Monday, June 27, 2022

Dispersion on a Sphere and Rainbow: UNIZOR.COM - Physics4Teens - Waves -...

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

Dispersion through a Sphere

Let's analyze how a rainbow is developed after a rain or above waterfalls.
We will model this process with a set of parallel rays of white light (from the Sun) going through air and falling onto a transparent sphere (a water droplet).

Generally speaking, when a ray of light falls on a border between two substances (like air and water in case of a rainbow), part of light goes through the border, refracting on its way, and part is reflected by a border surface.

Let's first examine the simplest case of a white light ray from the Sun going through a spherical water droplet without reflection.

Notice that the perpendicular to a surface of a sphere at its any point is a radius to this point.

Let α be the incident angle of the white light ray coming from the air onto a transparent sphere (a water droplet hanging in the air),
β will be a refraction angle (different for different colors because of the difference in the speed of light of different wavelengths),
β' will be an incident angle, as the light goes from inside of a sphere out into the air and
γ will be a refraction angle of the light that came out from the sphere into the air.

From the Law of Refraction:
sin(α)·nair = sin(β)·nwater
and
sin(β')·nwater = sin(γ)·nair

Incidentally, since OP=OQ=OR, ∠β = ∠β'.
Therefore,
sin(α)·nair = sin(γ)·nair
from which follows that
α = ∠γ.

So, the white light ray is split by the refraction on two surfaces of the sphere, on entry and on exit, and we definitely have a dispersion of white light, as it goes from the air through a sphere of the water droplet and into the air again.

As mentioned before, not all the light goes along the path on the picture above. On each sphere surface, on entry into or exit from a droplet of water, the light is partially reflected. Moreover, light internally partially reflected inside a sphere can be partially reflected multiple times before it exits the sphere, each time scattering rays in different directions.

We assume that the rays of white light from the Sun are parallel and fall similarly on all the water droplets in the air.
We can also assume that immediately after the rain with high humidity in the air or above the waterfall the concentration of water droplets is significant, so that significant amount of light would hit some droplets, refract, splitting the white light into its color components, and only then will reach the eyes of an observer.

With a fixed position of the Sun in the sky and an observer on the ground, for each given droplet of water, there is only a narrow set of sun rays that go through a droplet, disperse into a conical spectrum of colors and reaches the eyes of an observer.

Moreover, if a particular component of a particular white sun ray, like red, is directed by refractions into an eye of an observer, the other component of the same white sun ray, like green, will miss this eye; it's a green component of another (neighboring) white sun ray, that is refracted by another water droplet, will reach the same eye.
That green light will fall into the eye at a different angle than the red described above, and that's why we see red light in one point of a sky and green in another, all colors in the same sequence, in order of their refractive index in water.

The refraction without internal reflection described above will produce colorful rainbow, but an observer at a position presented on the picture above might not be able to see it because the Sun will brightly shine straight into his eyes.
More practical rainbow observation can be obtained if the Sun is behind or on the side from an observer and, even better, when an observer is shielded from the direct sun rays by trees or buildings or mountains.

Consider what happens if the sun rays after penetrating the water droplet reflect from the inside of a droplet and then refract on the way out going into the eyes of an observer.

The ray of white light hits a water droplet at point P, goes through the surface of water, refracts to points from Q (for red) to R (for violet).
Then the whole spectrum of these colorful rays reflects from the inside of a water surface to points from Q' to R'.
Here the rays penetrate the surface of water and come to the air, refracting again and going to the eyes of an observer.

In this scenario the Sun does not directly flashes into the eyes of an observer, and the rays of different colors can be seen.
All other considerations about how we see different colors explained above are applicable to this case as well.

In theory, the light can be internally reflected more than once, and, correspondingly, the rainbow will be visible in more than one place on the sky.
However, we always see a circularly formed rainbow, which requires an additional explanation.
So, let's talk about why a rainbow has its circular shape.

