Optical power of the lens. Which lens is stronger? Lenses

Lenses are bodies that are transparent to a given radiation and bounded by two surfaces of different shapes (spherical, cylindrical, etc.). The formation of spherical lenses is shown in Fig. IV.39. One of the surfaces limiting the lens can be a sphere of infinitely large radius, i.e., a plane.

The axis passing through the centers of the surfaces forming the lens is called the optical axis; For plano-convex and plano-concave lenses, the optical axis is drawn through the center of the sphere perpendicular to the plane.

A lens is called thin if its thickness is significantly less than the radii of curvature of the forming surfaces. In a thin lens, the displacement a of the rays passing through the central part can be neglected (Fig. IV.40). A lens is converging if it refracts the rays passing through it towards the optical axis, and diverging if it deflects the rays from the optical axis.

LENS FORMULA

Let us first consider the refraction of rays on one spherical surface of the lens. Let us denote the points of intersection of the optical axis with the surface under consideration through O, with the incident ray - through and with the refracted ray (or its continuation) - through the point is the center of the spherical surface (Fig. IV.41); denote the distance by the radius of curvature of the surface). Depending on the angle of incidence of the rays on the spherical surface, different locations of points relative to point O are possible. In Fig. IV.41 shows the path of rays incident on a convex surface at different angles of incidence and under the condition where is the refractive index of the medium from which the incident ray comes, and and the refractive index of the medium where the refracted ray goes. Let us assume that the incident beam is paraxial, i.e.

makes a very small angle with the optical axis, then the angles are also small and can be considered:

Based on the law of refraction at small angles a and y

From Fig. IV.41, and it follows:

Substituting these expressions into formula (1.34), we obtain after reduction to the formula for a refractive spherical surface:

Knowing the distance from the “object” to the refractive surface, you can use this formula to calculate the distance from the surface to the “image”

Note that when deriving formula (1.35), the value was reduced; this means that all paraxial rays emanating from a point, no matter what angle they make with the optical axis, will converge at the point

Carrying out similar reasoning for other angles of incidence (Fig. IV.41, b, c), we obtain, respectively:

From here we obtain the rule of signs (assuming the distance is always positive): if the point or lies on the same side of the refractive surface on which the point is located, then the distances

and should be taken with a minus sign; if the point or is located on the other side of the surface in relation to the point, then the distances should be taken with a plus sign. The same rule of signs will result if we consider the refraction of rays through a concave spherical surface. For this purpose, you can use the same drawings shown in Fig. IV.41, if only the direction of the rays is reversed and the designations for the refractive indices are changed.

Lenses have two refractive surfaces, the radii of curvature of which can be the same or different. Consider a biconvex lens; for a ray passing through such a lens, the first (input) surface is convex, and the second (output) is concave. The formula for calculation from the data can be obtained if we use formulas (1.35) for the input and (1.36) for the output surface (with the reverse path of the rays, since the ray passes from medium to medium

Since the “image” from the first surface is the “object” for the second surface, then from formula (1.37) we obtain, replacing by by

From this relationship it is clear that the value is constant, i.e., interconnected. Let us denote where the focal length of the lens is called the optical power of the lens and is measured in diopters). Hence,

If the calculation is carried out for a biconcave lens, we obtain

Comparing the results, we can come to the conclusion that to calculate the optical power of a lens of any shape, one should use one formula (1.38) in compliance with the sign rule: the radii of curvature of convex surfaces should be substituted with a plus sign, concave surfaces with a minus sign. Negative optical power i.e. negative focal length means that the distance has a minus sign, i.e. the “image” is on the same side where the “object” is located. In this case, the “image” is imaginary. Lenses with positive optical power are converging and give real images, while at distance they acquire a minus sign and the image turns out to be virtual. Lenses with negative optical power are divergent and always give a virtual image; for them and for no numerical values ​​it is impossible to obtain a positive distance

Formula (1.38) was derived under the condition that the same medium is located on both sides of the lens. If the refractive indices of the media bordering the surfaces of the lens are different (for example, in the lens of the eye), then the focal lengths to the right and left of the lens are not equal, and

where is the focal length on the side where the object is located.

Note that, according to formula (1.38), the optical power of a lens is determined not only by its shape, but also by the relationship between the refractive indices of the lens material and the environment. For example, a biconvex lens in a medium with a high refractive index has a negative optical power, i.e., it is a diverging lens.

