)=A·sin(ω·t). Here A is an amplitude and ω is the angular speed of oscillations.

The result of this analysis was a wave equation that function y(x,t) must satisfy
/T)·²y(x,t)/²t = ²y(x,t)/
where
μ is linear mass density of a rope (mass per unit of length), assumed to be uniform;
T is rope's tension;

The solution to this equation with initial condition y(0,t)=A·sin(ω·t) was suggested in the previous lecture as
y(x,t) = A·sin(ω·t−k·x)
where A and ω are defined above and constant k must satisfy the condition
k²/ω² = μ/T

For the purposes that will be clear later we would like to start time (t=0) not when a rope's end is at horizontal level (y=0), but when it's lifted up to a full extent of an amplitude A, that is our initial condition will be y(0,t)=A·cos(ω·t).
From the energy perspective this modification does not change anything because we will be dealing with an energy of a single wave from crest to crest or from trough to trough, which is the same, whether the wave is based on function sin() or cos(), as long as amplitude A and angular speed ω are the same.

Then the solution to a wave equation that satisfies our initial condition can be
y(x,t) = A·cos(ω·t−k·x)

The approach we will take is based on the analogy between sinusoidal (harmonic) movement up and down of each infinitesimal piece of a horizontally stretched rope and harmonic oscillation of the point-mass object on a vertically oriented spring. Assume that each infinitesimal piece of a rope of the length dx and mass μ·dx can be considered as an independent point-mass on some tiny spring, and all these pieces are oscillating in a manner resembling and totally identical by all parameters to movements of these pieces as when they are parts of a rope - the same oscillations in terms of the amplitude and angular speed as well as timing of these oscillations.

If all the motion parameters of each infinitesimal piece of a rope are the same as those of the corresponding point-mass on a spring, the forces must be the same and, consequently, the kinetic and potential energy are the same.

Since we have addressed the energy of a point-mass on a spring in earlier lectures (see "Energy of Waves"), we will use these results to calculate the energy of each infinitesimal piece of a rope, which, in turn, will allow to determine the energy of a single wave by integrating all the individual infinitesimal quantities of energy.

Let's start with the equation describing the vertical movement of a rope at each its point that satisfies the wave equation and the initial condition we have set
y(x,t) = A·cos(ω·t−k·x)
where
A is an amplitude (known, defined by external force that causes oscillations);
ω is an angular speed (known, defined by external force that causes oscillations),
t is time;
x is a known distance of the point of a rope that we analyze from its end that is forced to harmonically oscillate to make waves along the rope;
k is a coefficient that depends on the properties of a rope (linear mass density and tension) as well as the angular speed of oscillations ω and defined by an equation
k²/ω²=μ/T,
that is a known quantity equaled to ω·√μ/T .

The physical meaning of the coefficient k was discussed in the previous lecture, that derived the formula for a speed of propagation of a wave v:
v = ω/k.
At the same time speed of wave propagation v is the wavelength λ divided by period or, since a period is an inverse of frequency f, v=λ·f=λ·ω/(2π),
from which we derive
k = ω/v = 2π/λ
v = ω·λ/(2π)

From the formula for a speed of waves propagation we can derived another form of the solution to a wave equation
y(x,t) = A·cos(ω·t−k·x) =
= A·cos
[ω·(t−k·x/ω)] =
= A·cos
[ω·(t−x/v)]

The expression x/v is, obviously, the time it takes for the wave to go from the place of its origination (x=0) to any fixed distance x, that is a time delay of oscillations at point x from oscillations at point x=0.

So, oscillations at any point x of a rope are the same in terms of amplitude A and angular speed ω, as at point x=0, just with some delay x/v, which depends on a known distance from the wave origination x and a known speed of wave propagation v.

Obviously, we can make a tiny spring with infinitesimal piece of a rope attached to it that has the same oscillations as that same piece being a part of a rope.

We do know the mass attached to a spring, that is the mass of an infinitesimal piece of a rope dx, it's dm=μ·dx.

We can arrange the oscillations of a spring to have the same amplitude A as each piece of a rope. All we need to do is to initially stretch a spring by A.

We also need to have a spring with proper elasticity ke to make sure the angular speed of oscillations of a spring is equal to angular speed of oscillations of each piece of a rope ω.
NOTE: we use here symbol ke for elasticity to differentiate it from k used in the solution for a wave equation y(x,t)=A·cos(ω·t−k·x).

We know from previous lectures that for a spring the angular speed of oscillations relates to elasticity ke and mass of an object attached to it m as
ω² = ke/m.
In our case of a tiny spring with mass attached to it equaled to dm=μ·dx we can determine the elasticity from this mass and required angular speed ω:
ke = ω²·μ·dx
and choose a spring with this exact elasticity.

So, taking a tiny spring with the above elasticity and stretching it by initial length A from its neutral state will produce the oscillations with an amplitude A and angular frequency ω.

What remains to be done is to arrange a proper timing, so the oscillations of the tops of springs will go in waves similarly to waves on a rope.

To accomplish that, after initial stretching of all springs by an amplitude A, we have to release these springs with a time delay required for a wave to reach any particular point on a rope. For a distance x of a particular spring from the beginning, knowing the speed of wave propagation on a rope v, we will just release the spring after x/v time interval from the moment of releasing the first spring.

The movement of the point-mass attached to a spring with described above parameters and delayed start is described by an equation
y(x,t) = A·cos[ω·(t−x/v)]

In the above equation for a spring at distance x amplitude A and angular speed of oscillations ω are defined by initial condition of oscillating the rope
y(0,t) = A·cos(ω·t)

The speed of propagation v is a derived parameter and, therefore, is known as well.
As specified above,
v = √T/μ ,
where μ is a linear mass density of a rope and T is a rope's tension - all known parameters.