Imagine a line from the Sun to an observer as the Z-axis of some Cartesian coordinates in space with an observer being at coordinate z=0 and the Sun at some large positive Z-coordinate.
Assume that the cloud of water droplets is in space with also positive Z-coordinates, that is the cloud is in-between an observer and the Sun along the Z-axis.
If a particular component, like red, of a refracted white sun ray from the water droplet with coordinates (x,y,z) reaches the eye of an observer, then there will be the same color component, refracted into the eye of an observer by another droplet that is positioned at the same angle to an observer from the Z-axis and at the same distance from the Z-axis, as represented on this picture:

That makes all the places, where an observer sees red light, a circle in the sky.
If the Sun is not in the zenith, but positioned closer to a horizon, the circle will not be complete. Also, if water droplets do not fill an entire sky, the rainbow would not be a complete circle either.
But, some partial circle of a rainbow will be observed in the sky.

The above case is not very pactical, as the Sun flashes right in the eyes of an observer, preventing to see the rainbow.
In a more practical case, when the cloud of water droplets is not between an observer and the Sun along the Z-axis. Instead, an observer is in-between a cloud of water droplets and the Sun.

As we mentioned before, the sun rays are always partially go through a surface of the water and partially reflected. This reflection can be observed on the outside of a water droplet and inside it (internal reflection).

As the Sun sends its white light rays towards a cloud of water droplets in front of an observer from behind his back, some rays go through a surface of water droplets, refracting on the way, then they reflect back against the opposite inner surface of a droplet, go through another surface, refracting again, and into the eye of an observer.
This is how a rainbow is formed when the Sun is behind the back of an observer, as he looks at the water droplets in the sky.
Again, the same circular formation of the places in the sky that have the same color can be observed because of the same considerations as above.

Continuing this train of thought and taking into consideration the effect of partial reflection of light, the same ray of sunlight can produce one color component by going through two surfaces of a droplet, then, after partial reflection from the inside of a droplet and subsequent refraction, another color ray is produced directed differently then the first one.
That's why sometimes we see double rainbow concentric to the first one, but weaker in intensity.

Friday, June 24, 2022

Dispersion - Prism: UNIZOR.COM - Physics4Teens - Waves - Phenomena of Light

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

Dispersion through a Prism

Let's analyze now what happens with the ray of white light going through a side of a regular right triangular prism (with equilateral triangle as each base) on a trajectory parallel to prism's bases, as represented in a prism's section along a trajectory of this ray on the picture below.

Let α be the incident angle of the white light ray coming from the air onto a side surface of a prism parallel to the bases, hitting it at point P on the line AB of the prism section ΔABC.

Let β be the angle of refraction after this ray passed the border surface between the air and the glass. We will consider the fastest visible light in the glass, red, and its value of βr separately from the slowest color, violet, and its value of βv, while values of the refraction angle for other colors to be in between these two extremes.

Then, β' will be the group name for incident angles, when the rays of different colors reach the other side of a prism BC. We will differentiate points Q, where the red component of the ray of white light hits the side of a prism, and R, where the violet component hits the prism's side and distinguish β'r for red ray of light from β'v for violet one.

Finally, γ is a group name for refraction angles of red (γr) or violet (γv) rays after they pass the border from the glass into the air.

The Law of Refraction allows to calculate the refraction angles βr and βv, based on the incident angle α and refractive indices nr and nv of glass for each color, as described in the previous lecture:
sin(βi) = sin(α)·ni /nair
where we can safely assume that nair=1 for all colors and i index is either r for red or v for violet.

Since the light goes from the substance with a smaller refractive index (air) into a substance with a larger refractive index (glass), the refraction angle will be smaller than that of incident.

Simple geometry of ΔABC allows to calculate the angle of incident β' of the ray, as it goes from inside the prism out to the air through the side BC of a prism, based on the value of the angle β:
∠BPQ = 90° − β
∠PBQ = 60°
∠BQP = 180° − ∠BQP −
− ∠PBQ = 30° + β
β' = 90° − ∠BQP = 60° − β

Using the same Law of Refraction and knowing the incident angle of rays β', as they come out from the glass into the air, we can calculate the refraction angle γ for each color using the derived above formula
sin(β'i)·ni = sin(γi)·nair
sin(γi) = sin(β'i)·ni /nair
where we can safely assume that nair=1 for all colors and i index is either r for red or v for violet.