On the contrary, a biconcave lens in the same medium has a positive optical power, i.e., it is a converging lens.

Consider a system of two lenses (Fig. IV.42, a); Let us assume that the point object is at the focus of the first lens. The ray emerging from the first lens will be parallel to the optical axis and, therefore, will pass through the focus of the second lens. Considering this system as one thin lens, we can write Since then

This result is also true for a more complex system of thin lenses (if only the system itself can be considered “thin”): the optical power of a thin lens system is equal to the sum of the optical powers of its constituent parts:

(for diverging lenses, the optical power has a negative sign). For example, a plane-parallel plate composed of two thin lenses (Fig. IV.42, b) can be a collecting (if) or a diverging (if) lens. For two thin lenses located at a distance a from each other (Fig. IV.43) , optical power is a function of a and the focal lengths of the lenses and

(concave or dissipative). The path of rays in these types of lenses is different, but the light is always refracted, however, in order to consider their structure and principle of operation, you need to familiarize yourself with the same concepts for both types.

If we draw the spherical surfaces of the two sides of the lens to full spheres, then the straight line passing through the centers of these spheres will be the optical axis of the lens. In fact, the optical axis passes through the widest point of a convex lens and the narrowest point of a concave lens.

Optical axis, lens focus, focal length

On this axis there is a point where all the rays passing through the collecting lens are collected. In the case of a diverging lens, we can draw continuations of the diverging rays, and then we will get a point, also located on the optical axis, where all these continuations converge. This point is called the focus of the lens.

A converging lens has a real focus, and it is located on the opposite side of the incident rays; a diverging lens has an imaginary focus, and it is located on the same side from which the light falls on the lens.

The point on the optical axis exactly in the middle of the lens is called its optical center. And the distance from the optical center to the focal point of the lens is the focal length of the lens.

The focal length depends on the degree of curvature of the spherical surfaces of the lens. More convex surfaces will refract rays more strongly and, accordingly, reduce the focal length. If the focal length is shorter, then the lens will provide greater image magnification.

Lens optical power: formula, unit of measurement

To characterize the magnifying power of a lens, the concept of “optical power” was introduced. The optical power of a lens is the reciprocal of its focal length. The optical power of a lens is expressed by the formula:

where D is the optical power, F is the focal length of the lens.

The unit of measurement for the optical power of a lens is diopter (1 diopter). 1 diopter is the optical power of a lens whose focal length is 1 meter. The shorter the focal length, the greater the optical power, that is, the more the lens magnifies the image.

Since the focus of a diverging lens is imaginary, we agreed to consider its focal length to be a negative value. Accordingly, its optical power is also a negative value. As for the converging lens, its focus is real, therefore both the focal length and optical power of the converging lens are positive quantities.

Now we will talk about geometric optics. In this section, a lot of time is devoted to such an object as a lens. After all, it can be different. At the same time, the thin lens formula is one for all cases. You just need to know how to apply it correctly.

Types of lenses

It is always a transparent body that has a special shape. The appearance of the object is dictated by two spherical surfaces. One of them can be replaced with a flat one.

Moreover, the lens may have a thicker middle or edge. In the first case it will be called convex, in the second - concave. Moreover, depending on how concave, convex and flat surfaces are combined, lenses can also be different. Namely: biconvex and biconcave, plano-convex and plano-concave, convex-concave and concave-convex.

Under normal conditions, these objects are used in the air. They are made from a substance that is larger than air. Therefore, a convex lens will be converging, and a concave lens will be diverging.

General characteristics

Before we talk aboutthin lens formula, you need to decide on the basic concepts. You definitely need to know them. Because they will be constantly accessed by various tasks.

The main optical axis is straight. It is drawn through the centers of both spherical surfaces and determines the place where the center of the lens is located. There are also additional optical axes. They are drawn through a point that is the center of the lens, but do not contain the centers of spherical surfaces.

In the formula for a thin lens there is a quantity that determines its focal length. Thus, the focus is a point on the main optical axis. The rays running parallel to the specified axis intersect in it.

Moreover, each thin lens always has two focuses. They are located on both sides of its surfaces. Both focuses of the collector are valid. The scattering one has imaginary ones.

The distance from the lens to the focal point is the focal length (letterF) . Moreover, its value can be positive (in the case of collecting) or negative (for scattering).

Another characteristic associated with focal length is optical power. It is customary to denote itD.Its value is always the inverse of focus, that isD= 1/ F.Optical power is measured in diopters (abbreviated as diopters).

