Newton's formula for a thin lens. Optical lenses (physics): definition, description, formula and solution
"Lenses. Building an image in lenses"
Lesson Objectives:
Educational: we will continue the study of light rays and their propagation, introduce the concept of a lens, study the action of a converging and scattering lens; learn to build images given by the lens.
Developing: contribute to the development of logical thinking, the ability to see, hear, collect and comprehend information, independently draw conclusions.
Educational: cultivate attentiveness, perseverance and accuracy in work; learn to use the acquired knowledge to solve practical and cognitive problems.
Lesson type: combined, including the development of new knowledge, skills, consolidation and systematization of previously acquired knowledge.
During the classes
Organizing time(2 minutes):
greeting students;
checking the readiness of students for the lesson;
familiarization with the objectives of the lesson (the educational goal is set as a general one, without naming the topic of the lesson);
creation of psychological mood:
The universe, comprehending,
Know everything without taking away
What's inside - in the outside you will find,
What's outside, you'll find inside
So accept it without looking back
The world's intelligible riddles ...
I. Goethe
Repetition of previously studied material occurs in several stages.(26 min):
1. Blitz - poll(the answer to the question can only be yes or no, for better view students' answers, you can use signal cards, "yes" - red, "no" - green, it is necessary to specify the correct answer):
Does light travel in a straight line in a homogeneous medium? (Yes)
The angle of reflection is indicated by the Latin letter betta? (No)
Is reflection specular or diffuse? (Yes)
Is the angle of incidence always greater than the angle of reflection? (No)
At the boundary of two transparent media, does the light beam change its direction? (Yes)
Is the angle of refraction always greater than the angle of incidence? (No)
The speed of light in any medium is the same and equal to 3*10 8 m/s? (No)
Is the speed of light in water less than the speed of light in vacuum? (Yes)
Consider slide 9: “Building an image in a converging lens” ( ), using the reference abstract to consider the rays used.
Perform the construction of an image in a converging lens on the board, give its characteristics (performed by a teacher or student).
Consider slide 10: “Building an image in a diverging lens” ( ).
Perform the construction of an image in a diverging lens on the board, give its characteristics (performed by a teacher or student).
5. Checking the understanding of the new material, its consolidation(19 min):
Student work at the blackboard:
Construct an image of an object in a converging lens:
Advance task:
Independent work with a choice of tasks.
6. Summing up the lesson(5 minutes):
What did you learn in the lesson, what should you pay attention to?
Why is it not advised to water plants from above on a hot summer day?
Grades for work in the classroom.
7. Homework (2 minutes):
Construct an image of an object in a divergent lens:
If the object is beyond the focus of the lens.
If the object is between the focus and the lens.
Attached to the lesson , , and .
In this lesson, we will repeat the features of the propagation of light rays in homogeneous transparent media, as well as the behavior of the rays when they cross the border between the light separation of two homogeneous transparent media, which you already know. Based on the knowledge already gained, we will be able to understand what useful information about a luminous or light-absorbing object we can get.
Also, applying the laws of refraction and reflection of light already familiar to us, we will learn how to solve the main problems of geometric optics, the purpose of which is to build an image of the object in question, formed by rays falling into the human eye.
Let's take a look at one of the main optical devices- lens - and thin lens formulas.
2. Internet portal "CJSC "Opto-Technological Laboratory"" ()
3. Internet portal "GEOMETRIC OPTICS" ()
Homework
1. Using a lens on a vertical screen, a real image of a light bulb is obtained. How will the image change if the upper half of the lens is closed?
2. Construct an image of an object placed in front of a converging lens in the following cases: 1. ; 2.; 3.; four. .
Definition 1
Lens is a transparent body having 2 spherical surfaces. It is thin if its thickness is less than the radii of curvature of spherical surfaces.
The lens is an integral part of almost every optical device. Lenses are, by their definition, collecting and scattering (Fig. 3.3.1).
Definition 2
converging lens is a lens that is thicker in the middle than at the edges.
Definition 3
A lens that is thicker at the edges is called scattering.
Figure 3. 3 . one . Collecting (a) and diverging (b) lenses and their symbols.