Since our task is to determine an energy of a wavelength λ of a rope, we have to integrate the energy of all springs by x on a segment of the length λ. Periodicity of oscillations assures that all such segments will have the same energy.

Let's summarize the characteristics of tiny springs that should oscillate exactly like a rope.

1. Mass attached to each tiny spring equals to dm=μ·dx.
2. Each tiny spring is located at a distance x from the beginning.
3. All springs have the same elasticity ke=ω²·μ·dx
4. All springs are initially stretched by A, so their amplitude will be the same as that of the rope.
5. All springs will have the same angular speed ω, the same as waves on a rope.
6. After initial stretch the springs will be let go, but not simultaneously; the time delay of a spring at distance x from the beginning will correspond to time needed for a wave on a rope to reach that distance, that is the time delay will be equal to x/v=2π·x/(λ·ω).
7. As a result, a spring at distance x from the beginning oscillates according to a formula y(x,t)=A·cos[ω·(t−2π·x/(λ·ω))].

The above characteristics of the springs, that we have used to model the propagating waves on a rope, are sufficient to analyze the distribution of energy, using what we have already analyzed about springs.
Next lecture will use all this to find potential and kinetic energy of a single wave on an oscillating rope.

## Friday, July 22, 2022

### Rope Waves: UNIZOR.COM - Physics4Teens - Waves - Light Energy

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

Rope Waves

Waves on a rope are a more complex kind of waves than simple harmonic oscillations of a point mass on a spring.

Primary difference is that spring oscillations are longitudinal, that is each part of a spring moves along the direction of the wave propagation, while the oscillations of a rope are transverse, that is each piece of a rope moves perpendicularly to the propagation of waves on it.

To analyze these waves we will use the following model.
We assume that the rope we deal with is initially stretched along some line (we will call it "horizontal") and it has one end that is harmonically moving perpendicularly to an initial stretch of a rope. That is, its "vertical" deviation from the neutral position D(t) is a sinusoidal function of time t with certain amplitude A and angular frequency ω:
D(t) = A·sin(ω·t).

The other end of a rope we presume to be far away and fixed and there is some tension T in the rope.

We also assume that there are no other forces acting on this rope, including gravity.

Consider each infinitesimal piece of a rope as an independent point mass connected to both neighboring point masses by some links, like in a necklace.
When the edge of a rope moves up and down, each such a piece pulls its next neighbor in the same direction, as illustrated in the picture below Let's introduce a system of coordinates with X-axis coinciding with the rope in its neutral position (horizontal line on a picture above) and Y-axis perpendicular to it to measure the vertical deviation of the rope's pieces from their neutral position.

If some external force moves the rope's left end from its neutral position (y=0) up to the height y=A, then down by double A (so, the end is at y=−A) and up again to return the rope's end back in the neutral position just once, the left most rope's piece will force the connected to it next one to repeat this motion by pulling the link that connects them, then next one will pull the one after etc.

A single wave (crest and trough) will go along a rope with some speed, as one rope's piece after another will repeat the up and down movement of the previous one.

To describe the dynamics of movements of each rope's piece, we will introduce a function y(x,t) - vertical displacement of a rope's piece positioned at distance x from the left end of a rope at time t passed from the beginning of this experiment.

Consider three neighboring rope's pieces with links connecting them. Assume that the left most piece pulls up the middle one and the middle one pulls up the right most.
Let the X-coordinate of the piece B be x. We are analyzing the positions of these rope's pieces at time t. The most important force acting on all pieces is the rope's tension T, which is the same between pieces A and B as between B and C.

As piece A moves up, it pulls piece B, but to force B to move up, the piece A must be a little higher than B, so line AB should make some angle α with the horizontal level of rope's neutral position.

As piece B moves up, it pulls piece C, but to force C to move up, the piece B must be a little higher than C, so line BC should make some angle β with the horizontal level of rope's neutral position.

What's important is that angle α is greater than β because forces act in sequence, first A lifts up B and then, later, B lifts up C. So, piece A starts rising earlier than B and piece B starts earlier than C and at the same moment in time t piece A will go up more than B.

Since we are interested only in vertical component of motion of these rope's pieces, we have to consider only vertical components of the tension forces.

The vertical component of a tension T between pieces A and B pulls piece B up and equals to
Fup = T·sin(α)
The vertical component of a tension T between pieces B and C pulls piece B down and equals to
Fdown = T·sin(β)

Therefore, the resultant force pulling piece B upward is
FB = T·[sin(α)−sin(β)]

Since we found the force, it's time to use the Newton's Second Law to obtain the wave equation.

Firstly, we need a mass of a rope's piece B. Obviously, our model assumes that a rope consists of infinite number of infinitesimal pieces of the length dx each.
If the rope is uniform, the mass distribution is linear and the mass of a piece of the length dx is μ·dx, where μ is linear density of material a rope is made of (measured in kg/m in SI).

According to the Newton's Second Law, the force acting on the body equals to its mass times acceleration. Infinitesimal piece of a rope B has a length dx, its mass is μ·dx, its acceleration in the vertical direction is the second derivative of its vertical position y(x,t) by time for any fixed x.
Therefore, the force FB can be expressed as
FB = μ·d²y(x,t)/²t

Secondly, we have to connect expression sin(α)−sin(β) with the function y(x,t) describing the vertical location of a rope's piece at distance x from the rope's edge at a fixed moment of time t.
This is a purely technical calculus problem.
If piece B is, in mathematical terms, an infinitesimal interval with X-coordinates from x to x+dx, the value of sin(α)−sin(β) is an increment of a sin() of an angle between a tangent of y(x,t) and X-axis on this interval.