In the previous lecture we have calculated the values for different refraction angles β with the angle of incident α=30°.
 Color α n β Red 30° 1.520 19.205° Orange 30° 1.522 19.179° Yellow 30° 1.523 19.166° Green 30° 1.526 19.126° Blue 30° 1.531 19.062° Violet 30° 1.538 18.971°

Now, using these values of β and values of β'=60°−β, we can continue with calculating the final refraction angle γ.
Notice, the refraction angle γ will always be larger than incident angle β' because the ray of light goes from a substance with a higher refractive index (glass) into a substance with the lower one.

Another very interesting phenomenon can be observed in this case. The refraction angles for blue and violet colors cannot be calculated, because the sine of the refraction angle, as calculated based on the incident angle, refractive index of the glass and the Law of Refraction, becomes greater than 1. It indicates that rays, if falling at a sufficiently large incident angle from within a glass, will not go out, but will be internally reflected from the wall of the prism.
This phenomenon is called the total internal reflection and is the basis for fiber optics, where light signals are sent along a thin tube made of glass or similar material, so they internally reflected off the walls and propagate only inside the tube.

Here are the final results of the calculation of refraction angles on exit from the prism for different colors.

 Color β' == 60°−β n γ Red 40.795° 1.520 83.3° Orange 40.821° 1.522 84.2° Yellow 40.834° 1.523 84.8° Green 40.874° 1.526 87.0° Blue 40.938° 1.531 none Violet 41.029° 1.538 none

Let's analyze the angular deviation of the final rays of light after leaving the prism from the original direction of the white light.
The final angular deviation can be represented as a sum of two: the angular deviation on the border from air to glass and the angular deviation on the border from glass to air.
The first one is
Δ1 = α − β
The second one is
Δ2 = γ − β'
Since β'=60°−β, the final formula for total deviation Δ is
Δ = α − β + γ − β' =
= α − β + γ − 60° + β =
= α + γ − 60°

In our example of the original incident angle α=30° and taking into account average refractive index of glass for yellow color, the refraction angle γ=84.8°. This gives the deviation total to be
Δ=30°+84.8°−60=54.8°
Obviously, it's a little smaller for lights faster than yellow, like red or orange, and a little larger for lights slower than yellow, like green, but still the values are pretty close to this average one.

Thursday, June 23, 2022

Dispersion on Flat Surface: UNIZOR.COM - Physics4Teens - Waves - Phenome...

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

Dispersion on Flat Surface

Let's talk about rainbow.
Everybody saw it, it's beautiful, but what is the reason for rainbow to appear after rain or above the waterfall?
A short answer is - refraction.

As we know, when the light goes through a border between two different transparent substances, it deviates from the original direction (unless it hits the border perpendicularly to its surface).
The angle of refraction (an angle between an outgoing ray of light and a normal to a border surface) is related to an incident angle (an angle between an incoming ray of light and a normal to a border surface), according to the Law of Refraction, as:
sin(θ1)/sin(θ2) = V1/V2
where
θ1 is an incident angle
θ2 is a refraction angle
V1 is a speed of an incident ray of light (depends on the substance where the incoming ray propagates)
V2 is a speed of a refracted ray of light (depends on the substance where the outgoing ray propagates)

Instead of a speed of light in a particular substance, we can use the refractive index n of this substance, which is a ratio of the speed of light in vacuum c to a speed of light in this substance V:
n = c/V
Then the Law of Refraction would look like
sin(θ1)/sin(θ2) = n2 /n1
or
sin(θ1)·n1 = sin(θ2)·n2

As we see, the relationship between angles of incident and refraction depends on the speeds of light propagation in two bordering substances.

At this point it's important to notice from the equation above that, if the light moves from a substance, where it is faster, into a substance, where it is slower, that is n1 is less than n2 (like from the air into the glass) then the angle of incident must be greater than the angle of refraction.
Similarly, if the light moves from a substance, where it is slower, into a substance, where it is faster, that is n1 is greater than n2 (like from the glass into the air) then the angle of incident must be smaller than the angle of refraction.

Now we know that the visible white light is a combination of lights of different colors and wavelengths, with the longest wavelength being for red light and the shortest - for violet.