Definition 4
Main optical axis is a straight line that passes through the centers of curvature O 1 and O 2 of spherical surfaces.
In a thin lens, the main optical axis intersects at one point - the optical center of the lens O. The light beam passes through the optical center of the lens without deviating from its original direction.
Definition 5
Side optical axes are straight lines passing through the optical center.
Definition 6
If a beam of rays is directed to the lens, which are parallel to the main optical axis, then after passing through the lens the rays (or their continuation) will be concentrated at one point F.
This point is called main focus of the lens.
A thin lens has two main foci, which are located symmetrically on the main optical axis with respect to the lens.
Definition 7
Focus of a converging lens valid, and for the scattering imaginary.
Beams of rays parallel to one of the entire set of secondary optical axes, after passing through the lens, are also aimed at the point F "located at the intersection of the secondary axis with the focal plane Ф.
Definition 8
focal plane- this is a plane perpendicular to the main optical axis and passing through the main focus (Fig. 3.3.2).
Definition 9
The distance between the main focus F and the optical center of the lens O is called focal(F).
Figure 3. 3 . 2. Refraction of a parallel beam of rays in a converging (a) and diverging (b) lens. O 1 and O 2 are the centers of spherical surfaces, O 1 O 2 is the main optical axis, O – optical center, F is the main focus, F" is the focus, O F" is the secondary optical axis, Ф is the focal plane.
The main property of lenses is the ability to transmit images of objects. They, in turn, are:
- Real and imaginary;
- Straight and inverted;
- Enlarged and reduced.
Geometric constructions help determine the position of the image, as well as its nature. For this purpose, the properties of standard rays are used, the direction of which is defined. These are rays that pass through the optical center or one of the foci of the lens, and rays that are parallel to the main or one of the side optical axes. Drawings 3 . 3 . 3 and 3. 3 . 4 show construction data.
Figure 3. 3 . 3 . Building an image in a converging lens.
Figure 3. 3 . four . Building an image in a divergent lens.
It is worth highlighting that the standard beams used in Figures 3 . 3 . 3 and 3. 3 . 4 for imaging, do not pass through the lens. These rays are not used in imaging, but can be used in this process.
Definition 10
The thin lens formula is used to calculate image position and character. If we write the distance from the object to the lens as d, and from the lens to the image as f, then thin lens formula looks like:
1d + 1f + 1F = D.
Definition 11
Value D is the optical power of the lens, equal to the reciprocal focal length.
Definition 12
Diopter(d p t r) is a unit of measurement of optical power, the focal length of which is equal to 1 m: 1 d p t r = m - 1 .
The formula of a thin lens is similar to that of a spherical mirror. It can be derived for paraxial rays from the similarity of triangles in figures 3 . 3 . 3 or 3 . 3 . four .
The focal length of the lenses is written with certain signs: a converging lens F > 0, a diverging lens F< 0 .
The value of d and f also obey certain signs:
- d > 0 and f > 0 - in relation to real objects (that is, real light sources) and images;
- d< 0 и f < 0 – применительно к мнимым источникам и изображениям.
For the case in figure 3. 3 . 3 F > 0 (converging lens), d = 3 F > 0 (real object).
From the thin lens formula we get: f = 3 2 F > 0 , means that the image is real.
For the case in figure 3. 3 . 4F< 0 (линза рассеивающая), d = 2 | F | >0 (real object), the formula f = - 2 3 F< 0 , следовательно, изображение мнимое.
The linear dimensions of the image depend on the position of the object in relation to the lens.
Definition 13
Linear magnification of the lens G is the ratio of the linear dimensions of the image h "and the object h.
It is convenient to write the value h "with plus or minus signs, depending on whether it is direct or inverted. It is always positive. Therefore, for direct images, the condition Γ\u003e 0 is applied, for inverted Γ< 0 . Из подобия треугольников на рисунках 3 . 3 . 3 и 3 . 3 . 4 нетрудно вывести формулу для расчета линейного увеличения тонкой линзы:
G \u003d h "h \u003d - f d.