For infinitesimal interval the difference between sin() and tan() of this angle is an infinitesimal of a higher order, so we can replace sin(α)−sin(β) with tan(α)−tan(β).

But tan(α) is the first derivative of y(x,t) by x, that is y(x,t)/x.

An increment of the first derivative on an infinitesimal interval from x to x+dx is the second derivative times dx.
Therefore, as dx tends to zero,
sin(α)−sin(β) tends to
[²y(x,t)/]·dx

Now our original equation
FB = T·[sin(α)−sin(β)]
can be expressed as μ·d²y(x,t)/²t =
= T·
[²y(x,t)/]·dx

Canceling dx on both sides, we obtain the following
μ·²y(x,t)/²t = T·²y(x,t)/
which is equivalent to this differential equation of the second order - the wave equation
/T)·²y(x,t)/²t = ²y(x,t)/

Let's find a solution to this differential equation with initial condition describing the harmonic oscillations imposed on the rope's edge.
If we start from moving the rope's end up from a neutral position, the initial condition would be
y(0,t) = A·sin(ω·t)
If we start from moving the rope's end down from a neutral position, the initial condition would be
y(0,t) = −A·sin(ω·t)

With the initial condition of moving the rope's end up we can suggest the solution to an equation above:
y(x,t) = A·sin(ω·t−k·x)
For downward initial movement, a suitable solution can be
y(x,t) = A·sin(k·x−ω·t)
We will use the former in our calculations, but the latter one would be as good; in different textbooks you might see one or another.

Obviously, the suggested solution satisfies the initial condition:
y(0,t) = A·sin(ω·t−k·0) =
= A·sin(ω·t)

Let's check if this is a solution to a differential equation we have constructed.
We need to calculate ²y(x,t)/²t and ²y(x,t)/.

y(x,t)/t = A·ω·cos(ω·t−k·x)
²y(x,t)/²t =
= −A·ω²·sin(ω·t−k·x)
= −ω²·y(x,t)

y(x,t)/x = −A·k·cos(ω·t−k·x)
²y(x,t)/²x =
= −A·k²·sin(ω·t−k·x) =
= −k²·y(x,t)

From the above follows that for y(x,t)=A·sin(ω·t−k·x) it is true that
(k²/ω²)·²y(x,t)/²t = ²y(x,t)/

Therefore, thus chosen function y(x,t)=A·sin(ω·t−k·x) will be a solution to our differential equation, if we choose angular frequency ω and factor k such that
/ω² = μ/T

We can prove that the physical meaning of the ratio ω/k is the speed of a wave v, that is the speed of a crest of a wave moving along the X-axis as time t goes.

The crest of a wave A·sin(ω·t−k·x) is when function sin() reaches its maximum value of 1, which happens when its argument ω·t−k·x equals to 2πN, where N is any integer number.

Assume that for some moment in time t this crest is at point x, that is
ω·t−k·x = 2πN
Let a small amount of time Δt pass.
To find where the crest is now, that is what is an increment of the distance Δx a moving crest covered during the time Δt, we should solve the following equation for Δx:
ω·(t+Δt)−k·(x+Δx) = 2πN =
= ω·t−k·x

Opening all parenthesis, we obtain:
ω·t + ω·Δt − k·x − k·Δx =
= ω·t − k·x

Canceling identical members from left and right side, we obtain
ω·Δt − k·Δx = 0
From this follows
ω·Δt = k·Δx
Δx/Δt = ω/k

Obviously, when Δt→0,
lim Δx/Δt = v - a speed of a moving crest v.
Therefore,
v = ω/k
Consequently,
k = ω/v = (2π/T)/(λ/T) = 2π/λ

Above we have concluded that the requirement for
y(x,t) = A·sin(ω·t−k·x)
to be a solution to our differential equation is
/ω² = μ/T

Using the link between the speed of propagation of a wave v and the ratio of its angular frequency ω to a coefficient at the distance variable k, we can rewrite our original wave equation for a rope
/T)·²y(x,t)/²t = ²y(x,t)/
as
(1/v²)·²y(x,t)/²t = ²y(x,t)/

Physical characteristics of a rope, tension T and linear mass μ, determine the speed of wave propagation v:
v² = T/μ

The angular frequency of oscillation ω can be expressed in terms of physical characteristics as
ω = k·v = k·√T/μ
The more we tighten the string on a violin (that is, increasing tension T) - the higher the frequency of oscillation and the higher sound it makes.
The heavier the string (that is, increasing linear mass density μ) - the lower sound it makes.

The wave equation has many solutions of the form
y(x,t) = A·sin(ω·t−k·x)
The exact solution in any situation is defined by initial conditions.

Also, a bit more understandable from the physical standpoint way is to express this solution in a form
y(x,t) = A·sin[k·(v·t−x)] =
= A·sin
[(2π/λ)·(v·t−x)]

## Monday, July 18, 2022

### Energy of Waves: UNIZOR.COM - Physics4Teens - Waves - Light Energy

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

Energy of Waves

Any waves can do work. Water waves lift the ships up and down. Sound waves moves eardrums, so we hear sound. Light waves in a form of sun rays heat objects.
All these examples indicate that all waves carry some energy with them.

Light is the oscillations of electromagnetic field. But, prior to investigating these particular types of oscillations, let's spend some time discussing oscillations in general, starting from mechanical oscillation of a point mass on an ideal spring. The Hooke's Law for an ideal spring states that the force of a spring F is proportional to an amount of stretch or squeeze x and always directed towards a neutral position
F(x) = −k·x
where the factor k is a characteristic of a spring.

