The purpose of the lesson
To acquaint students with the laws of light propagation at the interface between two media, to give an explanation of this phenomenon from the point of view of the wave theory of light.
|
Homework: § 61, task No. 1035, 1036.
Check of knowledge
Test. reflection of light
new material
Observation of the refraction of light.
At the boundary of two media, light changes the direction of its propagation. Part of the light energy returns to the first medium, that is, light is reflected. If the second medium is transparent, then the light can partially pass through the boundary of the media, also changing, as a rule, the direction of propagation. This phenomenon is called refraction of light.
Due to refraction, an apparent change in the shape of objects, their location and size is observed. We can be convinced of this by simple observations. Let's put a coin or other small object on the bottom of an empty opaque glass. Let's move the glass so that the center of the coin, the edge of the glass and the eye are on the same straight line. Without changing the position of the head, we will pour water into the glass. As the water level rises, the bottom of the glass with the coin rises, as it were. The coin, which was previously only partially visible, will now be fully visible. Set the pencil obliquely in a vessel of water. If you look at the vessel from the side, you can see that the part of the pencil that is in the water seems to be shifted to the side.
These phenomena are explained by a change in the direction of the rays at the boundary of two media - the refraction of light.
The law of refraction of light determines the relative position of the incident beam AB (see figure), refracted by DB and the perpendicular CE to the media interface, restored at the point of incidence. The angle α is called the angle of incidence, and the angle β is angle of refraction.
Incident, reflected and refracted rays are easy to observe by making a narrow beam of light visible. The course of such a beam in air can be traced by blowing a little smoke into the air, or by placing a screen at a slight angle to the beam. The refracted beam is also visible in the fluorescein-stained aquarium water.
Let a plane light wave fall on a flat interface between two media (for example, from air into water) (see Fig.). The wave surface AC is perpendicular to the rays A 1 A and B 1 B . The surface MN will first reach the ray A 1 A . Beam B 1 B will reach the surface after a time Δt . Therefore, at the moment when the secondary wave at point B only begins to be excited, the wave from point A already has the form of a hemisphere with a radius
The wave surface of a refracted wave can be obtained by drawing a surface tangent to all secondary waves in the second medium, the centers of which lie on the interface between the media. In this case, this is the BD plane. It is the envelope of the secondary waves. The angle of incidence α of the beam is equal to CAB in the triangle ABC (the sides of one of these angles are perpendicular to the sides of the other). Consequently,
The angle of refraction β is equal to the angle ABD of the triangle ABD. That's why
Dividing the resulting equations term by term, we obtain:
|
|
where n is a constant value independent of the angle of incidence.
From the construction (see Fig.) it is clear that the incident ray, the refracted ray, and the perpendicular erected at the point of incidence lie in the same plane. This statement, together with the equation according to which the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant value for two media, represents law of refraction of light.
You can verify the validity of the law of refraction experimentally by measuring the angles of incidence and refraction and calculating the ratio of their sines at different angles of incidence. This relationship remains unchanged.
refractive index.
The constant included in the law of refraction of light is called relative refractive index or refractive index of the second medium relative to the first.
Huygens' principle not only implies the law of refraction. With the help of this principle, the physical meaning of the refractive index is revealed. It is equal to the ratio of the speeds of light in the media, at the boundary between which refraction occurs:
|
|
If the angle of refraction β is less than the angle of incidence α , then, according to (*), the speed of light in the second medium is less than in the first.
The refractive index of a medium with respect to vacuum is called the absolute refractive index of this medium. It is equal to the ratio of the sine of the angle of incidence to the sine of the angle of refraction during the transition of a light beam from vacuum to a given medium.
Using formula (**), one can express the relative refractive index in terms of the absolute refractive indices n 1 and n 2 of the first and second media.
Indeed, since
| and |
where c is the speed of light in vacuum, then
A medium with a lower absolute refractive index is called optically less dense medium.
The absolute refractive index is determined by the speed of propagation of light in a given medium, which depends on the physical state of the medium, that is, on the temperature of the substance, its density, and the presence of elastic stresses in it. The refractive index also depends on the characteristics of the light itself. As a rule, for red light it is less than for green, and for green it is less than for violet.
Therefore, in tables of refractive indices for different substances, it is usually indicated for which light a given value of n is given and in what state the medium is. If there are no such indications, then this means that the dependence on these factors can be neglected.
In most cases, it is necessary to consider the transition of light through the air-solid or air-liquid interface, and not through the vacuum-medium interface. However, the absolute refractive index n 2 of a solid or liquid substance differs slightly from the refractive index of the same substance relative to air. Thus, the absolute refractive index of air under normal conditions for yellow light is approximately 1.000292. Consequently,
Worksheet for the lesson
Sample Answers
"Light refraction"
At the interface between two transparent media, along with the reflection of light, its refraction is observed, passing into another medium, it changes the direction of its propagation.
The refraction of a light beam occurs when it falls obliquely on the interface (although do not always read further about total internal reflection). If the beam falls perpendicular to the surface, then there will be no refraction in the second medium, the beam will retain its direction and also go perpendicular to the surface.
4.3.1 Law of refraction (special case)
We will start with the particular case where one of the media is air. This situation is present in the vast majority of tasks. We will discuss the corresponding particular case of the law of refraction, and then we will give its most general formulation.
Suppose that a ray of light traveling through air falls obliquely on the surface of glass, water, or some other transparent medium. When passing into the medium, the beam is refracted, and its further course is shown in Fig. 4.11.
Wednesday O
Rice. 4.11. Refraction of a beam at the boundary ¾air-medium¿
At the point of incidence O, a perpendicular (or, as they say, normal) CD to the surface of the medium is drawn. The ray AO, as before, is called the incident ray, and the angle between the incident ray and the normal is the angle of incidence. Beam OB is a refracted beam; the angle between the refracted ray and the normal to the surface is called the angle of refraction.
Any transparent medium is characterized by the value n, which is called the refractive index of this medium. The refractive indices of various media can be found in the tables. For example, for glass n = 1;6, and for water n = 1;33. In general, any environment has n > 1; the refractive index is equal to unity only in vacuum. Air has n = 1; 0003, so for air it can be assumed with sufficient accuracy in problems n = 1 (in optics, air does not differ much from vacuum).
Law of refraction (transition ¾air-medium¿).
1) The incident ray, the refracted ray, and the normal to the surface drawn at the point of incidence lie in the same plane.