In the example with a converging lens in Figure 3. 3 . 3 for d = 3 F > 0 , f = 3 2 F > 0 .
Hence, Г = - 1 2< 0 – изображение перевернутое и уменьшенное в два раза.
In the diverging lens example in Figure 3. 3 . 4 for d = 2 | F | > 0 , the formula f = - 2 3 F< 0 ; значит, Г = 1 3 >0 - the image is straight and reduced by a factor of three.
The optical power D of the lens depends on the radii of curvature R 1 and R 2 , its spherical surfaces, and also on the refractive index n of the lens material. In the theory of optics, the following expression takes place:
D \u003d 1 F \u003d (n - 1) 1 R 1 + 1 R 2.
A convex surface has a positive radius of curvature, while a concave surface has a negative radius. This formula is applicable in the manufacture of lenses with a given optical power.
Many optical instruments are designed in such a way that light passes through 2 or more lenses in succession. The image of the object from the 1st lens serves as an object (real or imaginary) for the 2nd lens, which, in turn, builds the 2nd image of the object, which can also be real or imaginary. The calculation of the optical system of 2 thin lenses consists in
2-fold application of the lens formula, and the distance d 2 from the 1st image to the 2nd lens should be proposed equal to the value l - f 1, where l is the distance between the lenses.
The value f 2 calculated by the lens formula predetermines the position of the 2nd image, as well as its character (f 2 > 0 is a real image, f 2< 0 – мнимое). Общее линейное увеличение Γ системы из 2 -х линз равняется произведению линейных увеличений 2 -х линз, то есть Γ = Γ 1 · Γ 2 . Если предмет либо его изображение находятся в бесконечности, тогда линейное увеличение не имеет смысла.
Kepler's astronomical tube and Galileo's terrestrial tube
Let us consider a special case - the telescopic path of rays in a system of 2 lenses, when both the object and the 2nd image are located at infinitely large distances from each other. The telescopic path of the rays is carried out in the telescopes: Galileo's earthly tube and Kepler's astronomical tube.
A thin lens has some drawbacks that do not allow high resolution images to be obtained.
Definition 14
Aberration is the distortion that occurs during the imaging process. Depending on the distance at which the observation is made, aberrations can be spherical or chromatic.
The meaning of spherical aberration is that with wide light beams, rays that are at a far distance from the optical axis do not cross it at the focus. The thin lens formula only works for rays that are close to the optical axis. The image of a distant source, which is created by a wide beam of rays refracted by a lens, is blurry.
The meaning of chromatic aberration is that the refractive index of the lens material is affected by the light wavelength λ. This property of transparent media is called dispersion. The focal length of a lens is different for light with different wavelengths. This fact leads to blurring of the image when non-monochromatic light is emitted.
Modern optical devices are equipped not with thin lenses, but with complex lens systems in which it is possible to eliminate some distortion.
In devices such as cameras, projectors, etc., converging lenses are used to form real images of objects.
Definition 15Camera- this is a closed light-tight camera in which the image of the captured objects is created on the film by a system of lenses - lens. During the exposure, the lens is opened and closed using a special shutter.
The peculiarity of the operation of the camera is that on a flat film, rather sharp images of objects that are at different distances are obtained. Sharpness changes as the lens moves relative to the film. Images of points that do not lie in the plane of sharp pointing come out blurry in the images in the form of scattered circles. The size d of these circles can be reduced by lens aperture, that is, by reducing the aperture ratio a F , as shown in Figure 3. 3 . 5 . This results in increased depth of field.
Figure 3. 3 . 5 . Camera.
With the help of a projection device, it is possible to shoot large-scale images. The lens O of the projector focuses the image of a flat object (diapositive D) on the remote screen E (Figure 3.3.6). The lens system K (condenser) is used to concentrate the light source S on the slide. An enlarged inverted image is recreated on the screen. The scale of the projection device can be changed by zooming in or out of the screen and at the same time changing the distance between the aperture D and the lens O.
Figure 3. 3 . 6. projection device.
Figure 3. 3 . 7. thin lens model.
Figure 3. 3 . eight . Model of a system of two lenses.
