Thin lens: the formula and the derivation of the formula. Solving problems with the thin lens formula

Now we are talking about geometric optics. In this section, much time is devoted to an object such as a lens. After all, it can be different. In this case, the formula of a thin lens is one for all cases. Only you need to know how to apply it correctly.

Types of lenses

It is always a body transparent to light rays, which has a special shape. The appearance of the object dictates two spherical surfaces. One of them can be replaced by a flat one.

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

Under normal conditions, these objects are used in the air. They are made from a substance whose optical density is greater than that of air. Therefore, the convex lens will be collecting, and the concave lens will dissipate.

General characteristics

формуле тонкой линзы , нужно определиться с основными понятиями. Before talking about the formula of a thin lens , you need to decide on the basic concepts. They must be known. Because they will constantly handle a variety of tasks.

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

In the formula of a thin lens there is a quantity that determines its focal length. So, the focus is the point on the main optical axis. It intersects rays that run parallel to this axis.

And the foci of every thin lens is always two. They are located on both sides of its surfaces. Both focuses have a valid focus. In scattering - imaginary.

F ) . The distance from the lens to the point of focus is the focal length (letter F ) . And its value can be positive (in the case of collecting) or negative (for scattering).

With the focal length, another characteristic is associated: the optical force. D. Ее значение всегда - величина, обратная фокусу, то есть D = 1/ F. Измеряется оптическая сила в диоптриях (сокращенно, дптр). It is usually denoted D. Its value is always the inverse of the focus, that is, D = 1 / F. The optical force in diopters is measured (abbreviated, dpt).

What other notation is in the formula of a thin lens

In addition to the already indicated focal length, it will be necessary to know several distances and dimensions. For all types of lenses they are the same and are presented in the table.

Designation	Name
D	Distance to the object
H	Height of the studied subject
F	Distance to image
H	Height of the resulting image

All specified distances and heights are usually measured in meters.

In physics with the formula of a thin lens, there is also the notion of magnification. . It is defined as the ratio of the image dimensions to the height of the object, that is, H / h . It can be denoted by G.

What you need to build an image in a thin lens

It is necessary to know this in order to obtain the formula of a thin lens that collects or diffuses. The drawing begins with the fact that both lenses have their own schematic image. Both of them look like a piece. Only at the arrows that gather at its ends are directed outward, and at the scattering points - inside this segment.

Now to this segment it is necessary to draw a perpendicular to its middle. This will show the main optical axis. On it on both sides of the lens at the same distance, it is necessary to note foci.

The object, the image of which is 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 build an image in a thin lens

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

• Going from this point is parallel to the main optical axis. After contact with the lens, it goes through the main focus. If we are talking about a collecting lens, then this focus is behind the lens and the beam goes through it. When scattering is considered, the beam must be drawn 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 for her.

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

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

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

The derivation of the formula of a thin lens

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

The optics section is not for nothing called geometric. Knowledge will be required from this section of mathematics. ₁ ОВ ₁ . First, we need to consider the triangles AOB and A ₁ OB ₁ . They are similar, because they have two equal angles (straight and vertical). ₁ В ₁ и АВ относятся как модули отрезков ОВ ₁ и ОВ. From their similarity it follows that the modules of the segments A ₁ B ₁ and AB are referred to as the modules of the segments OB ₁ and OB.

COF и A ₁ FB ₁ . Similar (on the basis of the same principle at two angles) are two more triangles: COF and A ₁ FB ₁ . ₁ В ₁ с СО и FB ₁ с OF. In them, the ratios of the already modulated segments are equal: A ₁ B ₁ with CO and FB ₁ with OF. Proceeding from the construction, the segments AB and CD will be equal. Therefore, the left-hand sides of these relations are the same. Therefore, the right are equal. ₁ / ОВ равно FB ₁ / OF. That is, the OB ₁ / OB is FB ₁ / OF.

In this equality, the segments denoted by the points can be replaced by the corresponding physical concepts. ₁ — это расстояние от линзы до изображения. So OB ₁ is the distance from the lens to the image. OB is the distance from the subject to the lens. фокусное расстояние. OF is the focal length. FB ₁ равен разности расстояния до изображения и фокуса. And the segment FB ₁ is equal to the distance difference between the image and the focus. Therefore, it can be rewritten in a different way:

( f – F ) / F или Ff = df – dF. F / d = ( f - F ) / F or Ff = df - dF.

dfF. To derive the formula of a thin lens, the last equality must be divided into dfF. Then it turns out:

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

This is the formula of a thin collecting lens. In scattering, the focal length is negative. This leads to a change in the equality. True, it is insignificant. F. То есть: Just in the formula of a thin diffusing lens there is a minus before the ratio 1 / F. That is:

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

The problem of finding the lens magnification

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

Decision. It begins with the introduction of notation and the translation of units in C. d = 30 см = 0,3 м и F = 0,26 м. Теперь нужно выбрать формулы, основная из них та, которая указана для увеличения, вторая — для тонкой собирающей линзы. So, we know d = 30 cm = 0.3 m and F = 0.26 m. Now we need to choose the formulas, the main one is the one indicated for magnification, the second one for the thin collecting lens.