2) The ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the refractive index
environment:
Since n > 1, it follows from relation (4.1) that sin > sin, i.e. > the angle of refraction is less than the angle of incidence. Remember: passing from air to the medium, the beam after refraction goes closer to the normal.
The refractive index is directly related to the speed v of light propagation in a given medium. This speed is always less than the speed of light in vacuum: v< c. И вот оказывается,
Why this happens, we will understand when studying wave optics. In the meantime, combi-
Let us solve formulas (4.1 ) and (4.2 ): | |||||
Since the refractive index of air is very close to unity, we can assume that the speed of light in air is approximately equal to the speed of light in vacuum c. Taking this into account and looking at the formula (4.3), we conclude: the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the speed of light in air to the speed of light in a medium.
4.3.2 Reversibility of light rays
Now consider the reverse course of the beam: its refraction during the transition from the medium to the air. The following useful principle will help us here.
The principle of reversibility of light rays. The trajectory of the beam does not depend on whether the beam propagates in the forward or backward direction. Moving in the opposite direction, the beam will follow exactly the same path as in the forward direction.
According to the principle of reversibility, when passing from the medium to the air, the beam will follow the same trajectory as during the corresponding transition from air to the medium (Fig. 4.12) The only difference between Fig. 4.12 and Fig. 4.11 is that the direction of the beam has changed to opposite.
Wednesday O
Rice. 4.12. Ray refraction at the boundary ¾medium-air¿
Since the geometric picture has not changed, formula (4.1) will remain the same: the ratio of the sine of the angle to the sine of the angle is still equal to the refractive index of the medium. True, now the angles have changed roles: the angle has become the angle of incidence, and the angle has become the angle of refraction.
In any case, no matter how the beam goes from the air to the medium or from the medium to the air, the following simple rule works. We take two angles, the angle of incidence and the angle of refraction; the ratio of the sine of the larger angle to the sine of the smaller angle is equal to the refractive index of the medium.
We are now fully prepared to discuss the law of refraction in the very general case.
4.3.3 Law of refraction (general case)
Let light pass from medium 1 with refractive index n1 to medium 2 with refractive index n2. A medium with a higher refractive index is said to be optically denser; accordingly, a medium with a lower refractive index is said to be optically less dense.
Passing from an optically less dense medium to an optically denser one, the light beam after refraction goes closer to the normal (Fig. 4.13). In this case, the angle of incidence is greater than the angle of refraction: > .
Rice. 4.13. n1< n2 ) >
On the contrary, passing from an optically denser medium to an optically less dense one, the beam deviates further from the normal (Fig. 4.14). Here the angle of incidence is less than the angle of refraction:
Rice. 4.14. n1 > n2 )<
It turns out that both of these cases are covered by one formula by the general law of refraction, which is valid for any two transparent media.
The law of refraction.
1) Incident ray, refracted ray and normal to the media interface, drawn
in point of incidence lie in the same plane.
2) The ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive index of the second medium to the refractive index of the first medium:
It is easy to see that the previously formulated law of refraction for the transition ¾air-medium¿ is a special case of this law. Indeed, assuming in formula (4.4) n1 = 1 and n2 = n, we arrive at formula (4.1).
Recall now that the refractive index is the ratio of the speed of light in vacuum to the speed of light in a given medium: n1 = c=v1 , n2 = c=v2 . Substituting this into (4.4), we get:
Formula (4.5 ) naturally generalizes formula (4.3 ). The ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the speed of light in the first medium to the speed of light in the second medium.
4.3.4 Total internal reflection
When light rays pass from an optically denser medium to an optically less dense one, an interesting phenomenon is observed - total internal reflection. Let's see what it is.
Let us assume for definiteness that light goes from water to air. Let us assume that there is a point source of light S in the depth of the reservoir, emitting rays in all directions. We will look at some of these rays (Fig. 4.15).
S B 1
Rice. 4.15. Total internal reflection
Beam SO1 falls on the surface of the water at the smallest angle. This beam is partly refracted (beam O1 A1 ) and partly reflected back into the water (beam O1 B1 ). Thus, part of the energy of the incident beam is transferred to the refracted beam, and the rest of the energy is transferred to the reflected beam.
The angle of incidence of the SO2 beam is larger. This beam is also divided into two beams refracted and reflected. But the energy of the original beam is distributed between them in a different way: the refracted beam O2 A2 will be dimmer than the beam O1 A1 (that is, it will receive a smaller share of energy), and the reflected beam O2 B2 will be correspondingly brighter than the beam O1 B1 (it will receive a larger share of energy). energy).
As the angle of incidence increases, the same regularity can be traced: an increasing share of the energy of the incident beam goes to the reflected beam, and an ever smaller share to the refracted beam. The refracted beam becomes dimmer and dimmer, and at some point it disappears completely!
This disappearance occurs when the angle of incidence reaches 0 , which corresponds to a refraction angle of 90 . In this situation, the refracted ray OA would have to go parallel to the water surface, but there is nothing left to go. All the energy of the incident ray SO went entirely to the reflected ray OB.
With a further increase in the angle of incidence, the refracted beam will even be absent.
The described phenomenon is the total internal reflection. Water does not emit outward rays with angles of incidence equal to or greater than some value of 0, all such rays are entirely reflected back into the water. Angle 0 is called the limiting angle of total reflection.
The value 0 is easy to find from the law of refraction. We have:
sin 0 | ||||||||||
But sin 90 = 1, so | ||||||||||
sin 0 | ||||||||||
0 = arcsin | ||||||||||
So, for water, the limiting angle of total reflection is equal to:
0 = arcsin1; 1 33 48;8:
You can easily observe the phenomenon of total internal reflection at home. Pour water into a glass, raise it and look at the surface of the water slightly from below through the wall of the glass. You will see a silvery sheen on the surface due to total internal reflection, it behaves like a mirror.
The most important technical application of total internal reflection is fiber optics. Light rays launched into the fiber optic cable (optical fiber) almost parallel to its axis, fall on the surface at large angles and completely, without loss of energy, are reflected back into the cable. Repeatedly reflected, the rays go farther and farther, transferring energy over a considerable distance. Fiber-optic communication is used, for example, in cable television networks and high-speed Internet access.