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Topics of the USE codifier: building images in lenses, thin lens formula.
The rules for the path of rays in thin lenses, formulated in the previous topic, lead us to the most important statement.
Image theorem. If there is a luminous point in front of the lens, then after refraction in the lens, all rays (or their continuations) intersect at one point.
The point is called the point image.
If the refracted rays themselves intersect at a point, then the image is called valid. It can be obtained on the screen, since the energy of light rays is concentrated at a point.
If, however, not the refracted rays themselves intersect at a point, but their continuations (this happens when the refracted rays diverge after the lens), then the image is called imaginary. It cannot be received on the screen, because no energy is concentrated in the point. An imaginary image, we recall, arises due to the peculiarity of our brain - to complete the diverging rays to their imaginary intersection and see a luminous point in this intersection. An imaginary image exists only in our minds.
The image theorem serves as the basis for imaging in thin lenses. We will prove this theorem for both converging and diverging lenses.
Converging lens: real image of a point.
Let's look at a converging lens first. Let be the distance from the point to the lens, be the focal length of the lens. There are two fundamentally different cases: and (and also an intermediate case ). We will deal with these cases one by one; in each of them we
Let us discuss the properties of images of a point source and an extended object.
First case: . The point light source is located farther from the lens than the left focal plane (Fig. 1).
The beam passing through the optical center is not refracted. We will take arbitrary ray , construct a point at which the refracted ray intersects with the ray , and then show that the position of the point does not depend on the choice of the ray (in other words, the point is the same for all possible rays ). Thus, it turns out that all rays emanating from the point intersect at the point after refraction in the lens, and the image theorem will be proved for the case under consideration.
We will find the point by constructing the further course of the beam. We can do this: we draw a side optical axis parallel to the beam until it intersects with the focal plane in the side focus, after which we draw the refracted beam until it intersects with the beam at the point.
Now we will look for the distance from the point to the lens. We will show that this distance is expressed only in terms of and , i.e., it is determined only by the position of the source and the properties of the lens, and thus does not depend on a particular beam.
Let us drop the perpendiculars and onto the main optical axis. Let's also draw it parallel to the main optical axis, that is, perpendicular to the lens. We get three pairs of similar triangles:
, (1)
, (2)
. (3)
As a result, we have the following chain of equalities (the number of the formula above the equal sign indicates from which pair of similar triangles this equality was obtained).
(4)
But , so relation (4) is rewritten as:
. (5)
From here we find the desired distance from the point to the lens:
. (6)
As we see, it really does not depend on the choice of the ray . Therefore, any ray after refraction in the lens will pass through the point constructed by us, and this point will be a real image of the source
The image theorem is proved in this case.
The practical importance of the image theorem is this. Since all the rays of the source intersect after the lens at one point - its image - then to build an image it is enough to take the two most convenient rays. What exactly?
If the source does not lie on the main optical axis, then the following are suitable as convenient beams:
A beam passing through the optical center of the lens - it is not refracted;
- a ray parallel to the main optical axis - after refraction, it goes through the focus.
The construction of an image using these rays is shown in Fig. 2.
If the point lies on the main optical axis, then only one convenient ray remains - running along the main optical axis. As the second beam, one has to take the "uncomfortable" one (Fig. 3).
Let's look again at the expression ( 5 ). It can be written in a slightly different form, more attractive and memorable. Let's first move the unit to the left:
We now divide both sides of this equality by a:
(7)
Relation (7) is called thin lens formula(or just the lens formula). So far, the lens formula has been obtained for the case of a converging lens and for . In what follows, we derive modifications of this formula for other cases.
Now let's return to relation (6) . Its importance is not limited to the fact that it proves the image theorem. We also see that it does not depend on the distance (Fig. 1, 2) between the source and the main optical axis!
This means that whatever point of the segment we take, its image will be at the same distance from the lens. It will lie on a segment - namely, at the intersection of the segment with a ray that will go through the lens without refraction. In particular, the image of a point will be a point .