What other designations are there in the thin lens formula?

In addition to the focal length already indicated, you will need to know several distances and sizes. For all types of lenses they are the same and are presented in the table.

All indicated distances and heights are usually measured in meters.

In physics, the thin lens formula is also associated with the concept of magnification. It is defined as the ratio of the image size to the height of the object, that is, H/h. It can be designated by the letter G.

What is needed to construct an image in a thin lens

This is necessary to know in order to obtain the formula for a thin lens, converging or scattering. The drawing begins with both lenses having their own schematic representation. They both look like a segment. Only the collecting arrows at its ends are directed outward, and the scattering arrows are directed inward to this segment.

Now you need to draw a perpendicular to this segment to its middle. This will show the main optical axis. The focal points are supposed to be marked on it on both sides of the lens at the same distance.

The object whose image needs to be constructed is drawn in the form of an arrow. It shows where the top of the object is. In general, the object is placed parallel to the lens.

How to construct an image in a thin lens

In order to construct an image of an object, it is enough to find the points of the ends of the image and then connect them. Each of these two points can be obtained from the intersection of two rays. The simplest to construct are two of them.

Coming from a specified point parallel to the main optical axis. After contact with the lens, it goes through the main focus. If we are talking about a converging lens, then this focus is located behind the lens and the beam goes through it. When a diverging lens is considered, the beam must be directed so that its continuation passes through the focus in front of the lens.

Going directly through the optical center of the lens. He does not change his direction after her.

There are situations when an object is placed perpendicular to the main optical axis and ends on it. Then it is enough to construct an image of a point that corresponds to the edge of the arrow that does not lie on the axis. And then draw a perpendicular from it to the axis. This will be the image of the object.

The intersection of the constructed points gives an image. A thin converging lens produces a real image. That is, it is obtained directly at the intersection of the rays. An exception is the situation when an object is placed between the lens and the focus (as in a magnifying glass), then the image turns out to be virtual. For a scattering one, it always turns out to be imaginary. After all, it is obtained at the intersection not of the rays themselves, but of their continuations.

The actual image is usually drawn with a solid line. But the imaginary is dotted. This is due to the fact that the first is actually present there, and the second is only visible.

Derivation of the thin lens formula

This can be conveniently done on the basis of a drawing illustrating the construction of a real image in a converging lens. The designation of the segments is indicated in the drawing.

The branch of optics is not called geometric for nothing. Knowledge from this section of mathematics will be required. First you need to consider triangles AOB and A 1 OB 1 . They are similar because they have two equal angles (straight and vertical). From their similarity it follows that the modules of the segments A 1 IN 1 and AB are related as modules of segments OB 1 and OV.

Two more triangles turn out to be similar (based on the same principle at two angles):COFand A 1 FB 1 . In them the ratios of the following modules of segments are equal: A 1 IN 1 with CO andFB 1 WithOF.Based on the construction, the segments AB and CO will be equal. Therefore, the left sides of the indicated relational equalities are the same. Therefore, those on the right are equal. That is, OV 1 / OB equalsFB 1 / OF.

In the indicated equality, the segments indicated by dots can be replaced with the corresponding physical concepts. So OV 1 is the distance from the lens to the image. OB is the distance from the object to the lens.OF—focal length. And the segmentFB 1 is equal to the difference between the distance to the image and the focus. Therefore it can be rewritten differently:

f/d=( f - F) /ForFf = df - dF.

To derive the formula for a thin lens, the last equality must be divided bydfF.Then it turns out:

1/ d + 1/f = 1/F.

This is the formula for a thin converging lens. The diffuser has a negative focal length. This causes the equality to change. True, it is insignificant. It’s just that in the formula for a thin diverging lens there is a minus before the ratio 1/F.That is:

1/ d + 1/f = - 1/F.

The problem of finding the magnification of a lens

Condition. The focal length of the converging lens is 0.26 m. It is necessary to calculate its magnification if the object is at a distance of 30 cm.

Solution. It starts with introducing notation and converting units to C. Yes, they are knownd= 30 cm = 0.3 m andF= 0.26 m. Now you need to select formulas, the main one is the one indicated for magnification, the second one is for a thin converging lens.

They need to be combined somehow. To do this, you will have to consider a drawing of the construction of an image in a converging lens. From similar triangles it is clear that Г = H/h= f/d. That is, in order to find the magnification, you will have to calculate the ratio of the distance to the image to the distance to the object.

