Tag: space research and satellites

Questions Related to space research and satellites

In which of the following instruments is final image erect?

  1. Simple mircoscope

  2. Compound mircoscope

  3. Astronimical telescope

  4. None of the above


Correct Option: A
Explanation:

A simple microscope is magnifying glass, an ordinary double convex lens having a short focal length that produces virtual and erect image.

rather a Compound microscope and astronomical telescope produces virtual and final inverted image. hence correct option is A.

Large aperture objective is used in telescope. Which of the following is not a function of the objective?

  1. Increasing the brightness of image.

  2. Reducing image size.

  3. Increasing field of view.

  4. Increasing intensity by gathering more light.


Correct Option: B
Explanation:

Large aperture objective is used in telescope as it can help in increasing the brightness of image, increasing field of view and increasing intensity by gathering more light.

A gain telescope in an observatory has an objective of focal length $19\ m$ and an eye-piece of focal length $1.0\ cm$. What is the diameter of the image of moon formed by the objective in normal adjustment? The diameter of moon is $3.5\times {10}^{6}\ m$ and the radius of the lunar orbit round the earth is $3.8\times {10}^{8}\ m.$

  1. $10\ cm$

  2. $12.5\ cm$

  3. $15\ cm$

  4. $17.5\ cm$


Correct Option: D

A telescope of objective lens diameter $2m$ uses light of wavelength $5000 \mathring {A}$ for viewing starts. The minimum angular separation between two stars whose image is just resolved by their telescope is:

  1. $4\times 10^{-4}rad$

  2. $40.25times 10^{-6}rad$

  3. $0.31\times 10^{-6}rad$

  4. $5\times 10^{-3}rad$


Correct Option: C
Explanation:

Given that,

Diameter $d=2\,m$

Wave length $\lambda =5000\,\overset{\circ }{\mathop{A}}\,$

Now, minimum angular separation is

  $ \Delta \theta =\dfrac{1.22\lambda }{d} $

 $ \Delta \theta =\dfrac{1.22\times 5000\times {{10}^{-10}}}{2} $

 $ \Delta \theta =0.3\times {{10}^{-6}}\,rad $

Hence, the resolving power is $0.3\times {{10}^{-6}}\,rad$

A tower $100m$ tall at a distance of $3$km is seen through a telescope having objective of focal length $140$cm and eyepiece of focal length $5cm$. Then the size of final image if it is at $25$cm from the eye?

  1. 14 cm

  2. 28 cm

  3. 42 cm

  4. 56 cm


Correct Option: A

The angular resolution of a radio telescope is to be ${ 0.100 }^{ 0 }$ when the incident beam of wavelength $ 3.00mm$ is used. What is minimum diameter required for the telescope's receiving dish? 

  1. 2.0 m

  2. 4.20 m

  3. 2.20 m

  4. 3.20 m


Correct Option: A

The focal lengths of the objective and eye lens of telescope are respectively $200\ cm$ and $5\ cm$. The maximum magnifying power of the telescope will be.

  1. $-40$

  2. $-48$

  3. $-60$

  4. $-100$


Correct Option: B

An observer looks at a distant tree
of height $10$ m with a
telescope of magnifying power of $20$. To the
observer the tree appears :






.







  1. $10$ times taller.

  2. $10$ times nearer

  3. $20$ times taller.

  4. $20$ times nearer.


Correct Option: D

The angular resolution of a $10cm$ diameter telescope at a  wavelength of $5000A$ is of the order of 

  1. $ 10^{6} \mathrm{rad} $

  2. $ 10^{-2} \mathrm{rad} $

  3. $ 10^{-4} \mathrm{rad} $

  4. $ 10^{-6} \mathrm{rad} $


Correct Option: D

In a terrestrial telescope, the telescope, the focal length of objective is 90 cm of inverting lens is 5 cm and of eye lens is 6 cm. If the final image is at 3  cm, then the magnification will be

  1. 21

  2. 12

  3. 15

  4. 18


Correct Option: C