Tag: space travel

Questions Related to space travel

The final image formed by an astronomical telescope is

  1. real and erect

  2. virtual and erect

  3. real and inverted

  4. virtual and inverted


Correct Option: D
Explanation:

An astronomical telescope is an optical instrument which is used to see the magnified image of distant heavenly bodies. The final image formed by an astronomical telescope is always virtual, inverted and magnified.

For the astronomical telescope, the focal length of objective lens is ${f} {o}$ and the eye piece lens is ${f} _{e}$. Then the tube length of the telescope is ____

  1. $L\ge { f } _{ o }-{ f } _{ e }$

  2. $L\ge { f } _{ o }+{ f } _{ e }$

  3. $L<{ f } _{ o }+{ f } _{ e }$

  4. $L\le { f } _{ o }-{ f } _{ e }$


Correct Option: B
Explanation:

For astronomical telescope
$\left| { v } _{ 1 } \right| ={ f } _{ o }$
$\left| { u } _{ 2 } \right| \le { f } _{ e }$
Probable answer would be (c) conceptually correct

In an astronomical telescope, the intermediate image is

  1. virtual, erect and magnified

  2. real, erect and magnified

  3. real, inverted and reduced

  4. virual, inverted and reduced


Correct Option: C
Explanation:

Astronomical telescope in which the image is inverted$,$ is n of the two principle types of the telescopes$.$ Its primary function is to enlarge the retinal image of a distant object$.$

hence,
option $(C)$ is correct answer.

Magnifying power of an astronomical telescope is 15. Then ratio of the focal length of the objective to the focal length length of the eye piece is

  1. $15$

  2. $\cfrac 1 {15}$

  3. $11.5$

  4. $None$


Correct Option: B

In a terrestrial telescope the focal length erecting lens is $20cm$. The length of the telescope $96cm$ . If the magnifying power of the telescope $10$. Then the focal length of eye -piece and objective are respectively 

  1. $8cm,80cm$

  2. $\dfrac{96}{11}cm,\dfrac{960}{11}cm$

  3. $6cm,90cm$

  4. None of these


Correct Option: D

The aperture of the largest telescope in the world is $5 m,$ if the separation between the Moon and the Earth is $4 \times 10^5 km$ and the wavelength of the visible light is $5000 \overset {o}{A}$ then the minimum separation between the objects on the surface of the Moon which can be just resolve is approximately

  1. $1 m$

  2. $10 m$

  3. $50 m$

  4. $200 m$


Correct Option: A

If an astronomical telescope has objective and eye-pieces of focal length 200 cm and 4 cm respectively,then the magnifying power of the telescope for the normal vision is:

  1. 42

  2. 50

  3. 58

  4. 204


Correct Option: C
Explanation:

 

It is given that,

Focal length of eye- piece fe = 4 cm

Focal length of object is fo = 200 cm

Least distance of distinct vision is d = 25 cm

So, magnifying power of microscope is

$ M=\dfrac{-{{f} _{0}}}{{{f} _{e}}}\left( 1+\dfrac{{{f} _{e}}}{d} \right) $

$ M=\dfrac{-200}{4}\left( 1+\dfrac{4}{25} \right) $

$ =-58\,cm $

 

Focal lengths of the objective lens and eye-piece of an astronomical telescope are $2m$ and $0.05m$. Find the length of telescope in normal adjustment.

  1. $1.05 \mathrm { m }$

  2. $1.16 \mathrm { m }$

  3. $2.05 m$

  4. $2 \mathrm { gm }$


Correct Option: A

The diameter of moon is $3.5\times{10}^{3}km$ and its distance from the earth is $3.8\times{10}^{5}km$. The focal length of the objective and eyepiece are $4m$ and $10cm$ respectively. The angle subtended by the diameter of the image of the moon will be approximately

  1. ${2}^{o}$

  2. ${20}^{o}$

  3. ${40}^{o}$

  4. ${50}^{o}$


Correct Option: B

The magnifying power an astronomical telescope for normal adjustment is -

  1. $- \frac{f _0}{f _e}$

  2. $-f _0 \times f _e$

  3. $- \frac{f _e}{f _0}$

  4. $-f _0 + f _e$


Correct Option: A