Assume, we are stretching a spring from its initial position by a length A with future plans to let it go from that point to freely oscillate.
Stretching it from length x to length x+dx against the resistance of a spring requires work
dW = −F(x)·dx = k·x·dx
The total amount of work needed to stretch a spring by length A can be calculated by integrating this from 0 to A
W(A) = [0,A]k·x·dx = ½k·A²

If we have stretched a spring by length A doing ½k·A² amount of work and hold the spring in that position, obviously, a spring has a potential energy equal to that amount of work. So, potential energy depends only on how much we stretch or squeeze a spring and its physical characteristics that determine the coefficient of elasticity k.
If we let it go, it will start contracting, increasing the kinetic energy of a point mass on its end and decreasing its potential energy.
Let's calculate this kinetic energy, based on the spring force and the mass m of an object attached to its end.

According to the Newton's Second Law, if x(t) is a distance of the end of a spring with a mass m attached to it from its neutral position, as a function of time t,
F = m·d²x/d
On the other hand, according to the Hooke's law,
F = −k·x(t),
where a coefficient of elasticity of a spring k, as we mentioned, is a constant that characterizes the spring.

Therefore, we have a differential equation for x(t)
−k·x(t) = m·d²x/d
It's a second order differential equation. To exactly define a solution we need two initial conditions, which are:
x(0) = A since before letting a spring to oscillate we have stretched it at time t=0 by a distance A from a neutral position;
x'(t) = 0 because at time t=0 we just let it go without any initial speed.

Without getting into details of how to solve it (that was covered in details in the course "Math 4 Teens" on this Web site UNIZOR.COM for those interested), we offer a solution
x(t) = A·cos(ω·t),
where ω=√k/m.

Now we can determine the kinetic energy of a moving object on the end of a spring:
K(t) = ½m·v²(t)
Speed v(t) is the first derivative of a distance x(t), that is
v(t) = dx(t)/dt = −A·ω·sin(ω·t)

Therefore, kinetic energy is
K(t) = ½m·A²ω²·sin²(ω·t) =
= ½A²k·sin²(ω·t)

Potential energy at the same moment depends only on the distance x(t) of the spring's end from the neutral position. Its value for distance A we have determined as ½k·A². Analogously, for distance x(t) its value is
P(t) = ½k·x²(t) =
= ½k·A²·cos²(ω·t)

The total energy E at any moment of time t is a sum of potential energy of a spring and kinetic energy of a mass attached to its end:
E = K(t) + P(t) =
= ½A²k·sin²(ω·t) +
+ ½A²k·cos²(ω·t) = ½A²k

The last value equals to the work W(A) we have spent initially to stretch a spring by a distance A from its neutral position and is a constant independent of time t.
So, not surprisingly, we have come to a confirmation of the Energy Conservation Law for oscillating spring.

Initially, all the energy of a system spring + point mass at its end is a potential energy of a spring. As we release a spring to free oscillations, as it returns to a neutral position, its potential energy P(t)=½A²k·cos²(ω·t) diminishes, reaching zero when the end of a spring returns to a neutral position, while the kinetic energy of a point mass K(t)=½A²k·sin²(ω·t) increases, reaching its maximum at the same point.
After passing the neutral point the process reverses, potential energy grows to its maximum, when the distance of the end of a squeezed spring is at distance A from a neutral position, while the kinetic energy diminishes to zero at that point.
The oscillations continue along the same pattern to infinity.

Note that the oscillations propagate along the axis of a spring. An object at the and of a spring moves along the same axis as well.
That's why these oscillations are called longitudinal.

The next lecture will be dedicated to oscillations of a rope. The movements of a horizontally stretched rope with waves forced on it by periodic movement of its edge up and down are more complicated than those of an ideal spring. Not only each small part of a rope moves up and down similarly to oscillations of a spring, but there is a wave front moving along a rope. This complicated movement is close to oscillations of electromagnetic field.

## Wednesday, July 13, 2022

### Polarization: UNIZOR.COM - Physics4Teens - Waves - Phenomena of Light

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

Polarization of Light

The light is transversal electromagnetic waves.
Each individual ray of light has electric and magnetic components, oscillating perpendicularly to each other, each within its plane. (click the right button to open a larger image in a new tab)

The light from the Sun, from a lamp and from many other sources of electromagnetic waves constitute a mix of oscillations of electric and magnetic components in all directions perpendicular to the direction of light propagation, with different amplitudes and phases. There are some substances, called polaroids, naturally occurring crystals or artificially created, that, when placed on the trajectory of the light ray, selectively allow to pass through only oscillations in one particular plane.
This process of selecting only one particular plane of oscillations of electromagnetic waves to go through, while suppressing all others is polarization of light.

The following picture shows this process of polarization, where for simplicity only electric component is shown: As we see, while the light rays before the polaroid consist of oscillations in all different planes, there is only one plane of polarization, where oscillations of electric component of an electromagnetic field occur after the process of polarization.

Those oscillations before a polaroid, that are already in the plane of polarization, are going through a polaroid without change.
Oscillations in a plane perpendicular to a plane of polarization are not going through at all.
All other oscillations are weakened by the polarization for the following reason.

Assume, the plane of oscillations of electric component vector E of the electromagnetic wave is at angle θ relative to a plane of polarization of a polaroid.
We can then represent vector E as a sum of two vectors, one coinciding with a plane of polarization Ein, which the polaroid let through, and another perpendicular to it Eout, which will not go through a polaroid:
E = Ein + Eout =
= E·cos(θ) + E·sin(θ) The oscillations represented by electric component Ein=E·cos(θ) coinciding with the plane of polarization will go through the polaroid and on the other side the light would be represented only by polarized
Epol = Ein = E·cos(θ).

The brightness of the light will diminish after the polarization because almost all components of the incoming rays of light, having different amplitudes, phases and angles to a plane of polarization, will fall onto polaroid not within its plane of polarization, and only part of these rays would go through after their amplitude is multiplied by cos of the angle between the plane of polarization and the plane of oscillations of the electric component of the electromagnetic field.

