Site for children


J. Perelman
"Entertaining physics". Book 1.
Chapter 8. Reflection and refraction of light

WHEN A LONG PATH TRAVERSED FASTER THAN SHORT?

But is it really that broken path can quickly lead to the goal than direct? Yes, in those cases when the speed of movement in different parts of the path are different. Remember that you have to do to the villagers, located between two train stations in the vicinity of one of them. To get more on distant station, they go to the horse first in the reverse direction to the nearest station, where we sit on the train and go to the destination. Them shorter would be, of course, right to go there on a horse, but they prefer more a long way on a horse and wagon, because it leads to the goal sooner.

Take a minute attention to another example. Cavalry should come with a report from the point a to the tent of the commander at the point C (see Fig.). It is separated from tents band deep sand and strip meadows, delimited between a straight line EF, On sandy soil, the horse moves twice slower than in the meadow. Which way should choose trooper to reach the tents in the shortest time?


The problem of the trooper. To find the quickest way from A to C.

At first glance it seems that the fast path is a straight line drawn from A to C. But this is completely wrong, and I don't think that was the trooper, who would choose such a path. Slow motion over the sand would lead him to the correct idea to reduce this slow part of the way through the sandy strip along the lower oblique line; of course, extends to the second part of the path through the meadow; but as the meadow can move twice as fast, the lengthening of the path does not outweigh the benefits, and the total path will be done in less time. In other words, the path of the trooper should refracted at the boundary of both genera soil and, moreover, so that the path through the meadow was with the normal to the boundary of the greater angle than the path along the sandy soil.

Who is familiar with geometry, with the Pythagorean theorem, one can verify that the straight path AC really is not through the early and when those dimensions for the width of the bands and distances that we have here in mind, you can achieve the goal if to go, for example, in a broken AES.


The solution of the problem of the trooper. The quickest way AMC.

The first figure indicates that the width of the sand bar 2 km, meadow - 3 km, and the distance g - 7 km. Then the whole length of AC is equal, by the Pythagorean theorem, √(52 + 72) = √74 = 8,60 km Part'AN is the path on the sand - this cut is as easy to understand, 2/5 of this value, i.e., 3,44 km. as on sand movement occurs twice slower than the meadow, 3,44 km sandy path is equivalent, in terms of time required, to 6.88 km across the meadow. And, therefore, all mixed path on a straight line AC, is equal to at 8.60 km, corresponds 12,04 km path through the meadow.

Will do the same “cast to the meadow” and broken path AES. Part of AE = 2 km and corresponds to a 4 km path through the meadow. Part of the EU = √(32 + 72) = √(58) = to 7.61 km total all broken path AEC is responsible 4 + to 7.61 = to 11.61 km

So, short straight path corresponds 12,04 km movement across the meadow, and “long” broken - only to 11.61 km on the same soil. The “long” way, as you can see, gives the benefit in 12,04 - to 11.61 = 0,43, almost half a kilometer!

But we have not specified another very quick way. The fastest way teaches the theory would be one in which we have here to seek the services of trigonometry) the sine of angle b refers to the sine of angle A, as the speed in the meadow refers to the speed in the sand, i.e., as 2:1. In other words, you need to choose the direction so that sin b was twice as much sin as. For this you need to cross the border between bands in the point m, which is one kilometre from E. Indeed, then sin b = 6/√(32 + 62), sin a = 1/√(1 + 22 ), the ratio sin b / sin a= 6 / √45 / 1 / (3*√5) = (6/(3*√5) / (1/√5) = 2, i.e., just respect speeds.

And what will be in this case given to the meadow path length? Calculate: AM = √(22 + 12). what is responsible 4,47 km path through the meadow. MS = √45 = of 6.71 km Length all the way to 4.47% + of 6.71 = 11,18, i.e., at 860 km shorter rectilinear path, which, as we already know, corresponds 12,04 km

You see, what benefits it delivers under these conditions Islamiyya way. Light beam as time and shall elect such a speedy way because the law of refraction of light is strictly satisfies the requirement of mathematical solutions of the problem: the sine of angle of refraction refers to the sine of the angle of incidence, as the speed of light in the new environment to speed him leave the environment; on the other hand, this ratio is equal to the refractive index of light in these environments.

Bringing together in one rule features and reflection and refraction, we can say that the light beam in all cases should be on the fastest path, i.e., obeys the rule that physicists call “the principle as soon as possible” (Fermat's principle).

If the environment is heterogeneous and its refractive ability varies gradually, as, for example, in our atmosphere, and in this case, it is the fastest parish. This explains the small curvature of the rays of the heavenly bodies in the atmosphere, which in the language of astronomers called “atmospheric refraction”. In the atmosphere, gradually compactible down, a ray of light is bent so that its concavity faces the Earth. Then the beam remains longer in high layers, which are weaker slow his path, and spends less time in the “slow” low layers, in the end, he comes to the goal faster than on the way to strictly rectilinear.

The principle of the fastest ward (Fermat's principle) not fair for only one light phenomena: he fully submits to the dissemination of sound and all undulating movements, whatever the nature of these waves.

The reader, no doubt, would like to know why this property is undulating movements. Will therefore related considerations outstanding modern physicist schrödinger (In a paper read in Stockholm when receiving the Nobel prize in 1933). He comes from a friend already us the example of marching soldiers and has in mind the case of the movement of the light beam in the medium gradually changing density.

“Let them, " he writes, " in order to preserve the strict correctness of the front, the soldiers joined a long pole, which each of them firmly holds in his hands. The command reads: everyone to flee as soon as possible! If the nature of the soil is slowly changing from point to point, first, say, the right, and later the left wing front will move faster and the rotation of the front will be carried out by itself. We note that the path is not straight, but curved. What this way is strictly coincides with the shortest in terms of time of arrival at this point given the properties of the soil, is quite understandable, because after all, every soldier tried to move as quickly as possible”.

Entertaining physics J. Perelman

 




System Orphus

SUPPORT THE SITE!

Did you like our site and you would like to support it? It's very simple: tell your friends about us!
Read MORE

  © 2014 All children