The second is known. But the distance to the image should be derived from the formula indicated earlier. It turns out that

f= dF/ ( d- F).

Now these two formulas need to be combined.

G =dF/ ( d( d- F)) = F/ ( d- F).

At this point, solving the problem of the thin lens formula comes down to elementary calculations. It remains to substitute the known quantities:

G = 0.26 / (0.3 - 0.26) = 0.26 / 0.04 = 6.5.

Answer: the lens gives a magnification of 6.5 times.

A task where you need to find focus

Condition. The lamp is located one meter from the collecting lens. The image of its spiral is obtained on a screen spaced 25 cm from the lens. Calculate the focal length of the specified lens.

Solution. The following values ​​should be recorded in the data:d=1 m andf= 25 cm = 0.25 m. This information is enough to calculate the focal length from the thin lens formula.

So 1/F= 1/1 + 1/0.25 = 1 + 4 = 5. But the problem requires finding out the focus, not the optical power. Therefore, all that remains is to divide 1 by 5, and you get the focal length:

F=1/5 = 0, 2 m.

Answer: the focal length of a converging lens is 0.2 m.

The problem of finding the distance to an image

Condition. The candle was placed at a distance of 15 cm from the collecting lens. Its optical power is 10 diopters. The screen behind the lens is positioned so that it produces a clear image of the candle. What is this distance?

Solution. The following data should be recorded in a short entry:d= 15 cm = 0.15 m,D= 10 diopters The formula derived above needs to be written with a slight modification. Namely, on the right side of the equality we putDinstead of 1/F.

After several transformations, we obtain the following formula for the distance from the lens to the image:

f= d/ ( dD- 1).

Now you need to plug in all the numbers and count. This results in a value forf:0.3 m.

Answer: the distance from the lens to the screen is 0.3 m.

Problem about the distance between an object and its image

Condition. The object and its image are 11 cm apart from each other. A converging lens gives a magnification of 3 times. Find its focal length.

Solution. It is convenient to denote the distance between an object and its image by the letterL= 72 cm = 0.72 m. Increase G = 3.

There are two possible situations here. The first is that the object is behind the focus, that is, the image is real. In the second, there is an object between the focus and the lens. Then the image is on the same side as the object, and it is imaginary.

Let's consider the first situation. The object and the image are on opposite sides of the converging lens. Here you can write the following formula:L= d+ f.The second equation is supposed to be written: Г =f/ d.It is necessary to solve the system of these equations with two unknowns. To do this, replaceLby 0.72 m, and G by 3.

From the second equation it turns out thatf= 3 d.Then the first one is converted like this: 0.72 = 4d.It's easy to count from itd = 0,18 (m). Now it's easy to determinef= 0.54 (m).

All that remains is to use the thin lens formula to calculate the focal length.F= (0.18 * 0.54) / (0.18 + 0.54) = 0.135 (m). This is the answer for the first case.

In the second situation, the image is imaginary, and the formula forLthere will be another:L= f- d.The second equation for the system will be the same. Arguing similarly, we get thatd = 0,36 (m), af= 1.08 (m). A similar calculation of the focal length will give the following result: 0.54 (m).

Answer: The focal length of the lens is 0.135 m or 0.54 m.

Instead of a conclusion

The path of rays in a thin lens is an important practical application of geometric optics. After all, they are used in many devices, from simple magnifying glasses to precision microscopes and telescopes. Therefore, it is necessary to know about them.

The derived thin lens formula allows solving many problems. Moreover, it allows you to draw conclusions about what kind of image different types of lenses produce. In this case, it is enough to know its focal length and the distance to the object.

Problem 1. At what distance is the focus of a thin lens from its optical center if the optical power of the lens is 5 diopters? At what distance would the focus be if the optical power was − 5 diopters? − 10 diopters? Given: Solution: Lens optical power:

Task 2. The picture shows an object. Construct its images for a converging and diverging lens. Based on the drawing, estimate the linear magnification of the lens. Solution:

Task 3. The image of an object was formed at a distance of 30 cm from the lens. It is known that the optical power of this lens is 4 diopters. Find the linear increase. Given: SI: Solution: Lens optical power: Thin lens formula: Then

Task 3. The image of an object was formed at a distance of 30 cm from the lens. It is known that the optical power of this lens is 4 diopters. Find the linear increase. Given: SI: Solution: Then Linear increase:

Problem 4. An image of an object located at a distance of 40 cm from the lens is formed at a distance of 30 cm from the lens. Find the focal length of this lens. Also find at what distance the object must be placed so that the image appears at a distance of 80 cm. Given: SI: Solution: Thin lens formula: Answer:

Problem 5. An object is located at a distance of 10 cm from a thin converging lens. If it is moved away from the lens by 5 cm, then the image of the object will be twice as close to the lens. Find the optical power of this lens. Given: SI: Solution: Thin lens formula: Lens optical power: Then

The main application of the laws of light refraction is in lenses.