They need to somehow unite. To do this, we will have to consider the drawing of the image construction in the collecting lens. = f/d. From these triangles we see that Γ = H / h = f / d. That is, in order to find an increase, it is necessary 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 is supposed to be derived from the formula indicated earlier. It turns out that

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

Now these two formulas must be combined.

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

At this point, the solution of the problem for the thin lens formula reduces to elementary calculations. It remains to substitute the known quantities:

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

Answer: The lens gives an increase of 6.5 times.

The task in which to find the focus

Condition. The lamp is located one meter from the collecting lens. The image of its spiral is obtained on the screen, separated from the lens by 25 cm. Calculate the focal length of this lens.

Decision. d =1 м и f = 25 см = 0,25 м. Этих сведений достаточно, чтобы из формулы тонкой линзы вычислить фокусное расстояние. The data is supposed to record such values: d = 1 m and f = 25 cm = 0.25 m. This information is sufficient to calculate the focal length from the thin lens formula.

F = 1/1 + 1/0,25 = 1 + 4 = 5. Но в задаче требуется узнать фокус, а не оптическую силу. So 1 / F = 1/1 + 1 / 0.25 = 1 + 4 = 5. But in the task you need to know the focus, not the optical power. Therefore, it remains only to divide 1 by 5, and we get the focal length:

1/5 = 0, 2 м. F = 1/5 = 0, 2 m.

Answer: The focal length of the collecting lens is 0.2 m.

The problem of finding the distance to the image

Condition . The candle was placed at a distance of 15 cm from the collecting lens. Its optical strength is 10 dpt. The screen behind the lens is set so that it produces a clear image of the candle. What is this distance equal to?

Decision. d = 15 см = 0,15 м, D = 10 дптр. In the short record it is supposed to record such data: d = 15 cm = 0.15 m, D = 10 dpt. The formula derived above should be written with a slight change. D вместо 1/ F. Namely, in the right side of the equality, put D instead of 1 / F.

After several transformations, the following formula is obtained for the distance from the lens to the image:

= d / ( dD - 1). F = d / ( dD - 1).

Now it is necessary to substitute all the numbers and count them. f: 0,3 м. This is the value for f: 0.3 m.

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

The problem of the distance between an object and its image

Condition. The object and its image are separated by 11 cm. The collecting lens gives an increase of 3 times. Find its focal length.

Decision. L = 72 см = 0,72 м. Увеличение Г = 3. The distance between the object and its image is conveniently denoted by the letter L = 72 cm = 0.72 m. The increase in T = 3.

Here two situations are possible. The first - the object is behind the focus, that is, the image is real. In the second, the subject is between the focus and the lens. Then the image on the same side as the object, and imaginary.

Consider the first situation. The object and the image are on different sides of the collecting lens. L = d + f. Here we can write the following formula: L = d + f. f / d. The second equation is to write: Γ = f / d. It is necessary to solve the system of these equations with two unknowns. L на 0,72 м, а Г на 3. To do this, replace L by 0.72 m, and D by 3.

f = 3 d. It follows from the second equation that f = 3 d. d. Then the first is transformed so: 0,72 = 4 d. d = 0, 18 (м). From this it is easy to calculate d = 0, 18 (m). f = 0,54 (м). Now it is easy to determine f = 0.54 (m).

It remains to use the formula of a thin lens to calculate the focal length. = (0,18 * 0,54) / (0,18 + 0,54) = 0,135 (м). F = (0.18 * 0.54) / (0.18 + 0.54) = 0.135 (m). This is the answer for the first case.

L будет другой: L = f - d. In the second situation, the image is imaginary, and the formula for L is different: L = f - d. The second equation for the system will be the same. d = 0, 36 (м), а f = 1,08 (м). Similarly, arguing, we get that d = 0, 36 (m), and f = 1.08 (m). Such a 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 concluding

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

The derived formula of a thin lens allows us to solve many problems. And it allows you to draw conclusions about what image gives different types of lenses. It is sufficient to know its focal length and distance to the object.