What's extremely important to understand is that the speed of propagation in vacuum for lights of all wave lengths (that is, of all colors) is the same c, but, as soon as the light ray goes inside some substance, like glass or water, or even air, the speed of lights of different wavelengths (that is, of different colors) is different. The red light with the longest wavelength in the visible spectrum propagates faster than violet (the shortest length in the visible spectrum) with all intermediary light colors propagating within the interval between these two speeds.

Since refractive index of any substance depends on the speed of light propagation inside this substance, the refractive index depends on what color of light we use to measure the angle of refraction. For practical purposes, talking about refractive index of any substance without referencing a particular color, the yellow light is used, because yellow is somewhere in the middle between the fastest in the visible spectrum red color and the slowest violet.

Taking into consideration different speeds of different light colors, using the refractive index n=c/V as a measure of speed propagation in the glass, its value for red light is, approximately, 1.520, that is the red light in the glass is in 1.520 times slower than speed of light in the vacuum. The same glass' refractive index for violet light is, approximately, 1.538.

Here is a full table of glass refractive indices n for different visible colors (that is, different wavelengths λ in nanometers (nm)):
 Color Wave λ Refract n Red 660 1.520 Orange 610 1.522 Yellow 580 1.523 Green 550 1.526 Blue 470 1.531 Violet 410 1.538

Assume, for example, that the white light comes from the air onto a glass surface at the incident angle α=30°. This means that all color components of this white light are parallel to each other and each has exactly the same incident angle α=30°.
The air refractive index nair is, approximately, 1.000 for all colors, while the refractive index of the glass nglass is more significantly different for different colors, as represented in the table above.

Let's find the refraction angles β of different colors inside the glass using the Law of Refraction
sin(α)·nair = sin(β)·nglass,
from which we can find
sin(β) = sin(α)·nair /nglass
β = arcsin[sin(α)·nair /nglass]

Since sin(α)=sin(30°)=0.5 and nair=1.000, trivial calculations result in the following refraction angles for different colors, if the incident angle is 30° for each:
 Color α n β Red 30° 1.520 19.205° Orange 30° 1.522 19.179° Yellow 30° 1.523 19.166° Green 30° 1.526 19.126° Blue 30° 1.531 19.062° Violet 30° 1.538 18.971°

As you see, the angles of refraction are different for different colors, which causes the corresponding rays of different colors to differently deviate from the original direction of the white light. This angular deviation is called dispersion.

Picture below schematically represents the dispersion of the white light into its color components on the border between the air and the glass. All these color components deviate from the original direction of the white light (dotted line on the picture) by different deviation angle Δ=α−β.

Let's consider now what happens with a ray of white light when it passes through a window glass at some incident angle α. What's important is that the light undergoes two refractions, first going from the air into the glass, then from the glass into the air.

The picture below represents this process.

The refraction angle from the passing the top border from the air to the glass β will be an incident angle for the bottom border between the glass and the air, and the angle γ will be the refraction angle for the rays coming out from the glass into the air.

According to the Law of Refraction, the angle of incident β and angle of refraction γ are related as
sin(β)·nglass = sin(γ)·nair
Comparing this formula with the one for the top surface of the glass
sin(α)·nair = sin(β)·nglass,
we conclude that α = γ

The last equality, basically, states that all outgoing rays of different colors are parallel to the original incoming white light ray, but are at some distance from it caused by refraction, and this difference is different for different colors; the closest to original direction of the incoming white light after double refraction on two glass surfaces is the red component, and the farthest is the violet one.

We do not usually see much of a dispersion on window glass because the glass is relatively thin, so the component rays of different colors are not significantly deviated from the original direction after the first refraction, and after the second refraction they can mix together forming white light again. The only visible separation of colors can be observed when a relatively narrow ray of white light falls at an angle on a thick glass, which would allow the colors to visibly separate from each other, especially at the edges of the narrow white ray of light.

If the glass thickness is h, the maximum angle of refraction for red color is βmax and the minimum angle of refraction for violet color is βmin, the distance between outgoing parallel red and violet rays will be
[tan(βmax)−tan(βmin)]·cos(α)
Simple calculations for the glass of 2.5 mm thickness show that the white light coming at the incident angle of 30° will be split into different colors in such a way that the distance between outgoing parallel red and violet rays will be only 0.01 mm, which is not really noticeable.

