Subject: Physics

A telescope consists of two lenses, called the objective and eyepiece, separated by a distance. Both lenses are converging lenses but focal length of an objective is greater than that of the eyepiece.

The objective having larger aperture gathers as much light as possible from the distance object and produces a real inverted image of the object at its focus F_{o}. The eye piece magnifies this image producing a final image. The telescope is called in normal adjustment when the final image is formed at infinity. Of this adjustment, eye is completely relaxed and the image produced by objective acts as object for eyepiece that lies on the focus F_{e} of the eyepiece.

Let α be the angle subtended at the objective by the object. Since the object is very far, the angle subtended by the object at unaided eye is equal to the angle α. Let β be the angle subtended at the eyepiece by the final image. The angular magnification of the telescope is

\begin{align*} M &= \frac {\beta }{\alpha } \\ \text {Since angles are small,} \\ \alpha &= \frac {h}{f_o} \\ \beta &= \frac {h}{f_e} \\ \text {The angular magnification,} \: M &= \frac {\beta }{\alpha } \\ &= \frac {h}{f_e}/ \frac {h}{f_o} \\ &= \frac {f_o}{f_e} \\ \end{align*}

The separation between the lenses is f_{o} + f_{e}, the length of the telescope.

At this adjustment, the eyepiece is moved so that the image of the object is nearer than its focal length, f_{e}. The image is virtual, inverted and magnified that lies at the least distance of distinct vision, D from the eyepiece as shown in the figure.

The angle subtended at unaided eye by the object,\begin{align*}\alpha &= \frac {h}{f_o} \end{align*}and the angle subtended by the final image at the eyepiece, \begin{align*}\beta &= \frac {h}{u_e } \end{align*} where u_{e}the distance of final image from eyepiece.Angular magnification, \begin{align*}M &= \frac {\beta }{\alpha } \\ \frac {(h/ u_e)}{(h/f_o )} \\ &= \frac {f_o }{u_e } \end{align*}To express \begin{align*}\: u_e \: \text {in terms of D and} \: f_e, \end{align*}we have from the lens formula, \begin{align*}1/f = 1/u + 1/v \: \text {and} \: f = f_e, v = -D. \\ \therefore \frac {1}{f_e} &= \frac 1u - \frac 1D \\ \text {or,} \: \frac {1}{u_e} &= \frac {1}{f_e} - \frac 1D \\ \text {or,} \: \frac {1}{u_e} &= \left ( 1 + \frac {f_e}{D} \right ) \frac {1}{f_e} \\ \therefore M &= \frac {f_o}{f_e} \left ( 1 + \frac {f_e}{D} \right ) \\ \end{align*}

A terrestrial telescope produces a final erect image. This is obtained when a convex of suitable focal length between the objective and eyepiece of an astronomical telescope. The new lens is called the erecting lens as it only inverts the image. The ray diagram of a terrestrial telescope is shown in figure.

The objective forms the image AB of a distance object at the focus F_{o}. As the erecting lens lies at a distance 2f from AB, it forms as erect image XY at a distance 2f on the other side of the lens L. the image XY is erect with respect to distant object. Since the size XY of image produced by L is same as AB, the erecting lens does not change the magnifying power of the telescope. At normal adjustment,

\begin{align*} \: \text {angular magnification, M} &= \frac {f_o}{f_e} \\ \end{align*}and for image image at least distance of distinct vision,\begin{align*} \: \text {Angular magnification, M} &= \frac {f_o}{f_e} \left ( 1 + \frac {f_e}{D} \right ) \\ \end{align*}

The length of the telescope tube is f_{0} + 2f + 2f + f_{e} = f_{0} + f_{e} + 4f. So, length of tube increases by 4f in the terrestrial telescope.

A Galilean telescope forms an erect final image of a distance object with two lenses. In this telescope, the objective is a convex lens of focal length f_{o} and eye piece a concave lens of smaller focal length f_{e}. Light rays from a distance object fall on the objective which converges at focus F_{o }forming a real inverted image AB. The eye piece is so adjusted that the image AB lies at focus F_{e} of the eye-piece as shown in the figure and the final image is formed at the infinity.

In normal adjustment, angular magnification

\begin{align*} M &= \frac {\beta }{\alpha } \\ \text {and} \: \beta = \frac {h}{f_e} \: \text {and} \: \alpha = \frac {h}{f_o}, \text {so} \\ M &= \frac {\beta }{\alpha } = \frac {h/f_e}{h/f_o} = \frac {f_o}{f_e} \\ \end{align*}For higher magnification, an objective of higher focal length, f_{e}are required. For the final image at near point, we have\begin{align*}M &= \frac {f_o}{f_e} \left ( 1 - \frac {f_e}{D} \right ) \\ \end{align*}

The distance between two lenses is f_{o} – f_{e}. So Galilean telescope is shorter than terrestrial telescope but its field view is small.

The astronomical telescopes discussed above uses a set of lenses in which light is refracted. A reflecting telescope uses a curved mirror as an objective. Light reflected from it converged at the eyepiece with the help of the other mirrors and the telescope is free from chromatic aberration and spherical aberration, so, a brighter image is obtained in such telescope. There three types of reflecting telescope i.e Newtonian, Cassegrain and code arrangement.

The magnifying power of a reflecting telescope,

$$M = \frac {f_o}{f_e} $$

Modern telescopes are reflecting telescopes as the final image should be more intense.

A photographic camera consists of a converging lens system at one end of a light proof box and light-sensitive film at the other end. A real inverted image of the object is formed on the film. The amount of light entering the film is adjusted by changing the aperture of the lens. There is a shutter which opens and closes very fast and the film is exposed for a very short time to light entering through the lens.

**Exposure of time**

If the photograph is taken in bright light, the exposure of time is smaller and larger when the photograph is taken in dim light or indoors. The normal exposure times used in cameras are given below.

$$ 1/ 5000\: s, \: 1/250\: s, \: 1/125\: s, \:1/ 160\: s, \: 1/30 \: s $$

An f-number of f/2.8 means that diameter “d” of theaperture is f/2.8.

**Film speed**

The film speed is a measure of how fast the film will be exposed to light when it is in use. A fast film needs a relatively short time exposure while a slow speed film needs longer time. Fast films are used in poor lightning conditions and very slow films are used for still object photography.

**Exposure meter**

Many modern cameras have built-in exposure meters. There is a light sensitive surface in it. When the camera is exposed to an object, a current will flow in the meter which is proportional to the amount of light falling upon it. The meter indicates the exposure time and the aperture needed to take a good photo.

**Dept of focus**

When the aperture is larger, the image is not focused in a single plane, but it gets spread out. The depth of focus indicates the range of the object distance over which the focusing is reasonably good.

- A telescope consists of two lenses, called the objective and eyepiece, separated by a distance
- A terrestrial telescope produces a final erect image.
- The astronomical telescopes discussed above uses a set of lenses in which light is refracted.
- Modern telescopes are reflecting telescopes as the final image should be more intense.
- A photographic camera consists of a converging lens system at one end of a light proof box and light-sensitive film at the other end

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

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