Brightness of the visible light is related to the amount of energy the light carries.
Analysis shows that the amount of energy carried by a single ray of light is proportional to a square of its amplitude. We will address the logic behind this in the next lecture dedicated to energy of light.
Knowing this, we can say that the amount of energy of a single ray of light after going through the polaroid is less than the original amount by a factor cos²(θ).
So, the brightness of a single ray of light will diminish after polarization by cos²(θ). This is Malus Law.

This rule corresponds to the observable phenomena of light.
Consider a single ray of light with oscillations coinciding with a polarization plane. It means, θ=0 and cos(θ)=1, that is the ray goes through without any weakening.
On the other hand, if the plane of oscillations of the electromagnetic field is perpendicular to a polarization plane, θ=90° and cos(θ)=0, that is there is no light behind the polaroid.

## Saturday, July 9, 2022

### Diffraction: UNIZOR.COM - Physics4Teens - Waves - Phenomena of Light

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

Diffraction of Light

In the previous lecture about light interference we considered two slits that allow a flat wave front to go through and considered the light coming out from these two slits as two sources of light in phase with each other.
We assumed that these two slits were of infinitely small width, so we can only consider interaction between them, not paying attention to interaction of light rays coming into and from one single slit.

Since any slit has a finite width, according to the Huygens Principal, all points of a wave front reaching a slit are independent sources of light. Rays, coming in phase to a slit, go through, producing waves in all directions on exit from a slit, and these secondary waves of light will interfere with each other.

Generally, any wave front, according to the Huygens Principle, is a set of independent points that are sources of secondary light rays that, potentially, interfere with each other.
This process of interference between the neighboring rays of light is called diffraction.

Imagine a set of parallel rays of light going through a perpendicular to their direction slit and, after passing through it, falling onto a perpendicular to their direction screen positioned at some distance from a slit.

The first what we notice on a screen is a bright line that corresponds to a slit the light goes through.
But, if we look more attentively, we would notice the parallel lines on both sides of the bright line, which we can only attribute to the interference among light rays coming out of a slit.

The picture that might be observed would look similar to this: The lines parallel to the central bright line are the result of interference among light rays going through one and only slit of a small but finite width. This is called the diffraction of light.

Our task is to understand how the diffraction works, applying appropriate approximations to shorten the calculations, as physicists usually do.
For this purpose we choose a single point S on a screen at a distance x from a center of a main bright line and examine all light rays that go from a slit to this point, as presented on a picture below. Let the screen's distance from a slit be relatively large L (say, about 1 meter) and the width of a slit be relatively small d, like 1 micrometer or so, which is comparable to a visible light wave length.
Assume that monochromatic laser ray (for example, blue) goes through this slit.

Every point of observation on a screen S receives a set of light rays from all points of a wave front AB across a slit's width, that we can approximately consider as set of parallel rays because the distance L is significantly larger than d.

To determine the amount of light falling onto any particular point of observation, like O or S, we have to consider only these approximately parallel light rays coming from the slit to the point of observation and interactions between them, disregarding all other rays emitted to other directions from the slit.

The incident angle of the rays falling perpendicularly to a screen from a slit to point O equals, approximately, to zero. All the rays falling into this point from a slit are covering, approximately, the same distance L to reach this point.

Being in phase at a slit, all these light rays will be in phase at point O.
Therefore, all the rays falling into point O will interfere constructively with each other, they will all contribute light energy, and that's the reason for the bright image of a line to be present on the screen at this point.

The picture at other points, not on a perpendicular from a slit to a screen, is shown on the example of point S.
Different light rays falling from a slit onto point S will have to cover different distances because of finite width of a slit.

The distance AS is longer than BS by
Δ = √L²+(x+(d/2))²
− √L²+(x−(d/2))²

Usually, this complicated formula is approximated, assuming that the width of the slit d=AB is small relative to a distance between a slit and a screen L.
Let p=√L²+(x+(d/2))²
and q=√L²+(x−(d/2))².
Then the following equalities are valid:
Δ = p − q
Multiply and divide this expression by a sum of these two radicals p+q:
Δ = (p² − q²)/(p + q)
p² − q² = 2x·d
p + q ≅ 2√L²+x²

Therefore,
Δ ≅ 2x·d/(2√L²+x²) =
= d·(x/L²+x²)

But x/L²+x² is sin(θ), where θ is an incident angle of a ray coming from a slit to a point of observation.

Therefore, approximate value of the difference between the longest distance AS from a slit to point S and the shortest one BS is
Δ ≅ d·sin(θ)

All rays between AS and BS have to cover different distances to point S greater than BS, but less than AS.
This causes all these rays to be more or less out of phase and interfere with each other in a complex constructive or destructive ways.

There is one and only one condition, when point S would be relatively dark. This happens if all the rays coming to point S from all points on the wave front AB are interfering destructively with each other, that is for each ray of light reaching point S there is one light ray reaching point S in anti-phase.

If there are some light rays emitted by a wave front AB, for which there are no anti-phase pairs to destructively interfere with them, the point S will have some light, more or less, depending on how many rays of light are not paired with anti-phase rays and, therefore, not canceled by destructive interference.

Assume the wave length of the monochromatic light falling on the slit is λ.
The first case we consider is the one when the longest distance AS from a slit to point S is by Δ=d·sin(θ)=λ greater than the shortest distance BS.
That means, ray AS is longer than BS by the wavelength λ.

If we consider points of the wave front from A to B, the distance from these points to the point of observation S will be greater than the shortest distance BS by some value that monotonously changes from maximum λ to minimum 0.
Somewhere in between A and B there will be a point M such that MS is greater than BS by half the wavelength λ/2. Obviously, two initially in-phase rays, BS and MS that cover distances to the same point S that differ by half a wavelength λ/2, will be in anti-phase upon arriving to that point and should cancel each other.

