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 R_{dark_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) = R_{dark_1}/L,
R_{dark_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 2λ.

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 R_{dark_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) = R_{dark_2}/L,
R_{dark_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.

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 2π).
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 2π 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: r_M = √x²+(y−s)²
r_N = √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:
z_M(x,y,t) = A·cos(ω·t +
+ φ_M + 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:
z_N(x,y,t) = A·cos(ω·t +
+ φ_N + 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) = z_M(x,y,t) + z_N(x,y,t) =
= A·cos(ω·t+φ_M+2π·r_M/λ) +
+ A·cos(ω·t+φ_N+2π·r_N/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:
ω_M = ω_N = ω,
are at their top amplitude:
φ_M = φ_N = 0,
the same amplitudes:
A_M = A_N = 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 S_M and S_N 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.
S_M = √L²+(x+(d/2))²
S_N = √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.

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)·n_air = sin(β)·n_water
and
sin(β')·n_water = sin(γ)·n_air

Incidentally, since OP=OQ=OR, ∠β = ∠β'.
Therefore,
sin(α)·n_air = sin(γ)·n_air
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.

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 β_r separately from the slowest color, violet, and its value of β_v, 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 (γ_r) or violet (γ_v) rays after they pass the border from the glass into the air.

The Law of Refraction allows to calculate the refraction angles β_r and β_v, based on the incident angle α and refractive indices n_r and n_v of glass for each color, as described in the previous lecture:
sin(β_i) = sin(α)·n_i /n_air
where we can safely assume that n_air=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)·n_i = sin(γ_i)·n_air
sin(γ_i) = sin(β'_i)·n_i /n_air
where we can safely assume that n_air=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.

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 n_air is, approximately, 1.000 for all colors, while the refractive index of the glass n_glass 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)·n_air = sin(β)·n_glass,
from which we can find
sin(β) = sin(α)·n_air /n_glass
β = arcsin[sin(α)·n_air /n_glass]

Since sin(α)=sin(30°)=0.5 and n_air=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)·n_glass = sin(γ)·n_air
Comparing this formula with the one for the top surface of the glass
sin(α)·n_air = sin(β)·n_glass,
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 β_max and the minimum angle of refraction for violet color is β_min, the distance between outgoing parallel red and violet rays will be
h·[tan(β_max)−tan(β_min)]·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.

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 n_air and water n_water are known:
n_water·sin(α) = n_air·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) =
= n_air/n_water

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:
n_air ≅ 1.0003
n_water ≅ 1.333
Therefore, the ratio above that evaluates the visible decrease in depth of water is
n_air/n_water ≅ 0.75

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

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 t_a the touch time for this ray, we can say
t_a = t.

Next is an intermediate ray c that at time t goes through point C and later on at time t_c 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 t_b, which is greater than t_c, touches the border at point B'.

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

All this period of time Δt=t_b−t_a 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 t_a of radius
r_a=AA'=V₂·(t_b−t_a)
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 t_a to t_c, which is a part of the Δt=t_b−t_a period, and the rest of the time from t_c to t_b moves inside the glass with lower speed V₂.
So, we can build a sphere of radius
r_c=C'C"=V₂·(t_b−t_c)
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=t_b−t_a 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=t_b−t_a 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 = t_b−t_a = d·tan(θ₁)/V₁

During the time from t_a, when ray a crossed the border, to time t_b, 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,
r_a = AA' = V₂·(t_b − t_a) =
= 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
τ_c1=(d−x)·tan(θ₁)/V₁
in the air.
The remaining time τ_c2=Δt−τ_c1 it will travel through glass with a lower speed V₂, which will bring it on the distance
r_c = C'C"=V₂·τ_c2
from point on the border C'.
That gives
r_c = V₂·τ_c2 = V₂·(Δt−τ_c1) =
= 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
r_a = AA' = V₂·d·tan(θ₁)/V₁
Ray c will go inside the glass from point C' by a distance
r_c = 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
n_i = c/V_i, 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.