Sunday, June 19, 2022

Water Depth: UNIZOR.COM - Physics4Teens - Waves - Phenomena of Light

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

Water Depth Problem

If we look at some object at the bottom of a water from above, it looks closer than it really is.
Why?

Consider the process of measuring a distance to an object by looking at it. We feel that the object is closer, if we have to move the pupils of our eyes closer to each other pointing to an object using muscles that move our eyes. Another group of muscles make the lenses of our eyes more curved to focus the image on the retina at the back of the eyes, where optic nerves are ending.

The intensity with which our muscles work to focus on an object is somehow translated in the brain into a feeling of the distance the object is located at. Greater intensity of eye muscles contraction corresponds to a shorter distance to an object of interest.

Consider now what happens when we look at some object on the bottom of the water from above the water viewpoint.

Our eyes see an object, when the light is emitted by or reflected from it.

Let some object be at the bottom of the water at location A, as presented on the picture above.
Taking into consideration refraction of light from the object, when it crosses the border from water to air, we conclude that the rays of light from an object directed straight into our eyes will not reach the eyes along the yellow lines AL and AR, but will be refracted to the sides.
Instead, red rays AM and AN, correspondingly, refracted at the border between air and water, continue their path along lines ML and NR and reach our eyes.

Point B, an intersection of lines LM and RN, will be a perceived location of an object.
Obviously, the angle of vision ∠LBR is greater than ∠LAR and, therefore, it requires a greater strain on the eyes' muscles to focus on the image of an object, which entails our perception of an object to be at location B, which is closer to us than it really is at location A.

Let's do some calculations based on the laws of refraction that we know.

Assume, the depth of the water is AD=d and we are at the height CD=h above the water.
Also important is the distance between our eyes, which we will set to LR=s.

The real distance from our face to an object at the bottom of the pool is along the perpendicular AC from point A to line LR connecting the eyes and is equal to d+h.
Let's calculate the perceived distance, which is a distance from point B to line LR along the perpendicular BC.

Let's introduced the following variables of our problem:
incident angle α=MAD,
refraction angle β=MBD,
distance x=BD.
Now we will construct three equations with these unknown variables.

From triangle ΔBCL:
LC/BC = tan(LBC).
Therefore,
s/2 = (x+h)·tan(β)
This is our first equation.

From triangle ΔADM:
MD/AD = tan(MAD).
Therefore,
s/2 = h·tan(β) + d·tan(α)
This is our second equation.

The third equation is the Law of Refraction, assuming the refraction indices of air nair and water nwater are known:
nwater·sin(α) = nair·sin(β)

Subtracting the first equation from the second, we get
0 = (x+h)·tan(β) −
[h·tan(β) + d·tan(α)]
from which follows
x/d = tan(α)/tan(β)

At this time most physicists assume that the angles of incidents and refraction are small enough and just approximate the ratio of tangents with the ratio of sinuses (known from the Law of Refraction) and conclude that
x/d ≅ sin(α)/sin(β) =
= nair/nwater

Precise expression requires more involved calculations and we leave them to interested students to come up with an exact solution.

Indices of refraction for air and water are:
nair ≅ 1.0003
nwater ≅ 1.333
Therefore, the ratio above that evaluates the visible decrease in depth of water is
nair/nwater ≅ 0.75

So, the perceived depth of the water is smaller than the real one, approximately, by the factor nair/nwater ≅ 0.75.

Saturday, June 18, 2022

Angle Refraction of Light: UNIZOR.COM - Physics4Teens - Waves - Phenomen...

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

Angle Refraction of Light

Let's analyze what happens with a flat wave front of light, when its parallel rays fall at an angle onto a border between two transparent substances with incident rays coming from a substance with a smaller refraction index (and, therefore, higher speed of light in this substance, since refraction index is a ratio of the speed of light in the vacuum to a speed of light in the substance). For example, flat wave front of light from air falls on a glass surface.

As we know from the previous lectures, there is a dependency between refraction indices n1, n2 of (or speeds of light V1, V2 in) two bordering substances and angles θ1 of incidence and θ2 of refraction of a ray of light falling on the border:
sin(θ1)/V1 = sin(θ2)/V2
or sin(θ1)·n1 = sin(θ2)·n2
This was derived from the Fermat's Principle of the Least Time.