Without a doubt, you know how the heroes of Jules Verne's novel "The Mysterious Island", abandoned on uninhabited land, got fire without matches and flint. Lightning came to the aid of Robinson, lighting a tree, but the new Robinsons of Jules Verne were helped not by chance, but by the resourcefulness of a knowledgeable engineer and his solid knowledge of the laws of physics.Remember how surprised the naive sailor Pencroft was when, returning from a hunt, he found an engineer and a journalist in front of a blazing fire.
“But who lit the fire?” asked the sailor.
“The sun,” Spilett replied.
The journalist was not joking. Indeed, the Sun delivered the fire that the sailor admired so much. He could not believe his eyes and was so amazed that he could not even question the engineer.
“So you had a burning glass?” Herbert asked the engineer.
No, but I made it.
And he showed it. They were simply two glasses taken by the engineer from his watch and Spilett's. He connected their edges with clay, having previously filled it with water, and thus a real incendiary lentil was obtained, with the help of which, by concentrating the sun's rays on dry moss, the engineer made fire.
The reader will wish, I think, to know why it is necessary to fill the space between the watch glasses with water: does not a biconvex lentil filled with air concentrate rays?
Exactly no. The watch glass is limited by two parallel (concentric) surfaces - outer and inner; and it is known from physics that, passing through a medium bounded by such surfaces, the rays almost do not change their direction. Passing then through a second glass of the same type, they do not deviate here either, and therefore do not gather into focus. In order to concentrate the rays at one point, it is necessary to fill the space between the glasses with some transparent substance that would refract the rays more strongly than air. So did the engineer in Jules Verne's novel.
An ordinary carafe of water, if it is spherical, can also serve as incendiary lentils. This was already known by the ancients, who also noticed that the water itself remains cold. It even happened that a decanter of water standing on an open window lit the curtains, the tablecloth, and charred the table. Those huge spherical bottles with colored water, which, according to ancient custom, used to decorate the windows of pharmacies, could sometimes cause real disasters, causing the ignition of flammable substances located nearby.
With a small round flask filled with water, even if the flask is small, it is possible to bring water poured onto a watch glass to a boil: a flask of 12 centimeters in diameter is sufficient for this. At 15 cm at the focus [the focus is placed very close to the bulb] a temperature of 120° is obtained. It is as easy to light a cigarette with a flask of water as with glass lentils, about which Lomonosov wrote in his poem “On the Benefits of Glass”:
We get a flame of sunny glass here
And we emulate Prometheus comfortably.
Cursing the meanness of these clumsy lies,
We smoke tobacco with heavenly fire without sin.
It should be noted, however, that the incendiary effect of water lenses is much weaker than that of glass lenses. This is due, firstly, to the fact that the refraction of light in water is much less than in glass, and secondly, water strongly absorbs infrared rays, which play an important role in heating bodies.
It is curious that the incendiary effect of glass lentils was known to the ancient Greeks, more than a millennium before the invention of glasses and spotting scopes. Aristophanes mentions him in the famous comedy "Clouds". Socrates offers Streptias a task:
“If someone wrote an obligation on you in five talents, how would you destroy it?
Streptiad. I have found how to destroy an obligation, and in such a way that you yourself recognize it as cunning! Have you seen, of course, in pharmacies a beautiful, transparent stone that is lit?
Socrates. Fire glass?
Streptiad. Exactly.
Socrates. What's next?
Streptiad. While the notary is writing, I, standing behind him, will direct the rays of the Sun to the obligation, and the words will melt everything ... ”
Let me remind you for clarification that the Greeks of the time of Aristophanes wrote on waxed tablets, which easily melted from heat.
How to make fire with ice?
Ice can also serve as a material for a biconvex lens, and hence for making fire, if it is sufficiently transparent. At the same time, the ice, refracting the rays, does not heat up and does not melt. The index of refraction of ice is only slightly less than that of water, and if, as we have seen, it is possible to make fire with a ball of water, it is possible to do so with lentils of ice.
Ice lentils did a good job in Jules Verne's Voyage of Captain Hatteras. Dr. Clouboni lit the fire in this way when the travelers lost their flint and found themselves without fire, in a terrible frost of 48 degrees.
"It's a disaster," Hatteras told the doctor.
“Yes,” he replied.
“We don't even have a spyglass with which we could take lentils and make fire.
“I know,” replied the doctor, “and it’s a pity that I don’t: the sun’s rays are strong enough to light tinder.”
- What to do, you have to satisfy your hunger with raw bear meat, - said Hatteras.
“Yes,” said the doctor thoughtfully, “at the very least. But why don't we...
– What did you think? asked Hatteras.
“I came up with an idea…
- Thought? exclaimed the boatswain. - If you have a thought, then we are saved!
“I don’t know how it will be possible,” the doctor hesitated.
– What did you come up with? Hatteras asked.
We don't have lentils, but we'll make them.
- How? the boatswain asked.
- We grind from a piece of ice.
- Do you think...
- Why not? After all, it is only necessary that the rays of the Sun be brought to one point, and for this purpose ice can replace the best crystal for us. Only I would prefer a piece of freshwater ice: it is stronger and more transparent.
“Here, if I’m not mistaken, this block of ice,” the boatswain pointed out at an ice floe about a hundred paces from the travelers, “judging by its color, there is just what you need.”
- You're right. Take your axe. Let's go my friends.
All three went to the indicated ice block. Indeed, the ice turned out to be freshwater.
The doctor ordered that a piece of ice, a foot in diameter, be cut off, and he began to trim it with an axe. Then he trimmed it with a knife, and finally gradually polished it with his hand. It turned out transparent lentils, as if from the best crystal. The sun was quite bright. The doctor exposed the lentils to his beams and focused them on the tinder. A few seconds later, the tinder caught fire.”
Figure 113. "The doctor concentrated the rays of the Sun on tinder."
Jules Verne's story is not entirely fantastic: experiments on lighting a tree with ice-cold lentils, first successfully carried out in England with very large lentils as early as 1763, have since been repeatedly carried out with complete success. Of course it's hard to maketransparentice lentils using tools such as an ax, a knife and “just a hand” (in 48-degree frost!), but you can make ice lentils easier: pour water into a cup of the proper shape and freeze, and then, after slightly warming the cup, remove it from her cooked lentils.
Rice. 114. Cup for making ice lentils.
In making such an experiment, do not forget that it is possible only on a clear frosty day and in the open air, but not in a room behind a window glass: the glass absorbs a significant part of the energy of the sun's rays and not enough remaining to cause significant heating.