Our next step is to organize rays in pairs of mutually canceling rays of light and see if there are some remaining lights not canceled by a corresponding ray in anti-phase.

Let's pair a ray BS with a ray MS, where M, as mentioned above, is a point, from which the distance MS is by λ/2 greater than BS.
Since MS is longer than BS by λ/2 and these rays initially (at points M and B) were in-phase, they arrive to point S in anti-phase and cancel each other, as seen on a picture above.

Let's move slightly from point B to point B' along a wave front towards A. The distance B'S would be slightly longer than BS.
Since the distance to point S monotonously increasing as we move from B to A along the wave front from the length of BS to the length of AS, which is by λ longer, we can find a point M' slightly towards A from point M such that the distance M'S is greater than B'S by the same half wavelength λ/2.
Let's pair light rays B'S and M'S. Upon arriving to point S they will be in anti-phase and interfere destructively, that is cancel each other.

Actually, for every point B' from B (inclusive) to M (not inclusive) along the wave front we can find a corresponding point M' along a wave front from M (inclusive) to A (not inclusive) such that the difference between the distances B'S and M'S will be exactly λ/2, so the corresponding rays B'S and M'S going towards point S will arrive in anti-phase and cancel each other.
It means that all rays from every point on a wave front, except point A, will destructively interfere with each other.

The infinitesimal by amount of light ray AS, for which there is no pair in this logic, can be disregarded, as not delivering any noticeable energy to point S. There will be practically no light at point S positioned in such a way that
AS − BS = λ ≅ d·sin(θ).

The equality above allows to calculate the distance Rdark_1 of the first dark line from the bright image of the slot on the screen:
sin(θ) ≅ λ/d
Since for small θ
sin(θ) ≅ tan(θ) = Rdark_1/L,
Rdark_1 ≅ L·λ/d

Let's try to calculate the distance to the second dark line on a screen from a central bright line, using similar logic.
Imagine that
d·sin(θ) ≅ 2λ
Now, instead of breaking all the light rays into two groups, pairing each ray from the first group with the one from the second that travels the distance to point of observation S by λ/2 longer, which caused it to be in anti-phase with the first ray, effectively canceling both rays, we can divide all rays into four groups:

Group I are all those rays that go along paths longer than BS by no more than λ/2.
Group II are all those rays that go along paths longer than BS by more than λ/2, but no more than λ.
Group III are all those rays that go along paths longer than BS by more than λ, but no more than 3λ/2.
Group IV are all those rays that go along paths longer than BS by more than 3λ/2, but no more than .

All rays going through a slit would fall into one or another of these four groups.
Using the monotonic behavior of the length of a ray from a point on a wave front AB, from which it emitted, to the observation point S, we conclude that
(a) for each ray from the Group I there is one and only one ray from Group II, whose length is greater exactly by λ/2;
(b) for each ray from the Group III there is one and only one ray from Group IV, whose length is greater exactly by λ/2;

Therefore, each ray will be paired with another, whose length is longer by λ/2 and, correspondingly, which will come to point S in anti-phase, canceling both rays in a pair.

The condition d·sin(θ) ≅ 2λ above allows to calculate the distance Rdark_2 of the second dark line from the bright image of the slot on the screen:
sin(θ) ≅ 2λ/d
Since for small θ
sin(θ) ≅ tan(θ) = Rdark_2/L,
Rdark_2 ≅ 2L·λ/d

It's easy to generalize now that the condition for a dark line is
d·sin(θ) ≅ N·λ
where N - some integer number. For positive N we will get the location of dark lines on one side of the middle bright image of a slot, for negative N we will get symmetrical dark line on the other side from a center.
Obviously, this condition necessitates the maximum value for N to be such that sin(θ) does not exceed 1, that is
sin(θ) = N·λ/d < 1 or
N < d/λ

In-between the dark lines there must be bright lines. They appear, when the condition d·sin(θ) ≅ N·λ is not satisfied.
Depending on exact value of incident angle θ (or, which is equivalent, on a value of distance x from the center of picture to the observation point S), more or less rays will remain not canceled by another ray in anti-phase with it, or partially interfere destructively, retaining some light energy to light up point of observation S.

For example, consider situation:
d·sin(θ) ≅ 3λ/2
Divide all rays into three groups:
Group I are all those rays that go along paths longer than BS by no more than λ/2.
Group II are all those rays that go along paths longer than BS by more than λ/2, but no more than λ.
Group III are all those rays that go along paths longer than BS by more than λ, but no more than 3λ/2.

As before, for each ray from the Group I there is one and only one ray from Group II, whose length is greater exactly by λ/2, which will cause it to be in anti-phase with the first ray and cancel it at point S.
But for any ray in Group III there will be no pair in anti-phase. There will be rays out of phase, they will somehow interfere with each other in partially constructive or partially destructive way, but they will not completely cancel each other, some light from each ray of this group will reach point S and we will observe some light there.

Generally, the brightest lines will be observed when
d·sin(θ) ≅ (N+0.5)·λ
because for each ray of the length longer than the length of ray BS by no more than N·λ there will be a canceling ray coming to point S in anti-phase,
but for all rays of the length longer than the length of ray BS by more than N·λ there will be no canceling pairs, they will bring light to point S.

It should be noted that all the above calculations were intentionally simplified using typical approximations customary in physics.
It was sufficient for our purpose to demonstrate the qualitative picture of diffraction as the light phenomena.
More precise calculations could have been done, if more practical goals would be our aim.