Thus, we have established an important fact: the segment is puddles with the image of the segment. From now on, the original segment, the image of which we are interested in, we call subject and are marked with a red arrow in the figures. We need the direction of the arrow in order to follow whether the image is straight or inverted.
Converging lens: the actual image of an object.
Let's move on to the consideration of images of objects. Recall that while we are in the framework of the case . Three typical situations can be distinguished here.
one. . The image of the object is real, inverted, enlarged (Fig. 4; double focus is indicated). From the lens formula it follows that in this case it will be (why?).
Such a situation is realized, for example, in overhead projectors and film cameras - these optical devices give an enlarged image of what is on the film on the screen. If you have ever shown slides, then you know that the slide must be inserted into the projector upside down - so that the image on the screen looks right, and does not turn out upside down.
The ratio of the size of the image to the size of the object is called the linear magnification of the lens and is denoted by G - (this is the capital Greek "gamma"):
From the similarity of triangles we get:
. (8)
Formula (8) is used in many problems where the linear magnification of the lens is involved.
2. . In this case, from formula (6) we find that and . The linear magnification of the lens according to (8) is equal to one, i.e. the size of the image is equal to the size of the object (Fig. 5).
This situation is common for many optical instruments: cameras, binoculars, telescopes - in a word, those in which images of distant objects are obtained. As the object moves away from the lens, its image decreases in size and approaches the focal plane.
We have completely completed the consideration of the first case. Let's move on to the second case. It won't be as big anymore.
Converging lens: virtual image of a point.
Second case: . A point light source is located between the lens and the focal plane (Fig. 7).
Along with the ray going without refraction, we again consider an arbitrary ray. However, now two divergent beams and are obtained at the exit from the lens. Our eye will continue these rays until they intersect at a point.
The image theorem states that the point will be the same for all rays emanating from the point. We prove this again with three pairs of similar triangles:
Denoting again through the distance from to the lens, we have the corresponding chain of equalities (you can easily figure it out already):
. (9)
. (10)
The value does not depend on the ray, which proves the image theorem for our case. So, - imaginary image of the source. If the point does not lie on the main optical axis, then to construct an image, it is most convenient to take a beam passing through the optical center and a beam parallel to the main optical axis (Fig. 8).
Well, if the point lies on the main optical axis, then there is nowhere to go - you have to be content with a beam that falls obliquely on the lens (Fig. 9).
Relation (9) leads us to a variant of the lens formula for the considered case . First, we rewrite this relation as:
and then divide both sides of the resulting equality by a:
. (11)
Comparing (7) and (11) , we see a slight difference: the term is preceded by a plus sign if the image is real, and a minus sign if the image is imaginary.
The value calculated by formula (10) also does not depend on the distance between the point and the main optical axis. As above (remember the reasoning with a dot), this means that the image of the segment in Fig. 9 will be a segment.
Converging lens: a virtual image of an object.
With this in mind, we can easily build an image of an object located between the lens and the focal plane (Fig. 10). It turns out to be imaginary, direct and enlarged.
You see such an image when you look at a small object in a magnifying glass - a magnifying glass. The case is completely disassembled. As you can see, it is qualitatively different from our first case. This is not surprising - because between them lies an intermediate "catastrophic" case.
Converging lens: An object in the focal plane.
Intermediate case: The light source is located in the focal plane of the lens (Fig. 11).
As we remember from the previous section, the rays of a parallel beam, after refraction in a converging lens, will intersect in the focal plane - namely, at the main focus if the beam is incident perpendicular to the lens, and at the side focus if the beam is incident obliquely. Using the reversibility of the path of the rays, we conclude that all the rays of the source located in the focal plane, after leaving the lens, will go parallel to each other.
 |
| Rice. 11. a=f: no image |
Where is the image of the dot? There are no images. However, no one forbids us to assume that parallel rays intersect at an infinitely distant point. Then the image theorem remains valid and in this case - the image is at infinity.
Accordingly, if the object is entirely located in the focal plane, the image of this object will be located at infinity(or, what is the same, will be absent).
So, we have completely considered the construction of images in a converging lens.
Converging lens: virtual image of a point.