With the help of the sun's rays
Do another experiment, also easy to do in winter. Lay on the snow, flooded with sunlight, two pieces of cloth of the same size, light and black. After an hour or two, you will see that the black patch has sunk into the snow, while the light one has remained at the same level. It is not difficult to find the reasons for such a difference: under a black patch, the snow melts more strongly, since the dark fabric absorbs most of the sun's rays falling on it; light, on the contrary, scatters most of them and therefore heats up less than black.
This instructive experiment was first performed by the famous fighter for the independence of the United States, Benjamin Franklin, who immortalized himself, as a physicist, by inventing a lightning rod. “I took from the tailor several square pieces of cloth of various colors,” he wrote. “In between were: black, dark blue, light blue, green, purple, red, white, and various other colors and shades. One bright sunny morning I put all these pieces on the snow.After a few hours, the black piece, which had heated up more than the others, sunk so deep that the rays of the Sun could no longer reach it; the dark blue sunk almost as much as the black; the light blue much less; the other colors sank the less, the lighter they were, while White remained on the surface, i.e., did not sink at all.”
“What good would a theory be if no use could be made of it?” he exclaims on this occasion and continues: “Can’t we deduce from this experience that a black dress in a warm sunny climate is less suitable than white because in the sun it heats up our body more, and if we still make movements that warm us in themselves, then excessive heat is formed? which causes sunstroke in some?... Further, can not the blackened walls absorb so much solar heat during the day as to remain warm to some extent at night and protect the fruit from frost? importance?
What these conclusions and useful applications can be is shown by the example of the German south polar expedition of 1903 on the ship Gauss. The ship was frozen in ice, and all the usual methods of release did not lead to any results. , removed only a few hundred cubic meters of ice and did not free the ship. Then they turned to the help of the sun's rays: they made a strip of dark ash and coal on the ice 2 km long and ten meters wide; it led from the ship to the nearest wide gap in the ice. the long, clear days of a polar summer, and the sun's rays did what dynamite and a saw could not do: the ice melted and broke along the piled strip, and the ship was freed from ice.
Old and new about mirages
Probably everyone knows what the physical cause of an ordinary mirage is. The hot sand of the desert acquires mirror properties because the heated layer of air adjacent to it has a lower density than the overlying layers. An inclined beam of light from a very distant object, having reached this air layer, bends its path in it so that in its further movement it again moves away from the ground and hits the observer's eye, as if reflected from a mirror at a very large angle of incidence. And it seems to the observer that a water surface spreads out in front of him in the desert, reflecting coastal objects (Fig. 115).
Rice. 115. How a mirage arises in the desert. This drawing, commonly reproduced in textbooks, represents the path of the light beam slanting towards the ground in an exaggeratedly steep manner.
It would be more correct, however, to say that the heated layer of air near the hot soil reflects the rays not like a mirror, but like a water surface viewed from the depths of the water. What is happening here is not a simple reflection, but what in the language of physics is called "internal reflection". For this, it is necessary that the light beam enter the air layers very gently - more gently than shown in our simplified Fig. 115; otherwise it will not be surpassed " limiting angle” of incidence of the beam, and without this, internal reflection is not obtained.
We note in passing one point of this theory, which can give rise to a misunderstanding. The above explanation requires such an arrangement of the air layers, in which the denser layers would be higher than the less dense ones. We know, however, that dense, heavy air tends to sink and force the underlying light layer of gas upwards. How can there be that arrangement of layers of dense and rarefied air, which is necessary for the appearance of a mirage?
Rice. 116. Mirage on the tarmac.
The answer lies in the fact that the required arrangement of air layers is not in still air, but in air in motion. The layer of air heated by the soil does not rest on it, but is continuously forced upward and is immediately replaced by a new layer of heated air. Continuous change determines that a certain layer of rarefied air always adjoins the hot sand, even if not the same, but this is already indifferent to the course of the rays.
The kind of mirage that we are considering has been known since antiquity. In modern meteorology, it is called the "lower" mirage (as opposed to the "upper" mirage, generated by the reflection of light rays by layers of rarefied air in the upper atmosphere). Most people are convinced that this classic mirage can only be observed in the sultry air of the southern deserts and does not occur in more northern latitudes.
Meanwhile, the lower mirage often happens to be observed in our area. Such phenomena are especially frequent in the summer on asphalt and tarmac roads, which, due to their dark color, become very hot in the sun. The matte surface of the road then seems from afar as if poured with water and reflects distant objects. The path of light rays in this mirage is shown in Fig. 116. With some observation, such phenomena can be seen not as rarely as is commonly thought.
There is another kind of mirage - a mirageside, whose existence is usually not even suspected. This is a reflection from a heated sheer wall. Such a case is described by one French author. Approaching the fort of the fortress, he noticed that the even concrete wall of the fort suddenly shone like a mirror, reflecting the surrounding landscape, soil, sky. Taking a few more steps, he noticed the same change in the other wall of the fort. It seemed as if the gray uneven surface was suddenly replaced by a polished one. It was a sultry day, and the walls must have become very hot, which was the key to their specularity. On fig. 117 shows the location of the walls of the fort (F and F") and the location of the observer (A and A"). It turned out that a mirage is observed whenever the wall is heated enough by the sun's rays. We even managed to photograph this phenomenon.
On fig. 118 shows (on the left) wall F of the fort, first matte, and then shiny (on the right), like a mirror (taken from point A"). The left picture shows ordinary gray concrete, in which, of course, the figures of two to the right - the same wall for the most part has acquired mirror properties, and the nearest figure of a soldier gives its symmetrical image in it. Of course, it is not the surface of the wall itself that reflects the rays, but only the layer of heated air adjacent to it.
Rice. 117. Plan of the fort where the mirage was observed. Wall F seemed mirrored from point A, wall F" - from point A"
Rice. 118. A gray uneven wall (left) suddenly becomes as if polished, reflective (right).
On hot summer days, one should pay attention to the heated walls of large buildings and look for mirage phenomena. No doubt, with some attention, the number of observed cases of mirage should noticeably increase.
"Green Beam"
"Have you ever observed the Sun setting below the horizon of the sea? Yes, without a doubt. Have you followed it until the moment when the upper edge of the disk touches the horizon and then disappears? Probably yes. But have you noticed the phenomenon what happens at the moment when the luminous luminary throws its last ray, if the sky is cloud-free and completely transparent? a color such that no artist can get on his palette and that nature itself does not reproduce either in various shades of vegetation, or in the color of the most transparent sea.