Now we will use the Huygens principle to analyze this process from the wave theory viewpoint and demonstrate the same result.
Let's recall a simple illustration to Huygens Principle

The picture above shows how a wave front propagates through space by assuming that each point of this wave front at time t is a source of oscillations propagated in all directions, reaching during the next interval of time Δt a surface of a small sphere of radius r=c·Δt around this point, where c is the speed of light. The resulting new wave front at time t+Δt will be a surface enveloping all these small spheres, that is tangent to each and every one of them.

Situation becomes more complex, when the speed of light is not the same at different points of the wave front because the wave front falls onto a different transparent substance, for example, it falls from the air onto a glass surface.

If, for example, a flat wave front falls from the air perpendicularly to a flat glass surface, the speed of light at different points of a wave front at the same time is the same, faster for wave front in the air and slower in the glass. The light will propagate in the same direction in the glass, as it was in the air.

But, if the flat wave front falls from the air at some acute incident angle onto flat glass surface or a spherical wave front falls on a flat glass surface, those point of the wave front that reached the glass earlier will emit light at a slower speed, so the propagation of the wave front will not be the same as on the picture above.

Let's examine the behavior of the flat wave front falling from the air onto a flat glass surface at an incident angle θ1 and analyze the shape and direction of propagation of the wave front at moment in time t+Δt, knowing its position and direction at time t.
The picture below represents a section of a set of synchronous parallel rays of light falling onto a flat air/glass border at an acute incident angle θ1.
(for a clearer view click the right mouse button on the picture and open it in another browser tab)

The wave front of light, consisting of parallel rays synchronously emitted by flat plane source, can be obtained by connecting points on different rays, where light comes at the same time. As the rays of this light come from some flat source and move in the uniform environment (air), the wave front will always be a plane perpendicular to rays.

There are three rays presented on the above picture out of the whole set of rays - those going through points A, C and B. We will call these rays a, c and b to correspond to points they pass.
Let's assume that at time t the wave front goes through these points A, C and B perpendicularly to the propagation in the air.

Because the incident angle of all those rays is not zero, different parallel among themselves rays will reach the border between air and glass at different time.
The first ray that touches the air/glass border at point A is ray a. Representing as ta the touch time for this ray, we can say
ta = t.

Next is an intermediate ray c that at time t goes through point C and later on at time tc touches the border at point C'.

Finally, ray b at time t goes through point B and later on (later than ray c) at time tb, which is greater than tc, touches the border at point B'.

Let's examine the wave front at time t+Δt, where Δt is the time difference between moments tb (the last ray to touch the glass surface) and ta (the first ray touching the glass surface).

All this period of time Δt=tb−ta ray a moved inside the glass with lower speed V2.
We don't know its direction, but can build a sphere around last known location (point A) at time ta of radius
ra=AA'=V2·(tb−ta)
centered at A.
The new wave front at time t+Δt will be tangent to this sphere.

Ray c moved in the air to point C' with higher speed V1 during time from ta to tc, which is a part of the Δt=tb−ta period, and the rest of the time from tc to tb moves inside the glass with lower speed V2.
So, we can build a sphere of radius
rc=C'C"=V2·(tb−tc)
centered at C', which the new wave front at time t+Δt should be a tangent to.

Ray b during the entire period of time Δt=tb−ta moved in the air with higher speed V1 along a known trajectory from point B to point B'.
The new wave front at time t+Δt should go through its end point B'.

First, let's find the interval of time Δt=tb−ta for ray b to travel from point B to touch the border at point B' or for ray a to travel from point A on the border into the glass to point A' (which we don't know) or for ray c to travel from point C in the air to, first, point C' on the border and then to C" in the glass (which we don't know).

Let the distance between the earliest to touch the border ray a and the latest b (that is, the length of AB) be d. Then the length BB' will be d·tan(θ1).
From this, taking into account the speed of light in the air V1,
Δt = tb−ta = d·tan(θ1)/V1

During the time from ta, when ray a crossed the border, to time tb, when ray b touched the border, ray a moves inside the glass to point A' with slower speed V2, while ray B moves in the air from point B to B' with higher speed V1.