Fortunately, there is not such a variety of situations as for a converging lens. The nature of the image does not depend on how far the object is from the diverging lens, so there will be only one case here.
Again we take a ray and an arbitrary ray (Fig. 12). At the exit from the lens, we have two divergent beams and , which our eye builds up to the intersection at the point .
We again have to prove the image theorem - that the point will be the same for all rays. We act with the help of the same three pairs of similar triangles:
(12)
. (13)
The value of b does not depend on the ray span
, so the extensions of all refracted rays span
intersect at a point - the imaginary image of the point. The image theorem is thus completely proved.
Recall that for a converging lens we obtained similar formulas (6) and (10) . In the case of their denominator vanished (the image went to infinity), and therefore this case distinguished fundamentally different situations and .
But for formula (13), the denominator does not vanish for any a. Therefore, for a diverging lens there is no qualitatively different situations source location - the case here, as we said above, there is only one.
If the point does not lie on the main optical axis, then two beams are convenient for constructing its image: one goes through the optical center, the other is parallel to the main optical axis (Fig. 13).
If the point lies on the main optical axis, then the second beam has to be taken arbitrary (Fig. 14).
Relation (13) gives us another version of the lens formula. Let's rewrite first:
and then divide both sides of the resulting equality by a:
(14)
This is how the lens formula for a diverging lens looks like.
Three lens formulas (7) , (11) and (14) can be written in the same way:
subject to the following sign convention:
For a virtual image, the value is considered negative;
- for a diverging lens, the value is considered negative.
This is very convenient and covers all the considered cases.
Divergent lens: a virtual image of an object.
The value calculated by formula (13) again does not depend on the distance between the point and the main optical axis. This again gives us the opportunity to construct an image of the object, which this time turns out to be imaginary, direct and reduced (Fig. 15).
 |
| Rice. 15. The image is imaginary, direct, reduced |
There are objects that are capable of changing the density of the electromagnetic radiation flux incident on them, that is, either increasing it by collecting it at one point, or decreasing it by scattering it. These objects are called lenses in physics. Let's consider this question in more detail.
What are lenses in physics?
This concept means absolutely any object that is capable of changing the direction of propagation of electromagnetic radiation. it general definition lenses in physics, which includes optical glasses, magnetic and gravitational lenses.
In this article, the main attention will be paid to optical glasses, which are objects made of a transparent material and limited by two surfaces. One of these surfaces must necessarily have curvature (that is, be part of a sphere of finite radius), otherwise the object will not have the property of changing the direction of propagation of light rays.
The principle of the lens
The essence of this simple optical object is the phenomenon of refraction of sunlight. At the beginning of the 17th century, the famous Dutch physicist and astronomer Willebrord Snell van Rooyen published the law of refraction, which currently bears his last name. The formulation of this law is as follows: when sunlight passes through the interface between two optically transparent media, then the product of the sine between the beam and the normal to the surface and the refractive index of the medium in which it propagates is a constant value.
To clarify the above, let's give an example: let the light fall on the surface of the water, while the angle between the normal to the surface and the beam is equal to θ 1 . Then, the light beam is refracted and begins its propagation in the water already at an angle θ 2 to the normal to the surface. According to Snell's law, we get: sin (θ 1) * n 1 \u003d sin (θ 2) * n 2, here n 1 and n 2 are the refractive indices for air and water, respectively. What is the refractive index? This is a value showing how many times the propagation speed of electromagnetic waves in vacuum is greater than that for an optically transparent medium, that is, n = c/v, where c and v are the speeds of light in vacuum and in the medium, respectively.
The physics of refraction lies in the implementation of Fermat's principle, according to which light moves in such a way as to overcome the distance from one point to another in space in the shortest time.
The type of optical lens in physics is determined solely by the shape of the surfaces that form it. The direction of refraction of the beam incident on them depends on this shape. So, if the curvature of the surface is positive (convex), then, upon exiting the lens, the light beam will propagate closer to its optical axis (see below). Conversely, if the curvature of the surface is negative (concave), then passing through the optical glass, the beam will move away from its central axis.