A similar note in an English newspaper led the young heroine of Jules Verne's novel "The Green Ray" into an enthusiastic state and prompted her to undertake a series of trips with the sole purpose of seeing the green beam with her own eyes. The young Scot was not able, as the novelist tells, to observe this beautiful phenomenon of nature. But it still exists.The green ray is not a legend, although there are many legendary things associated with it.It is a phenomenon that every nature lover can admire if he searches for it with due patience.
Why does the green beam appear?
You will understand the cause of the phenomenon if you remember in what form objects appear to us when we look at them through a glass prism. Do this experiment: hold the prism near the eye horizontally with the wide side down and look through it at a piece of paper pinned to the wall. You will notice that the leaf, firstly, has risen much higher than its true position, and secondly, it has a violet-blue border at the top, and a yellow-red one at the bottom. Raising depends on the refraction of light, colored borders - ondispersionglass, i.e. glass propertiesunequallyrefract rays of differentcolors.Violet and blue rays are refracted more strongly than others, so we see a violet-blue border at the top; red ones are refracted the weakest, and therefore the lower edge of our paper sheet has a red border.
For a better understanding of what follows, it is necessary to dwell on the origin of these colored borders. The prism decomposes the white light emanating from the paper into all the colors of the spectrum, giving many color images of the paper sheet, arranged, partly superimposed on one another, in the order of refraction. From the simultaneous action of these superimposed. On top of each other of color images, the eyes get a sense of white color (the addition of spectral colors), but at the top and bottom there are rims of immiscible colors. The famous poet Goethe, who went through this experiment and did not understand its meaning, imagined that he had thus exposed the falsity of Newton's doctrine of colors, and then wrote his own "Science of colors", which is almost entirely based on false ideas. The reader, presumably, does not will repeat the delusions of the great poet and will not expect that the prism will recolor all objects for him. The Earth's atmosphere is for our eyes like a huge air prism turned base down. Looking at the Sun at the horizon, we look at it through a gas prism. The disk of the Sun receives at the top there is a border of blue and green, at the bottom - red-yellow. While the Sun is above the horizon, the light of the disk interrupts much less bright colored stripes with its brightness, and we do not notice them at all. But at the moments of sunrise and sunset, when almost all of its disk hidden under the horizon, we can see the blue border of the upper edge.It is two-colored: above is a blue stripe, below - blue, from a mixture of blue and green beams whose. When the air near the horizon is completely clear and transparent, we see a blue border - the "blue beam". But more often the blue rays are scattered by the atmosphere and only one green border remains: the phenomenon of the "green beam". Finally, in most cases, blue and green rays are also scattered by the cloudy atmosphere - then no border is noticed: the Sun sets in a crimson ball.
Pulkovo astronomer G. A. Tikhov, who devoted a special study to the “green beam”, reports some signs of the visibility of this phenomenon. “If the Sun has a red color at sunset and it is easy to look at it with a simple eye, then we can say with confidence that there will be no green beam ". The reason is clear: the red color of the solar disk indicates a strong scattering of blue and green rays by the atmosphere, i.e., the entire upper rim of the disk. “On the contrary,” the astronomer continues, “if the Sun has little changed its usual whitish-yellow color and sets very bright (that is, if the absorption of light by the atmosphere is small. –I.P.), then we can most likely expect a green beam. But here it is just important that the horizon be a sharp line, without any irregularities, nearby forest, buildings, etc. These conditions are best met at sea; that is why the green beam is so well known to sailors.”
So, in order to see the “green beam”, you need to observe the Sun at the time of sunset or sunrise with a very clear sky. In southern countries, the sky near the horizon is more transparent than ours, so the “green beam” phenomenon is observed there more often. But in our country it is not as rare as many people think, probably under the influence of Jules Verne's novel. Persistent searches for the “green beam” are sooner or later rewarded with success. It happened to capture this beautiful phenomenon even with a telescope. Two Alsatian astronomers describe such an observation as follows:
"... In the last minute before sunset, when, therefore, a noticeable part of it is still visible, the disk, which has a wavy moving, but sharply defined border, is surrounded by a green rim. Until the Sun has completely set, this rim is not visible to the naked eye. It becomes visible only at the moment of the complete disappearance of the Sun behind the horizon.If you look through a telescope with a sufficiently strong magnification (about 100 times), you can trace in detail all the phenomena: the green border becomes noticeable no later than 10 minutes before sunset; it limits the upper part disk, while from the bottom there is a red border.The width of the border, at first very small (only a few seconds of arc), increases as the Sun sets; it sometimes reaches up to half a minute of the arc.Green protrusions are often observed above the green rim, which, with the gradual disappearance of the Sun as if they slide along its edge to the highest point; sometimes they come off the rim and glow separately for several seconds, until they go out” (Fig. 119).
Rice. 119. Long-term observation of the "green beam"; the observer saw the "green beam" behind the mountain range for 5 minutes. Above on the right is the "green beam" seen through a telescope. The disk of the Sun has irregular contours. In position 1, the brilliance of the solar disk blinds the eye and prevents the green border from being seen with the naked eye. In position 2, when the disk of the Sun almost disappears, the "green beam" becomes accessible to the naked eye.
Usually the phenomenon lasts a second or two. But under exceptional circumstances, its duration is noticeably lengthened. A case was noted when a "green beam" was observed for more than 5 minutes! The sun was setting behind a distant mountain, and a fast-paced observer saw a green border of the solar disk, as if sliding along the mountainside (Fig. 119).
Very instructive cases of observing the "green beam" atsunriseSun, when the upper edge of the luminary begins to appear from under the horizon. This refutes the often expressed conjecture that the “green beam” is an optical illusion that the eye succumbs to when weary of the bright brilliance of the just setting Sun.
The sun is not the only luminary that sends out a "green ray". It happened to see this phenomenon generated by the setting Venus [On mirages and a green ray, you can learn from the excellent book by M. Minnart "Light and Color in Nature". Fizmatgiz, 1958Note. ed.].
Refraction of light is a change in the direction of the beam at the boundary of two media of different density.
Explanation: a ray of light, falling into the water, changes its direction at the border of two media (that is, on the surface of the water). The beam is literally refracted. This phenomenon is called the refraction of light. This happens because water and air have different densities. Water is denser than air, and the speed of a beam of light falling on its surface slows down. Thus, water is an optically denser medium.
