Tag: space travel

Questions Related to space travel

The length of an astronomical telescope for normal vision (relaxed eye) will be:

physics observing space: telescopes space research and satellites space travel space exploration and forms of light

  1. $f _0 - f _e$

  2. $f _0 / f _e$

  3. $f _0 \times f _e$

  4. $f _0 + f _e$

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Correct Option: D

The focal length of the objective of a terrestrial telescope is $80cm$ and it is adjusted for parallel rays, then its power is $20$. If the focal length of erecting lens is $20cm$, then full length of the telescope will be

physics observing space: telescopes space research and satellites space travel space exploration and forms of light

  1. $164cm$

  2. $124cm$

  3. $100cm$

  4. $84cm$

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Correct Option: A
Explanation:

Magnification for parallel rays
$m=\cfrac { { f } _{ o } }{ { f } _{ e } } $
$\Rightarrow 20=\cfrac { 80 }{ { f } _{ e } } $
or ${ f } _{ e }=4cm$
If the focal length of erecting lens is $20cm$ then the length of the telescope
${ L } _{ \infty  }={ f } _{ o }+4f+{ f } _{ e }\quad $
[where $f$ is the focal length of erecting lens]
$=80+4\times 20+4=164cm$

An astronomical refractive telescope has an objective of focal length 20 m and an eyepiece of focal length 2 cm. then

physics observing space: telescopes space research and satellites space travel space exploration and forms of light

  1. the magnification is 1000

  2. the length of the telescope tube is 20.02 m

  3. the image formed is inverted

  4. all of these

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Correct Option: D
Explanation:

Here, $f _0 = 20 \,m \,\, and \,\, f _e = 2 \, cm = 0.02 \, m$ In normal adjustment, Length of telescope tube, $L = f _0 + f _e = 20 +0.02 = 20.02m$
and magnification, $m = \dfrac{f _0}{f _e} = \dfrac{20}{0.02} = 1000$
The image formed is inverted with respect to the object.