# What is the name of binoculars with a lens

How is a simple lens telescope (Kepler's telescope) constructed?

In its simplest form, the so-called Kepler telescope consists of two converging lenses (or - to avoid aberrations - lens combinations that each act like a converging lens), the objective (which faces the observed object) and the eyepiece through which the observer looks .

What kind of images do diverging lenses produce? What is the virtual focus of a diverging lens?

In contrast to converging lenses, because of their concave curvature, diverging lenses widen an incident light bundle, to a certain extent "diffuse" the light bundle. Light rays that hit a diverging lens parallel to the optical axis are deflected as if they came from one and the same point on the optical axis. In analogy to the focal point of a converging lens, this point is called virtual focus of the diverging lens. Each diverging lens has two virtual focal points equidistant from the lens. Since the two focal points are on the opposite side as in the case of a converging lens, the focal length of a diverging lens is given a negative sign.

The same can be used for the graphic representation of an image through a diverging lens Main rays use as with a converging lens. These rays belong to a bundle of light that emanates from a point on the object. However, the beam paths differ for both lens shapes - except for the central beam, which goes through the lens with both lens types without being deflected:

With a diverging lens, a parallel beam runs behind the lens as if it came from the virtual focal point. A focal point beam is directed to the virtual focal point lying on the other side and runs behind the lens parallel to the optical axis. These principles can be used to construct images with diverging lenses:

The image produced by a diverging lens can only be observed if one looks into the lens, so it is virtual, also scaled down and - like all virtual images - upright.

A diverging lens usually creates a virtual, upright and reduced image of an object.

Annotation: With the help of the relationship obtained from the lens formula for the image distance as a function of the object and focal length, it can be seen that the image must be virtual. The following applies: b = g · f / (g - f);
Since the focal length f of a diverging lens is negative, the numerator g · f of the fraction becomes negative. The denominator is always positive, so the quotient value is negative overall. This means that the image distance is negative and the image is therefore on the same side as the object - which only applies to virtual images.

A candle with a flame 5 cm high stands 30 cm in front of a lens with a focal length of f = - 10 cm. What can be said about the image of the candle flame created by the lens?

Since the focal length of the lens is negative, it is a diverging lens, so the image is virtual (it cannot be made visible on a screen), but can only be seen when looking into the lens; it is scaled down and upright. Image width and image size result from the relationships known from the converging lens
b = gf / (g - f)
Image width: b = g f / (g - f)
b = 5 cm · (-10 cm) / [30 cm - (-10 cm)]
b = -150 cm² / 40 cm = -7.5 cm
Image size: B / G = b / g; B = G x (b / g) = 5 cm x (-7.5 cm: 30 cm) = -1.25 cm

The minus sign means that it is a virtual, upright image. The diverging lens creates a 1.25 cm high, upright virtual image of the candle flame at a distance of 7.5 cm in front of the lens.

Can a diverging lens produce a real image of an object?

This is possible in conjunction with a converging lens.
Example: The sun is imaged with a converging lens with high refractive power. Your image lies in the focal plane of the lens. If you hold a diverging lens of lower refractive power between the converging lens and its focal point, you get a real image of the sun at a greater distance. The diverging lens can indeed expand the strongly convergent, incident light bundle, but it remains convergent, so that the light rays falling through the diverging lens intersect behind the lens despite their diverging effect.

A lens creates an image of an object 2.50 m away at 50 cm from the lens on the other side. Is the picture real or virtual? Calculate the focal length of the lens!

It is a converging lens, since the object and the image are on different sides of the lens, so the image distance b is positive. The focal length f of the lens results from the lens formula
1 / f = 1 / g + 1 / b
1 / f = 1/250 cm + 1/50 cm = 0.024 cm
f = 1 / 0.024 cm = 41.7 cm

You want to build a telescope out of the lenses of two reading glasses with a power of +2 and +5 diopters.
Which of the two lenses should you use as a lens? What is the maximum magnification that a telescope constructed in this way could achieve?

Note: With the help of the diopter values ​​(which correspond to the refractive power D) you can calculate the focal lengths according to the formula: f = 1 / D m.

Since the refractive powers of the two lenses are marked with a +, both lenses are convergent lenses (the glasses were therefore intended for farsighted people). The specified refractive powers in diopters show the focal lengths of the two lenses:

Lens 1: +2 diopters correspond to a focal length of f = ½ m = 50 cm
Lens 2: +5 diopters correspond to a focal length of f = 1/5 m = 20 cm

The lens with the larger focal length, i.e. the one with a focal length of 50 cm, is used as the lens.

The (maximum) magnification v that can be achieved with a telescope results from the focal lengths of the objective and the eyepiece:

v = flens / feyepiece (for converging lenses)

In our case, the magnification is: v = 50 cm / 20 cm = 2.5, that is: The telescope composed of the spectacle lenses has a (maximum) magnification of the factor 2.5.

Why are powerful refractors (lens telescopes) - like the telescope shown in the film with which the planet Neptune was discovered - exceptionally long?

The magnification of a refractor is greater, the greater the focal length of the objective and the smaller that of the eyepiece. The length of the refractor is equal to the sum of the focal lengths of the two lenses, since the focal points of the objective and eyepiece must coincide for observation with "relaxed eyes" (the observer then sees the image "at infinity"). A high magnification is achieved above all with a large focal length of the lens. The focal length of the eyepiece must not be too small, as otherwise the image quality would deteriorate. Powerful refractors have lens focal lengths of up to a few meters and are therefore correspondingly long.

How do the images in a Galilean telescope differ from those produced by a Kepler telescope?

A Kepler's telescope provides reversed images. The opposite orientation of image and object does not interfere with astronomical observations, but it does interfere with observing objects on earth, since the image of the object observed through the eyepiece is upside down.

Annotation: With a trick, the Kepler telescope can also produce upright images. To do this, a third converging lens, known as a Field lens build in. When observing the earth, the distance between the object being viewed and the lens is so great that the real intermediate image appears upside down in its focal plane. Now it is necessary to position the field or erecting lens in such a way that the reversal is canceled: This is the point at twice the distance of the focal length of the field lens (measured from the focal point of the objective). If the field lens is placed at this point, it maps the real intermediate image onto an upright real image of the same size that can be seen through the eyepiece. By reversing the image twice, the end image is upright. Telescopes with an erecting lens were used, for example, as opera glasses before the development of prismatic telescopes.

Galilean telescope

A Galilean telescope using a diverging lens as an eyepiece does not have this disadvantage: it is installed between the objective and its focal point, so that the diverging lens catches the beams bundled by the objective in front of their merging point and thus directly creates an upright virtual image of the object.