The optical density of the medium is characterized by different speeds of light propagation.
refraction angle (ϒ) is the angle formed by the refracted beam and the perpendicular to the point of incidence of the beam at the interface between two media.
Explanation:
The beam fell on the surface of the water at a certain point and was refracted. Let's draw a perpendicular from this point in the same direction in which the refracted ray "left" - in our case, the perpendicular is directed towards the bottom of the reservoir. The angle formed by this perpendicular and the refracted beam is called the angle of refraction.
If light travels from an optically less dense medium to an optically denser medium, then the angle of refraction is always less than the angle of incidence.
For example, light falling into water has an angle of incidence greater than the angle of refraction. The reason is that water is a denser medium than air.
For any two media with different optical density, the following formula is true:
sin α
---
= n
sinϒ
where n is a constant value independent of the angle of incidence.
Explanation:
Let's take three rays falling into the water.
Their angles of incidence are 30°, 45° and 60°.
The angles of refraction of these rays will be respectively 23°, 33° and 42°.
If we make the ratio of the angles of incidence and the angles of refraction, we get the same number:
sin 30° sin 45° sin 60°
--- = --- = ---
≅ 1,3
sin 23° sin 33° sin 42°
Thus, if we divide the angle of incidence of the beam into the water and the angle of its refraction, we get 1.3. This is a constant ( n ), which is found using the above formula.
The incident beam, the refracted beam and the perpendicular drawn from the point of incidence of the beam lie in the same plane.
REFRACTION OF LIGHT DURING THE TRANSITION FROM WATER TO AIR
A stick dipped into water, a spoon in a glass of tea, due to the refraction of light on the surface of the water, seem to us to be refracted.
Place a coin on the bottom of an opaque vessel so that it is not visible. Now pour water into the vessel. The coin will be visible. The explanation of this phenomenon is clear from the video.
Look at the bottom of the pond and try to estimate its depth. Most of the time, it doesn't work right.
Let us trace in more detail how and how much the depth of the reservoir seems to us to be reduced if we look at it from above.
Let H (Fig. 17) be the true depth of the reservoir, at the bottom of which lies a small object, such as a pebble. The light reflected by it diverges in all directions. A certain beam of rays falls on the surface of the water at the point O from below at an angle a 1 , is refracted on the surface and enters the eye. According to the law of refraction, we can write:
but since n 2 \u003d 1, then n 1 sin a 1 \u003d sin ϒ 1.
The refracted ray enters the eye at point B. Note that not one ray enters the eye, but a beam of rays, the cross section of which is limited by the pupil of the eye.
In Figure 17, the beam is shown as thin lines. However, this beam is narrow, and we can neglect its cross section, taking it for the AOB line.
The eye projects A to point A 1, and the depth of the reservoir seems to us equal to h.
It can be seen from the figure that the apparent depth of the reservoir h depends on the true value of H and on the observation angle ϒ 1 .
Let's express this dependence mathematically.
From triangles AOC and A 1 OS we have:
Excluding OS from these equations, we get:
Given that a \u003d ϒ 1 and sin ϒ 1 \u003d n 1 sin a 1 \u003d n sin a, we get:
In this formula, the dependence of the apparent depth of the reservoir h on the true depth H and the observation angle does not appear explicitly. For a clearer representation of this dependence, let us express it graphically.
On the graph (Fig. 18), along the abscissa axis, the values of the observation angles are plotted in degrees, and along the ordinate axis, the apparent depths corresponding to them h in fractions of the actual depth H. The resulting curve shows that at small viewing angles, the apparent depth
is about ¾ of the real value and decreases as the viewing angle increases. At an observation angle a = 47°, total internal reflection occurs and the ray cannot escape from the water.
MIRAGES
In an inhomogeneous medium, light does not propagate in a straight line. If we imagine a medium in which the refractive index changes from bottom to top, and mentally divide it into thin horizontal layers,
then, considering the conditions for the refraction of light during the transition from layer to layer, we note that in such a medium the light beam should gradually change its direction (Fig. 19, 20).
Such a curvature of the light beam undergoes in the atmosphere, in which, for one reason or another, mainly due to its uneven heating, the refractive index of the air changes with height (Fig. 21).
The air is usually heated by the soil, which absorbs the energy of the sun's rays. Therefore, the air temperature decreases with height. It is also known that air density decreases with height. It has been established that with increasing height, the refractive index decreases, so the rays passing through the atmosphere are bent, bending down to the Earth (Fig. 21). This phenomenon is called normal atmospheric refraction. Due to refraction, the celestial bodies seem to us somewhat "raised" (above their true height) above the horizon.
It is calculated that atmospheric refraction "raises" objects at a height of 30° by 1"40", at a height of 15° - by 3"30", at a height of 5° - by 9"45". For bodies on the horizon, this value reaches 35 ". These figures deviate in one direction or another depending on the pressure and temperature of the atmosphere. However, for one reason or another, air masses with a temperature higher than the lower layers. They can be brought by winds from hot countries, for example, from a hot desert area. If at this time the cold, dense air of an anticyclone is in the lower layers, then the phenomenon of refraction can significantly increase and the rays of light coming out of terrestrial objects upward at a certain angle to the horizon, they can return back to the ground (Fig. 22).
However, it may happen that at the surface of the Earth, due to its strong heating, the air warms up so much that the refractive index of light near the soil becomes less than at a certain height above the soil. If at the same time there is calm weather, then this state can persist for quite a long time. Then the rays from objects falling at some rather large angle to the Earth's surface can be bent so much that, having described an arc near the Earth's surface, they will go from bottom to top (Fig. 23a). The case shown in Figure 236 is also possible.
The states described above in the atmosphere explain the occurrence of interesting phenomena - atmospheric mirages. These phenomena are usually divided into three classes. The first class includes the most common and simple in origin, the so-called lake (or lower) mirages, which cause so many hopes and disappointments among desert travelers.
The French mathematician Gaspard Monge, who participated in the Egyptian campaign of 1798, describes his impressions of this class of mirages as follows:
“When the surface of the Earth is strongly heated by the Sun and is only just beginning to cool before the onset of twilight, the familiar terrain no longer extends to the horizon, as during the day, but passes, as it seems, about one league into a continuous flood.
The villages further away look like islands in a vast lake. Under each village there is its overturned reflection, only it is not sharp, small details are not visible, like a reflection in the water swayed by the wind. If you start approaching a village that seems to be surrounded by a flood, the bank of imaginary water is moving away, the water branch that separated us from the village gradually narrows until it disappears completely, and the lake ... now begins behind this village, reflecting the villages located further” (Fig. 24).