By the time ray B reaches the border surface at point B' ray A will move inside the glass by a distance AA' that is shorter than BB' because it's speed in the glass is slower than that of ray B in the air.
Therefore,
ra = AA' = V2·(tb − ta) =
= V2·d·tan(θ1)/V1

During the same time ray c will partially travel through air with speed V1 and partially through glass with slower speed V2, which will bring it to a distance C'C" from the border inside the glass.

Assuming the distance between parallel rays b and c is x, the ray c will travel the time
τc1=(d−x)·tan(θ1)/V1
in the air.
The remaining time τc2=Δt−τc1 it will travel through glass with a lower speed V2, which will bring it on the distance
rc = C'C"=V2·τc2
from point on the border C'.
That gives
rc = V2·τc2 = V2·(Δt−τc1) =
= V2·x·tan(θ1)/V1

Since initially rays a, c and b were parallel to each other and they sustain the same refraction on the border between air and glass, the refracted rays will be parallel as well.

Consider now positions of our three rays at time t+Δt.
Ray a will go inside the glass from point A by a distance
ra = AA' = V2·d·tan(θ1)/V1
Ray c will go inside the glass from point C' by a distance
rc = C'C" = V2·x·tan(θ1)/V1
Ray b will be on the border at point B', that is will go inside the glass by a distance zero.

As we see, the distance inside the glass is changing from its maximum for ray a to zero at ray b. Considering variable x as changing from zero (when ray c coincides with ray b) to d (when ray c coincides with ray a), we see that the distance of penetration inside the glass, as a function of x, is linear.

Let's apply the Huygens Principle.
What follows from this is that, if we will make spheres around points of touching the glass for each ray, the radii of these spheres will linearly change from maximum for ray a to zero for ray b. Therefore, a surface that envelopes all these spheres will be a flat plane, which in a section presented on the picture above will be represented by a line A'B'.

From the above follows:
sin(θ2) = AA'/AB' =
= AA'·cos(θ1)/d =
= V2·d·tan(θ1)·cos(θ1)/(d·V1) =
= V2·sin(θ1)/V1

Therefore,
V1/sin(θ1) = V2/sin(θ2)
or
sin(θ1)/sin(θ2) = V1/V2

If we use refraction indices
ni = c/Vi, where c is the speed of light in the vacuum, the above formula is equivalent to
n1·sin(θ1) = n2·sin(θ2)
or
sin(θ1)/sin(θ2) = n2/n1

As you see, these formulae are identical to those derived before using Fermat's Principle of the Least Time.

Sunday, June 12, 2022

Huygens Principle: UNIZOR.COM - Physics4Teens - Waves - Phenomena of Light

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

Huygens Principle

The history of science knows two different approaches to understanding of what light is: corpuscular and wave theories. While the former was pioneered by such an authority as Isaac Newton and explained quite satisfactorily such properties of light as reflection, it was unable to properly explain some other observable properties, like changing the direction of light rays during refraction on a surface of glass or water.

The 17th century Dutch mathematician and physicist Cristiaan Huygens has suggested an interesting theory of light propagation that seems to explain properties of light from the wave theory viewpoint. This theory is called Huygens Principle or Huygens-Fresnel Principle, adding the name of French physicist Augustin Fresnel, who contributed a lot to develop this theory.

The first point of Huygens was that light is longitudinal mechanical oscillations of some substance called ether (or aether) that exists everywhere and penetrates everything, including vacuum in the outer space. These oscillations are analogous to sound waves in the air - periodic change of density of the air.
Based on this idea, Huygens introduced a principle that states: when any wave front (light waves, sound waves, water waves etc.) reaches any point in whatever environment it propagates, it becomes a source of new waves propagating from this point in all directions.

To better understand the logic behind this principle, imagine two rooms with a door in between them. The music is played in one room, while you are sitting in another. You hear the music, as it comes from the door, not from the original player in another room. The reason is that the sound propagates in all direction from the original source, reaches the door and propagates from the door in all directions, as if it is produced inside the door. All other sounds from all other directions you cannot hear, they are absorbed by the walls. So, the door plays a role of a secondary source of sound.

