Optical instruments : telescope - class-IX

The focal length of objective mirror $f _0 = \dfrac{R}{2} = \dfrac{80}{2} = 40 ,cm$
and focal length of eye piece = 1.6 cm
$\therefore $ magnifying power, $m = \dfrac{f _0}{f _e} = \dfrac{40}{1.6 } = 25$

Aperature of the telescope, $D = \dfrac{1.22\lambda}{d\theta}$ 

A refracting telescope uses two convex lenses: an objective lens to gather light and an eyepiece lens to magnify the image.

Quartz makes an ideal mirror. Its high hardness results in the smoothest micropolished surface possible, and its thermal dimensional stability allows the optician to produce a flat surface with almost no edge defects.

To make a Telescope,
$L= f _{0}+f _{e}$

$= 0.3+0.05$

$= 0.35m$

An astronomical telescope of refracting type has objective of greater focal length to have more surface area for incoming light from celestial bodies. 

In astronomical telescope, magnification is inversely proportional to focal length of eye piece. Hence, it will be minimum i.e. +15cm as in question.

Unlike convex mirrors, concave mirrors show different image types depending on the distance between the object and the mirror. That is why a reflecting telescope utilizes a concave mirror. If you observe the design of a reflecting telescope, it uses the incoming rays to focus through a concave mirror to focus on the second mirror which in turn sends the rays to the eye-piece. So, the concave mirror is intrinsic to the design of a reflecting telescope.

Large telescopes are reflecting telescopes because mirrors are easier to support from the back, whereas lenses must be supported at the edges, which causes sagging in large sizes.

The amount of light gathered by a telescope objective is proportional to its area, which scales with the square of its diameter (A proportional to D^2). Since the diameter of telescope A is 3 times that of telescope B, the light gathering power is 3^2 = 9 times greater.

The magnifying power of a telescope in normal adjustment is given by m = f_o / f_e = 60 / 5 = 12. The angle subtended by the image at the eyepiece is equal to the magnification times the angle subtended by the object at the objective, so beta = m * alpha = 12 * 2 degrees = 24 degrees.

A). Astronomical telescope - Contains a convex lens of large focal length in the objective and a convex lens of small focal length in the eye-piece

Answer is A.

The astronomical telescope makes use of two positive lenses: the objective, which forms the image of a distant object at its focal length, and the eyepiece, which acts as a simple magnifier with which to view the image formed by the objective. Its length is equal to the sum of the focal lengths of the objective and eyepiece, and its angular magnification is -fo/fe , giving an inverted image.
Hence, the sum of the focal lengths of the objective and an eyepiece, In case of an astronomical telescope is equal to the length of the telescope.

Hubble Telescope is a device used to collect space information.

$f _{0}=20cm$

We know, $ \dfrac {1}{f} = \dfrac {1}{v} - \dfrac {1} {u} $

$f _o = 60cm$

The final image of the camera and projector is real because it is obtained or projected on a screen or a photographic film or a chip.

A simple microscope uses convex lens and the object is placed between the focus and the optical center, so that, an erect, virtual and enlarged image of the object is formed.

A telescope uses two co-axially placed convex lenses in such a way that the focus of objective lens is past the focus of the eye piece. The first lens or the objective lens produces a real and inverted image of the object to be observed and this real image formed acts as an object for the eye piece convex lens and is between the focus and the optical center, so that, an inverted, virtual and enlarged image of the object is formed.

Hence, the correct answers are OPTIONS C and D.  

A telescope uses two
co-axially placed convex lenses in such a way that the focus of objective lens
is past the focus of the eye piece as evident here from the focal lengths of the objective lens and the eye piece. $ \therefore $ The objective lens is larger than eye piece.

Length of the tube is given as 
$ L = f _o + f _e = 16\ m + 2\ cm = 16.02\ m $

Angular magnification is given as:
$ m = \dfrac{f _o}{f _e} = \dfrac{1600}{2} = 800 $

Angular magnification is proportional to the focal length of the objective, while angular resolution is proportional to the diameter of the objective. Therefore, both a large focal length and a large diameter are required.