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The sum of the focal lengths of the objective and an eyepiece, In case of an astronomical telescope is equal to : ( final image is at $\infty$)

physics observing space: telescopes optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

  1. The length of the telescope

  2. Half the length of the telescope

  3. Double the length of the telescope

  4. None of these

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Correct Option: A
Explanation:

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.

By which instrument we collect the space information:

physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

  1. Convex lens

  2. Microscope

  3. Hubble Telescope

  4. None of these

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Correct Option: C
Explanation:

Hubble Telescope is a device used to collect space information.

The focal lengths of the objective and the eyepiece of an astronomical telescope are 20 cm and 5 cm respectively. If the final image is formed at a distance of 30 cm from the eyepiece, find the separation between the lenses required for distinct vision

physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

  1. 32.4 cm

  2. 42.3 cm

  3. 24.3 cm

  4. 30.24 cm

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Correct Option: C
Explanation:

$f _{0}=20cm$


$f _{e}=5cm$

$V _{e}=30cm = D$

$L=?$

$L _{D}=f _{0}+ \dfrac{Df _e}{D+f _e}$ $=20+\dfrac{5\times 30}{35}=24.3 cm$

The focal length of the objective of an astronomical telescope is 1 m and it is in normal adjustment.Initially the telescope is focussed to a heavenly body. If the same telescope is to be focussed to an object at a distance of 21 m from the objective,then identify the correct choice

physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

  1. eye piece should be displaced by 2 cm away from the objective

  2. eye piece should be displaced by 2 cm towards the objective

  3. eye piece should be displaced by 5 cm towards the objective

  4. eye piece should be displaced by 5 cm away from the objective

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Correct Option: D
Explanation:

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


We seeing heavenly body,  $u =  \infty$
$v = f = 1 m$

When seeing 21 m far
$u = - 21 m $
$f = 1 m$
$v = \dfrac {f u} {(f+u)} = 21 / 20 = 1.05 m$

So, eye piece need to move $1.05-1 = 0.05 m$ further away from objective

Answer. D) eye piece should be displaced by $5 cm$ away from the objective

The focal lengths of the objective and eyepiece of a telescope are 60cm and 5cm respectively.The telescope is focused on an object 360cm from the objective and the final image is formed at a distance of 30cm from the eye of the observer. The length of the telescope is

physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

  1. 66.3 cm

  2. 86.3 cm

  3. 76.3 cm

  4. 96.3 cm

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Correct Option: C
Explanation:

$f _o = 60cm$


$f _e = 5cm$

$r _e = 30cm$

$L = ?$

$u _o = 360cm$

$\dfrac {1}{v _o} + \dfrac {1}{360} = \dfrac {-1}{60}$

$\dfrac {1}{v _o} = \dfrac {-1}{60} + \dfrac {1}{360}$

In which of the following instruments is the final image virtual?

physics observing space: telescopes optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

  1. Projector

  2. Camera

  3. Microscope

  4. Telescope

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Correct Option: C,D
Explanation:

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 planet is observed by an astronomical reflecting telescope having an objective of focal length $16 m$ and an eye-piece of focal length $2 cm$. Then :

physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

  1. the distance between the objective and the eye - piece is $16.02 m$

  2. the angular magnification of the planet is $800$

  3. the image of planet is erect

  4. the objective is larger than eye - piece

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Correct Option: A,B,D
Explanation:

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 $


Since, 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, an inverted, virtual and enlarged image of the object is formed. $ \therefore $ The final image is inverted. 

Hence, the correct answers are OPTIONS A,B and D.

An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of

physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

  1. large focal length and small diameter

  2. large focal length and large diameter

  3. small focal length and large diameter

  4. small focal length and small diameter

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Correct Option: B

Which of the following is not a fact about Hubble Telescope?

physics universe and space science xx radio telescope and space telescopes launching vehicles optical instruments : telescope

  1. The Hubble Space Telescopes captures and transmits colorful images of objects in space such as dying stars and other galaxies.

  2. Although the Hubble Space Telescope can observe many things in space, it cannot observe the Sun or Mercury.

  3. The Hubble Space Telescope orbits about 353 miles above Earth

  4. None of the above

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Correct Option: D
Explanation:

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.

Which of the following is not a part of HST?

physics universe and space science xx radio telescope and space telescopes launching vehicles optical instruments : telescope

  1. Advanced Camera for Surveys (ACS).

  2. Faint Object Camera (FOC)

  3. Computer-controlled polishing machines

  4. Faint Object Spectrograph (FOS).

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Correct Option: C
Explanation:

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.