We will use the term "wave front" to characterize a set of points reached by light emitted by some source at the same time. If the source of light is a point and light propagates in the uniform environment, the wave front is spherical; if the source of light is a straight line, the wave front in the uniform environment is cylindrical; if the source of light is a flat plane, the wave front in the uniform environment is a flat plane.
Below picture illustrates the Huygens Principle for two kinds of waves - flat plane wave front and spherical wave front represented in two-dimensional section.

Let's assume that at time t the wave front is a certain surface in space (sphere, cylinder, plane or any complex surface).
The Huygens Principle states that at any time every point on this wave front emits secondary spherical waves of light around it. During the next small time period Δt the light emitted by each point will reach a surface of a tiny sphere of radius r=c·Δt around this point, where c is a speed of light. Then the wave front at time t+Δt is a surface enveloping all these tiny spheres.
That means that the wave front at time t+Δt is tangent to all tiny spheres of radius r=c·Δt around each point on the wave front at time t.

In the uniform environment all tiny spheres around all points on the wave front at time t will have the same radius and the wave front at time t+Δt will resemble the wave front at time t, as seen on the picture above.

In the non-uniform environment, when the speed of light is not the same for all points on the wave front, the picture is different. The tiny spheres around different points on the wave front will have different radius, and the new wave front will have a different shape, like on the picture below.

When the radius r=c·Δt of a secondary spherical wave of light is different for different points of a wave front at time t because the speed of light c is different, the wave front at time t+Δt might not resemble original wave front.
It usually happens, when the light goes from one transparent substance into another, like from air to glass.

The retaining of the shape of the wave front in case of a uniform environment might need some explanation.
Let's consider the propagation of light in details for a flat wave front, when the speed of light is the same at all points on the wave front.
Consider two neighboring points on the wave front (green) and light rays emitted by them (blue and red).

The rays of light perpendicular to a wave front (thick blue rays on the picture) go always in the direction of the wave front propagation.
The symmetrical red rays emitted by two neighboring points on the wave front at an angle to the wave front propagation towards each other superimpose each on the other and, as a result, direct the combined rays also in the direction of the wave front propagation.
For the spherical or cylindrical shape of the wave front the considerations are similar.
That's why, according to Huygens Principle, the light emitted by any source retains the shape of the wave front in the uniform environment.

We should note that an explanation above is very primitive and requires some sophisticated mathematics to support. Many scientists contributed to this process, and the Huygens Principle was tested both theoretically and experimentally. Not always it was firmly confirmed, but for our purposes it's a good tool to explain certain light phenomena, which will be addressed in the following lectures.

As an example of one particular wave propagation phenomena, not properly explainable by a corpuscular theory and where the Huygens Principle shines, consider a ray of parallel light rays, emitted by a flat plane source, going through a small opening and falling on the screen.
According to a corpuscular theory, the screen should show exactly the same light spot as the opening is. Experiments show, however, that the light spot on the screen is wider than the opening with intensity of the light spot on a screen rapidly diminishing, as we go further from the center of a spot.

Picture below illustrates this phenomena from the position of Huygens Principle.

Since every point of a flat wave front that has reached the opening is a source of secondary waves emitted in all directions, the light after the opening is widening with the brightest part along the previous direction of the light rays, but surrounding space also obtaining its share of light, diminishing as the distance from the original direction increases.

Actually, the above picture illustrates even more than just gradual dissipation of light. It illustrates the phenomena of diffraction, which will be considered later.

Finally, the presentation of the Huygens Principle would not be complete without mentioning its drawbacks.

1. Ether does not exist, as experiments show. It does not mean that light is not oscillations, it's just not mechanical oscillation of some substance, but oscillations of electro-magnetic field.

2. Propagation of light along a straight line is not properly explained in the framework of Huygens Principle; the above explanation of combination of neighboring symmetrical rays is far from rigorous.

3. Huygens Principle cannot explain the polarization of light.

4. Photoelectric effect cannot be explained within the framework of Huygens Principle.

5. Backward propagation of light seems to follow from the Huygens Principle, which is not the way light goes.

Regardless of these drawbacks, Huygens has introduced an extremely important idea that light is oscillations. This led to a substantial flow of discoveries and theories.

Monday, June 6, 2022

A few exams were added to Geometry part of the "Math 4 Teens" course on UNIZOR.COM. I strongly recommend not to bypass exams, if they are available. Good luck!