The explanation for this phenomenon is simple. The lower layers of air, warmed up by the soil, have not had time to rise up; their refractive index is less than the upper ones. Therefore, rays of light emanating from objects (for example, from point B on a palm tree, Fig. 23a), bending in the air, enter the eye from below. The eye projects a beam to point B 1 . The same happens with rays coming from other points of the object. The object appears to the observer to be overturned.
Where is the water from? Water is a reflection of the sky.
To see a mirage, there is no need to go to Africa. It can be observed on a hot, quiet summer day and over the heated surface of an asphalt highway.
Mirages of the second class are called superior or distant vision mirages. The “unheard-of miracle” described by N.V. Gogol most of all resembles them. We give descriptions of several such mirages.
From the Cote d'Azur of France, in the early clear morning, from the waters of the Mediterranean Sea, from the horizon, a dark chain of mountains rises, in which the inhabitants recognize Corsica. The distance to Corsica is more than 200 km, so a line of sight is out of the question.
On the English coast, near Hastings, one can see the French coast. As the naturalist Niedige reports, “near Reggio in Calabria, opposite the Sicilian coast and the city of Messina, whole unfamiliar areas with grazing herds, cypress groves and castles are sometimes visible in the air. After staying in the air for a short time, the mirages disappear.
Far vision mirages appear if the upper layers of the atmosphere turn out to be especially rarefied for some reason, for example, when heated air gets there. Then the rays emanating from terrestrial objects are more strongly bent and reach the earth's surface, going at a large angle to the horizon. The observer's eye projects them in the direction in which they enter it.
Apparently, in that a large number of mirages of distant vision are observed on the coast of the Mediterranean Sea, the Sahara desert is to blame. Hot air masses rise above it, then are carried away to the north and create favorable conditions for the occurrence of mirages.
Superior mirages are also observed in northern countries when warm southerly winds blow. The upper layers of the atmosphere are heated, and the lower layers are cooled due to the presence of large masses of melting ice and snow.
Sometimes both direct and reverse images of objects are observed. Figures 25-27 show precisely such phenomena observed in the Arctic latitudes. Apparently, above the Earth there are alternating denser and more rarefied layers of air, bending the rays of light approximately as shown in Figure 26.
Mirages of the third class - ultra-long vision - are difficult to explain. Let's describe a few of them.
“Based on the testimonies of several persons who are trustworthy,” writes K. Flamarion in the book “Atmosphere”, “I can report on a mirage that was seen in the city of Verviers (Belgium) in June 1815. One morning, the inhabitants of the city saw an army in the sky, and it was so clear that they could make out the suits of artillerymen, a cannon with a broken wheel that was about to fall off ... It was the morning of the Battle of Waterloo! The distance between Waterloo and Verviers in a straight line is 105 km.
There are cases when mirages were observed at a distance of 800, 1000 or more kilometers.
Here is another amazing case. On the night of March 27, 1898, in the middle of the Pacific Ocean, the crew of the Bremen ship Matador was frightened by a vision. Around midnight, the crew spotted a ship about two miles (3.2 km) away, which was battling a severe storm.
This was all the more surprising because the surroundings were calm. The ship crossed the course of the Matador, and there were moments when it seemed that a collision of ships was inevitable ... The crew of the Matador saw how, during one strong impact of a wave on an unknown ship, the light went out in the captain's cabin, which was visible all the time in two windows . After a while, the ship disappeared, taking the wind and waves with it.
The matter was clarified later. It turned out that all this happened with another ship, which at the time of the "vision" was from the "Matador" at a distance of 1700 km.
In what ways does light travel in the atmosphere so that distinct images of objects are preserved at such great distances? There is no exact answer to this question yet. There were suggestions about the formation of giant air lenses in the atmosphere, the delay of a secondary mirage, i.e., a mirage from a mirage. It is possible that the ionosphere* plays a role here, reflecting not only radio waves, but also light waves.
Apparently, the described phenomena have the same origin as other mirages observed on the seas, called the “Flying Dutchman” or “Fata Morgana”, when sailors see ghostly ships that then disappear and inspire fear in superstitious people.
RAINBOW
Rainbow - this beautiful celestial phenomenon - has always attracted the attention of man. In the old days, when people still knew very little about the world around them, the rainbow was considered a "heavenly sign." So, the ancient Greeks thought that the rainbow is the smile of the goddess Irida.
The rainbow is observed in the direction opposite to the Sun, against the background of rain clouds or rain. A multi-colored arc is usually located at a distance of 1-2 km from the observer, sometimes it can be observed at a distance of 2-3 m against the background of water drops formed by fountains or water sprayers.
The center of the rainbow is on the continuation of the straight line connecting the Sun and the observer's eye - on the anti-solar line. The angle between the direction to the main rainbow and the antisolar line is 41-42° (Fig. 28).
At the time of sunrise, the antisolar point (point M) is on the horizon line and the rainbow looks like a semicircle. As the sun rises, the antisolar point falls below the horizon and the size of the rainbow decreases. It is only part of a circle. For an observer who is high, for example on. aircraft, the rainbow is seen as a complete circle with the observer's shadow in the center.
Often there is a secondary rainbow, concentric with the first, with an angular radius of about 52 ° and the reverse arrangement of colors.
At a Sun altitude of 41°, the main rainbow ceases to be visible and only a part of the secondary rainbow appears above the horizon, and at a Sun height of more than 52°, the secondary rainbow is not visible either. Therefore, in the middle and equatorial latitudes, this natural phenomenon is never observed during the near noon hours.
The rainbow, like the spectrum, has seven primary colors that smoothly transition into one another. The shape of the arc, the brightness of the colors, the width of the stripes depend on the size of the water droplets and their number. Large drops create a narrower rainbow, with sharply prominent colors, small drops create an arc that is blurry, faded and even white. That is why a bright narrow rainbow is visible in the summer after a thunderstorm, during which large drops fall.
For the first time the theory of the rainbow was given in 1637 by R. Descartes. He explained the rainbow as a phenomenon associated with the reflection and refraction of light in raindrops.
The formation of colors and their sequence were explained later, after unraveling the complex nature of white light and its dispersion in a medium. The diffraction theory of the rainbow was developed by Airy and Pertner.