What is a lens?

The word "lens" itself means "lentil".

A lens is a transparent body bounded on both sides by spherical surfaces.

Let's look at how a lens works on the principle of light refraction.

Rice. 1. Biconvex lens

The lens can be broken into several separate parts, each of which is a glass prism. Let's imagine the upper part of the lens in the form of a triangular prism: falling on it, light is refracted and shifted towards the base. Let us imagine all the following parts of the lens as trapezoids, in which a beam of light passes in and comes out again, shifting in the direction (Fig. 1).

Types of lenses(Fig. 2)

Rice. 2. Types of lenses

Converging lenses

1 - biconvex lens

2 - plano-convex lens

3 - convex-concave lens

Diffusing Lenses

4 - biconcave lens

5 - flat-concave lens

6 - convex-concave lens

Lens designation

A thin lens is a lens whose thickness is much less than the radii that bound its surface (Fig. 3).

Rice. 3. Thin lens

We see that the radius of one spherical surface and the other spherical surface is greater than the thickness of the lens α.

The lens refracts light in a certain way. If the lens is converging, then the rays are concentrated at one point. If the lens is divergent, then the rays are scattered.

A special drawing has been introduced to indicate different lenses (Fig. 4).

Rice. 4. Schematic representation of lenses

1 - schematic representation of a converging lens

2 - schematic representation of a diverging lens

Lens points and lines:

1. Optical center of the lens

2. Main optical axis of the lens (Fig. 5)

3. Focus lens

4. Lens power

Rice. 5. Main optical axis and optical center of the lens

The main optical axis is an imaginary line that passes through the center of the lens and is perpendicular to the plane of the lens. Point O is the optical center of the lens. All rays passing through this point are not refracted.

Another important point of the lens is the focus (Fig. 6). It is located on the main optical axis of the lens. At the focal point, all rays that fall on the lens parallel to the main optical axis intersect.

Rice. 6. Focus lens

Each lens has two focal points. We will consider an equifocal lens, that is, when the foci are at the same distance from the lens.

The distance between the center of the lens and the focus is called the focal length (segment in the figure). The second focus is located on the back side of the lens.

The next characteristic of a lens is the optical power of the lens.

The optical power of a lens (denoted by ) is the ability of a lens to refract rays. The optical power of the lens is the reciprocal of the focal length:

Focal length is measured in units of length.

For the unit of optical power, the unit of measurement chosen is one at which the focal length is equal to one meter. This unit of optical power is called a diopter.

For converging lenses, a “+” sign is placed in front of the optical power, and if the lens is diverging, then a “-” sign is placed in front of the optical power.

The unit of diopter is written as follows:

There is one more important concept for each lens. This is an imaginary trick and a real trick.

The actual focus is the focus formed by rays refracted in the lens.

An imaginary focus is a focus that is formed by continuations of rays passing through the lens (Fig. 7).

The imaginary focus, as a rule, is that of a diverging lens.

Rice. 7. Imaginary focus of the lens

Conclusion

In this lesson you learned what a lens is and what types of lenses there are. We got acquainted with the definition of a thin lens and the main characteristics of lenses and learned what an imaginary focus is, a real focus, and what is their difference.

References

1. Gendenshtein L.E., Kaidalov A.B., Kozhevnikov V.B. /Ed. Orlova V.A., Roizena I.I. Physics 8. - M.: Mnemosyne.
2. Peryshkin A.V. Physics 8. - M.: Bustard, 2010.
3. Fadeeva A.A., Zasov A.V., Kiselev D.F. Physics 8. - M.: Enlightenment.

1. Tak-to-ent.net ().
2. Tepka.ru ().
3. Megaresheba.ru ().

Homework

1. Task 1. Determine the optical power of a converging lens with a focal length of 2 meters.
2. Task 2. What is the focal length of a lens whose optical power is 5 diopters?
3. Task 3. Can a biconvex lens have negative optical power?