Telescopes and Optical Instruments - Class XII

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.

All the facts mentioned about Hubble Telescope are true. The Hubble Telescope orbits 353 miles above earth and transmits colorful images of objects in space and through Hubble telescope we can observe many things in space except the sun or mercury.

Computer-controlled polishiing machines is not a part of HST. This is used mostly in fabrication purposes. In fabrication process automation techniques like computer-controlled polishing is used to achieve high level of fabrication.

As the early drawback of Hubble space telescope, Hubble filed to achieve a final sharp focus and the best image quality obtained was drastically lower. Also the image of point sources spread out over a radius of more than one arc second.

The hubble space Telescope (HST) has accurately determined rate of expansion of the universe. It has contributed to studies of the behaviour of graphite composite structures in vacuum and has observed the optical contamination from residual gases.

The average speed of Hubble space telescope is $17000$ $mph$ or $5$ miles per second. Hubble space telescope takes 95 minutes to complete one orbit.

The Hubble space telescope was launched in 1990 to observe stellar objects It has gathered spectacular images and has helped advanced knowledge in the field of cosmology. After it immense contribution it retired in the year 2014.

Hubble data was initially stored on the spacecraft in reel-to-reel tape recorders but these were replaced by solid state data storage facilities during serving missions 2 and 3A. Hubble transmits about 140 gigabits of raw science data every week.

The Corrective Optics Space Telescope Axial Replacement(COSTAR) system was designed to correct the spherical aberration for light focused at the FOC, FOS and GHRS. Spherical aberration is the condition when there is a loss in the definition of the image arising from the surface geometry of a spherical mirror or lens. Hence to solve this  condition COSTAR was designed.

Astronomical data taken with CCD (charged couple devices) must undergo several calibration steps before they are suitable for astronomical analysis.

The unique windows on the universe enabled by Hubble are Hubble deep field, Hubble Ultra-Deep Field and Extreme Deep Field.

All images from Hubble are monochromatic gray scale, in which its cameras incorporate a variety of filters each sensitive to specific wavelengths of light.

The GHRS was a spectrograph designed to operate in ultraviolet light. It was installed on the Hubble Space Telescope during its original constructing and it was launched into space as a part of the space telescope abroad the space shuttle on April 24,1990.

To fit the COSTAR system onto the telescope one of the other instruments had to be removed, and astronomer selected the High Speed Photometer (HSP) to be sacrificed.

The instruments of hubble that analyses light from the universe such as cameras, spectrographs and inter ferometers helps in observing the near UV (ultraviolet range) to IR (infrared range).

Clearly Telescope instrument used for seeing details of celestial bodies.

Hubble space telescope is reflecting telescope it is designed with TWO hyperbolic mirrors. A curved primary mirror is the basic optical element that creates an image at the focal plane. And secondary mirror was added to modify the optical characteristics for visual observation.

We know that Earth has water, blue planet, 71% of Earth's surface is covered in water. Hubble telescope was definitely not the first to found the traces of water on Earth. 

Magnifying power of a telescope relates to the angular magnification. It makes the object appear closer by a factor equal to the magnifying power, effectively reducing the perceived distance.

For a telescope where the final image is at the least distance of distinct vision (D = 25 cm), the magnification M = f_o / f_e * (1 + f_e / D). Plugging in f_o = 200, f_e = 5, and D = 25 gives M = (200/5) * (1 + 5/25) = 40 * 1.2 = 48.