We note again that the surface of any curvature refracts the rays in the same way (according to Stella's law), but the normals to them have a different slope relative to the optical axis, resulting in a different behavior of the refracted beam.
A lens bounded by two convex surfaces is called a converging lens. In turn, if it is formed by two surfaces with negative curvature, then it is called scattering. All other views are associated with a combination of the indicated surfaces, to which a plane is also added. What property the combined lens will have (diffusing or converging) depends on the total curvature of the radii of its surfaces.
Lens elements and ray properties
To build in lenses in imaging physics, it is necessary to get acquainted with the elements of this object. They are listed below:
- Main optical axis and center. In the first case, they mean a straight line passing perpendicular to the lens through its optical center. The latter, in turn, is a point inside the lens, passing through which the beam does not experience refraction.
- Focal length and focus - the distance between the center and a point on the optical axis, in which all rays incident on the lens parallel to this axis are collected. This definition is true for collecting optical glasses. In the case of divergent lenses, it is not the rays themselves that will converge to a point, but their imaginary continuation. This point is called the main focus.
- optical power. This is the name of the reciprocal of the focal length, that is, D \u003d 1 / f. It is measured in diopters (diopters), that is, 1 diopter. = 1 m -1.
The following are the main properties of rays that pass through a lens:
- the beam passing through the optical center does not change the direction of its movement;
- rays incident parallel to the main optical axis change their direction so that they pass through the main focus;
- rays falling on optical glass at any angle, but passing through its focus, change their direction of propagation in such a way that they become parallel to the main optical axis.
The above properties of rays for thin lenses in physics (as they are called, because it does not matter what spheres they are formed and how thick they are, only the optical properties of the object matter) are used to build images in them.
Images in optical glasses: how to build?
The figure below shows in detail the schemes for constructing images in the convex and concave lenses of an object (red arrow) depending on its position.
Important conclusions follow from the analysis of the circuits in the figure:
- Any image is built on only 2 rays (passing through the center and parallel to the main optical axis).
- Converging lenses (denoted with arrows at the ends pointing outward) can give both an enlarged and reduced image, which in turn can be real (real) or imaginary.
- If the object is in focus, then the lens does not form its image (see the lower diagram on the left in the figure).
- Scattering optical glasses (denoted by arrows at their ends pointing inward) always give a reduced and imaginary image regardless of the position of the object.
Finding the distance to an image
To determine at what distance the image will appear, knowing the position of the object itself, we give the lens formula in physics: 1/f = 1/d o + 1/d i , where d o and d i are the distance to the object and to its image from the optical center, respectively, f is the main focus. If we are talking about a collecting optical glass, then the f-number will be positive. Conversely, for a diverging lens, f is negative.
Let's use this formula and solve a simple problem: let the object be at a distance d o = 2*f from the center of the collecting optical glass. Where will his image appear?
From the condition of the problem we have: 1/f = 1/(2*f)+1/d i . From: 1/d i = 1/f - 1/(2*f) = 1/(2*f), i.e. d i = 2*f. Thus, the image will appear at a distance of two foci from the lens, but on the other side than the object itself (this is indicated by the positive sign of the value d i).
Short story
It is curious to give the etymology of the word "lens". It comes from the Latin words lens and lentis, which means "lentil", since optical objects in their shape really look like the fruit of this plant.
The refractive power of spherical transparent bodies was known to the ancient Romans. For this purpose, they used round glass vessels filled with water. Glass lenses themselves began to be made only in the 13th century in Europe. They were used as a reading tool (modern glasses or a magnifying glass).
The active use of optical objects in the manufacture of telescopes and microscopes dates back to the 17th century (at the beginning of this century, Galileo invented the first telescope). Note that the mathematical formulation of Stella's law of refraction, without knowing which it is impossible to manufacture lenses with desired properties, was published by a Dutch scientist at the beginning of the same 17th century.
Other types of lenses
As noted above, in addition to optical refractive objects, there are also magnetic and gravitational objects. An example of the former are magnetic lenses in an electron microscope, a vivid example of the latter is the distortion of the direction of the light flux when it passes near massive cosmic bodies (stars, planets).