## Saturday, July 2, 2022

### Interference of Light: UNIZOR.COM - Physics4Teens - Waves - Phenomena of...

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

Interference of Light

Before examining the phenomena of interference of light (recall that light is transverse waves in the electromagnetic field), to clarify this issue, let's start with a simple experiment of transverse waves on a surface of water, produced by two independent sources of oscillation positioned close to each other.

Each source of oscillations, by itself, produces concentric waves on the surface of water with a circular wave front, gradually expanding its radius with some speed of propagation.

Let's introduce a Cartesian system of coordinates with XY-plane coinciding with the water surface in its neutral (without waves) state, Y-axis going along a line connecting two sources of oscillation that have XY-coordinates (0,s) and (0,−s), and Z-axis going perpendicularly to the water surface through X,Y-coordinates (0,0).

The picture below schematically represents the surface of water with the position of the wave front from each source of oscillations (M and N) at times t, 2t and 3t, where t is some time interval: The periodic motion of water molecules, when a wave front and subsequent waves from a single source of oscillations go through them, is, approximately, up and down. Their X and Y coordinates remain, approximately, constant, while Z coordinate oscillates from some maximum positive value A (called amplitude) to negative −A.

This is usually represented as harmonic oscillations in the following form:
z = A·cos(ω·t+φ(r))
where
z is the distance up or down along the Z-axis from the neutral position of a molecule on the XY plane, that is its Z coordinate,
A is the amplitude of these oscillation up and down,
ω is the angular frequency of oscillations of the source that caused the waves on the water surface (number of oscillations per unit of time f multiplied by ).
φ(r) is the angular phase shift, which depends on the distance of an oscillating water molecule from the source of oscillations r=√x²+y².

Assume,
T is the time period of the wave (the time one molecule of water takes to go from one highest position all the way down to the lowest point and up to the highest point again),
f is a frequency of oscillations, that is, the number of full oscillation cycles per unit of time,
v is a speed of propagation of the wave front and
λ is the wave length (the distance between two consecutive crests or two consecutive troughs of waves).

Obviously,
v = λ/T because speed is distance divided by time it takes for an object to move along this distance,
f = 1/T because, if one oscillation cycle lasts T units of time, the number of cycles per unit of time is 1/T,
ω = 2π·f = 2π·v/λ because, if one full cycle is equivalent to radians rotation, f rotations per unit of time are equivalent to 2π·f angular rotation per unit of time.

For a water molecule at distance r from the source of oscillations the number of waves between it and the source of oscillations is r/λ (of course, it is not necessarily an integer number). The angular phase shift of oscillations of that molecule relative to the source of oscillation is, therefore, 2π·r/λ.
The absolute phase shift of this molecule's oscillation is then
φ(r) = φ0 + 2π·r/λ
where φ0 is the initial phase shift of the source of oscillation at the initial time t=0, which is important to take into consideration when two sources of simultaneous oscillations are not synchronized.

Therefore, we can write an equation that describes the oscillations of a molecule at coordinates (x,y,z) as
z = A·cos(ω·t + φ0 + 2π·r/λ)

Let's now return to a case with two independent sources of oscillations at points M(0,s,0) and N(0,−s,0) produce waves on the surface of water. For simplicity, assume the frequency of oscillation f (and, therefore, angular frequency ω and wavelength λ) is the same for both sources. Also, assume their amplitudes are the same and equal to A.

For any point P(x,y,0) on a surface of water the distances to both sources of oscillations are: rM = √x²+(y−s)²
rN = √x²+(y+s)²

The source of oscillations M produces the oscillations at point P(x,y,0) at time t, according to this formula:
zM(x,y,t) = A·cos(ω·t +
+ φM + 2π·√x²+(y−s)²/λ)

The source of oscillations N produces the oscillations at point P(x,y,0) at time t, according to this formula:
zN(x,y,t) = A·cos(ω·t +
+ φN + 2π·√x²+(y+s)²/λ)

The resulting oscillations at point P(x,y,0) at time t are the combinations of the above two:
z(x,y,t) = zM(x,y,t) + zN(x,y,t) =
= A·cos(ω·t+φM+2π·rM/λ) +
+ A·cos(ω·t+φN+2π·rN/λ)

The waves on the water produced by one source of oscillations are simple concentric ones going symmetrically in all directions with the same speed.
The waves from two sources of oscillations, especially with different amplitudes, angular frequencies and initial phase shifts produce a much more complicated picture of interference between two waves.

Just to demonstrate the complexity of the movement of a water molecule responding to two independent sources of oscillations in a more complicated case of different frequencies, amplitudes and phase shifts, here is a graph of periodic oscillations at some point:
z = 0.6·cos(10t+2)+2·cos(2t+6) In this case the water molecule at any point will make periodic motion, but within each period the movements up and down will have many intermediate local crests and troughs of different height and depth, so the surface of water will not look like nice concentric waves moving symmetrically from some center, but rather like pretty chaotic set of crests and troughs.

The location of crests and troughs on the water depends very much on two main factors:
(a) the difference in distance from both sources of oscillation and
(b) the wave length.

We will consider only one simple case of wave propagation.
Let's choose the initial time t=0, and assume that both waves produced by both oscillations at their initial position have the same angular frequencies:
ωM = ωN = ω,
are at their top amplitude:
φM = φN = 0,
the same amplitudes:
AM = AN = A,
and the distances from these sources to a point on the water surface that we observe are
r1 and r2.

If the difference between these distances from sources to a point of observation is a multiple of the wave length, that is
Δr = r1 − r2 = N·λ,
where N is an integer number and
λ is the wave length on the water,
the waves coming from two sources of oscillation will be in phase and will enhance each other.
If one wave comes to an observation point at its top crest, another will also come up at its top crest. The result of their superposition will be a crest of a double height.
If one wave comes to an observation point at its bottom trough, another will also come up at its trough. The result of their superposition will be a trough of a double depth.
All intermediate states of the waves will also be in phase and enhance their appearance (double up or double down).