Consider the simplest case: let a beam of parallel solar rays fall on a drop having the shape of a ball (Fig. 29). A beam incident on the surface of a drop at point A is refracted inside it according to the law of refraction: n 1 sin a \u003d n 2 sin β, where n 1 \u003d 1, n 2 ≈ 1.33 - refractive indices of air and water, respectively, a - angle incidence, β is the angle of light refraction.
Inside the drop, the beam travels in a straight line AB. At point B, the beam is partially refracted and partially reflected. Note that the smaller the angle of incidence at point B, and hence at point A, the lower the intensity of the reflected beam and the greater the intensity of the refracted beam.
The beam AB after reflection at point B passes at an angle β 1 "= β 1 hits point C, where partial reflection and partial refraction of light also occur. The refracted beam leaves the drop at an angle y2, and the reflected beam can go further, to point D and etc. Thus, a ray of light in a drop undergoes multiple reflection and refraction. With each reflection, a certain part of the light rays goes out and their intensity inside the drop decreases. The most intense of the rays emerging into the air is the ray that emerged from the drop at point B. However, it is difficult to observe it, since it is lost against the background of bright direct sunlight... The rays refracted at point C, together, create a primary rainbow against the background of a dark cloud, and the rays refracted at point D
give a secondary rainbow, which, as follows from what has been said, is less intense than the primary.
For the case K=1 we get Θ = 2 (59°37" - 40°26") + 1 = 137° 30".
Therefore, the viewing angle of the first order rainbow is:
φ 1 \u003d 180 ° - 137 ° 30 "= 42 ° 30"
For the ray DE" giving a rainbow of the second order, i.e. in the case of K = 2, we have:
Θ = 2 (59°37" - 40°26") + 2 = 236°38".
Second order rainbow viewing angle φ 2 = 180° - 234°38" = - 56°38".
From this it follows (this can also be seen from the figure) that in the case under consideration, a second-order rainbow is not visible from the ground. In order for it to be visible, the light must enter the drop from below (Fig. 30, b).
When considering the formation of a rainbow, one more phenomenon must be taken into account - the unequal refraction of light waves of different lengths, that is, light rays of different colors. This phenomenon is called dispersion. Due to dispersion, the angles of refraction ϒ and the angles of deflection of rays Θ in a drop are different for rays of different colors. The course of three rays - red, green and purple - is schematically shown in Figure 30, a for the first order arc and in Figure 30, b for the second order arc.
It can be seen from the figures that the sequence of colors in these arcs is opposite.
Most often we see one rainbow. It is not uncommon for cases when two iridescent stripes appear simultaneously in the sky, located one above the other; they observe, however, quite rarely, and an even greater number of iridescent celestial arcs - three, four and even five at the same time. This interesting phenomenon was observed by Leningraders on September 24, 1948, when four rainbows appeared among the clouds over the Neva in the afternoon. It turns out that a rainbow can occur not only from direct sunlight; often it appears in the reflected rays of the sun. This can be seen on the coast of sea bays, large rivers and lakes. Three or four such rainbows - ordinary and reflected - sometimes create a beautiful picture. Since the rays of the Sun reflected from the water surface go from bottom to top, the rainbow formed in these rays can sometimes look completely unusual.
You should not think that a rainbow can be observed only during the day. It happens at night, however, always weak. You can see such a rainbow after a night rain, when the moon looks out from behind the clouds.
Some semblance of a rainbow can be obtained in the following experiment. Take a flask of water, shine it with sunlight or a lamp through a hole in the whiteboard. Then a rainbow will become clearly visible on the board (Fig. 31, a), and the angle of divergence of the rays compared to the initial direction will be about 41-42 ° (Fig. 31.6). Under natural conditions, there is no screen, the image appears on the retina of the eye, and the eye projects this image onto the clouds.
If a rainbow appears in the evening before sunset, then a red rainbow is observed. In the last five or ten minutes before sunset, all the colors of the rainbow, except for red, disappear, it becomes very bright and visible even ten minutes after sunset.
A beautiful sight is a rainbow on the dew.
It can be observed at sunrise on the grass covered with dew. This rainbow is shaped like a hyperbola.
halos
When looking at a rainbow in a meadow, you will involuntarily notice an amazing uncolored halo of light - a halo surrounding the shadow of your head. This is not an optical illusion or a phenomenon of contrast. When the shadow falls on the road, the halo disappears. What is the explanation for this interesting phenomenon? Dew drops certainly play an important role here, for when the dew disappears, the phenomenon disappears.
To find out the cause of the phenomenon, do the following experiment. Take a spherical flask filled with water and expose it to sunlight. Let her represent a drop. Place a sheet of paper behind the flask close to it, which will act as grass. Look at the flask at a small angle with respect to the direction of the incident rays. You will see it brightly lit by the rays reflected from the paper. These rays go almost exactly towards the rays of the Sun falling on the flask. Take your eyes a little to the side, and the bright illumination of the flask is no longer visible.
Here we are dealing not with a scattered, but with a directed beam of light emanating from a bright spot on paper. The bulb acts like a lens that directs light towards us.
A beam of parallel solar rays, after refraction in the bulb, gives on paper a more or less focused image of the Sun in the form of a bright spot. In turn, quite a lot of the light emitted by the spot is captured by the bulb and, after refraction in it, is directed back towards the Sun, including our eyes, since we are standing with our backs to the Sun. The optical shortcomings of our lens - the flasks give some scattered light flux, but still the main stream of light coming from a bright spot on paper is directed towards the Sun. But why is the light reflected from the blades of grass not green?
It actually has a slight greenish tint, but is mostly white, much like light reflected directionally from smooth painted surfaces, such as reflections from a green or yellow chalkboard, or stained glass.
But dew drops are not always spherical. They may be distorted. Then some of them direct the light to the side, but it passes by the eyes. Other droplets, as, for example, shown in Figure 33, have such a shape that the light falling on them, after one or two reflections, is directed back towards the Sun and enters the eyes of the observer standing with his back to him.
Finally, one more ingenious explanation of this phenomenon should be noted: only those leaves of grass on which the direct light of the Sun falls, i.e., those that are not obscured by other leaves from the side of the Sun, reflect light directionally. If we take into account that the leaves of most plants always turn their plane towards the Sun, then it is obvious that there will be quite a lot of such reflective leaves (Fig. 33, e). Therefore, halos can also be observed in the absence of dew, on the surface of a smoothly mowed meadow or compressed field.