An opposite situation occurs if
Δr = (N+0.5)·λ
In this case two waves are coming to an observation point with opposite amplitudes, which sometimes is called anti-phase.
When a wave from one source of oscillation comes to such a point at its top crest, a wave from another source comes at the bottom trough, and they neutralize each other, the water at such a point will be still.

In other points of observation the waves of two sources will be partially out of phase and the oscillations will make a different picture, not the one as in phase waves, nor as anti-phase, but some mixture of both.

All the oscillations of water can be easily observed. We can see that at one point water goes up and down to a greater extent than in another point because waves of two sources are coming to the first point in phase, while in another point they are in anti-phase.

With this picture of the interference on the water surface in mind let's move to the visible light, taking into consideration that it is transverse oscillations of electromagnetic field, that is electromagnetic waves.

Everything we said about interference of the water waves is applicable to the electromagnetic waves, including such concepts as
amplitude A,
frequency f,
angular frequency ω=2π·f,
period T=1/f,
wavelength λ,
speed of light c=λ/T=λ·f,
angular phase shift φ,
waves being in phase, anti-phase, out of phase.
As another useful formula derived from above definitions, consider this:
ω = 2π·f = 2π/T = 2π·c/λ

Amplitude of the light waves is related to intensity, brightness of light.
Frequency of light, its angular frequency and wavelength are related to the color of light.
So, if different light rays come to the same point (for example, they fall into our eyes or on the flat screen), the interference picture of light of different intensity and color will be visible.

We will consider a simple case of monochromatic light of some wavelength λ, emitted by some source of light, going through two parallel slits, not far from each other, and falling on a screen positioned parallel to slits, as on the picture below that represents a two-dimensional section of this experiment by a plane perpendicular to slits. As the flat wave front goes simultaneously through both slits, we can assume that, according to the Huygens principle, two independent sources of light exist at points M and N, sending light rays in all directions, and that electromagnetic waves sent by these slits are of the same wavelength, amplitude and are in phase.

On the screen we will observe certain number of light lines parallel to slits with brightness diminishing as they are further and further from the slits.
The wavy red curve on a picture represents the amplitude (brightness, intensity) of the light on a screen.

We will not see on a screen actual light oscillations (like waves on a water surface) because light oscillations have very high frequency and our eye cannot transmit signals individually to a brain with this frequency. So, the spot on a screen where light oscillates with higher amplitude is visible as a bright spot, and the spot with lower amplitude is visible as dark.
Amplitude is, obviously, a result of interference between rays from our two slits, exactly as on a water surface.

Point A on the screen is on a perpendicular bisector of segment MN and, therefore, is on the same distance from both slits. Therefore, the light rays from both slits M and N, as they reach point A, are in phase and strengthen each other, causing the light line at point A to be the brightest.

For our calculations we will use the distance x on the screen from a middle point A to evaluate the brightness of the light line on a screen.
The other variables are the distance from the slits to our screen L, distance between sources of light d=MN and the wavelength of the light λ.

For any point of observation x on a screen we can calculate the difference Δ between distances SM and SN of this point from sources of light M and N to evaluate phase synchronization between two rays coming to this point from these two sources.
SM = √L²+(x+(d/2))²
SN = √L²+(x−(d/2))²
Δ = √L²+(x+(d/2))²
− √L²+(x−(d/2))²

If this difference in distances is equal to an integer number of wavelengths, that is if Δ is equal to n·λ, where n is any integer number, both rays from M and N coming into this point will be in phase, enhance each other and we will have a bright light line on a screen at this point.
If this difference in distances is equal to an integer number of wavelengths plus half of the wavelength, that is if Δ is equal to (n+0.5)·λ, where n is any integer number, both rays from M and N coming into this point will be in anti-phase, weaken each other and we will have a dark line on a screen at this point.

Obviously, there are intermediate values of Δ, which result in gradual change of brightness from maximum, when Δ=n·λ, to minimum, when Δ=(n+0.5)·λ.

In addition, the general strength of light rays from slits M and N will weaken, as the distance of our point of observation to them increases because these rays will fall at higher and higher incident angle to a screen surface.

Finally, we should mention that physicists usually simplify the formulas, using approximation. Let's follow this example and simplify the results above.
We will assume that distance between the slits d=MN is small relative to a distance between slits and a screen L. Also, the distance x from the point of observation and the middle point A also small relative to L.

Let p=√L²+(x+(d/2))² and q=√L²+(x−(d/2))².
Then the following approximate equalities are valid:
Δ = p − q
Multiply and divide this expression by a sum of these two radicals p+q:
Δ = (p² − q²)/(p + q)
p² − q² = 2x·d
p + q ≅ 2√L²+x²

Therefore,
Δ ≅ 2x·d/(2√L²+x²) =
= d·(x/L²+x²)

But x/L²+x² is sin(θ), where θ is an incident angle of a ray coming from a midpoint between the slits to a point of observation.

Therefore, if the distance between two sources of light is d and our screen of observation is sufficiently far from the sources of light (L is significantly larger than d), and an incident angle θ of a ray of light falling into an observation point from a midpoint between the slits is such that d·sin(θ) equals to n·λ, where n is any integer number and λ is the wavelength of the light, we will observe the bright light lines.

In a case when d·sin(θ) equals to (n+0.5)·λ we will get a dark lines with graduate change from bright to dark in intermediary points on a screen.

If we use the formula d·sin(θ)=n·λ and determine the possible values of n, we see that
n = d·sin(θ)/λ
Considering sin() cannot be greater than 1 by absolute value, we conclude that |n| is less than d/λ.