Tag: space travel

Quizzes (7) Questions (83)

Questions Related to space travel

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

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

  1. $f _0 - f _e$
  2. $f _0 / f _e$
  3. $f _0 \times f _e$
  4. $f _0 + f _e$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a relaxed eye, the image is formed at infinity, which corresponds to the normal adjustment of the telescope where the length is the sum of the focal lengths of the objective and eyepiece.

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

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

  1. $164cm$
  2. $124cm$
  3. $100cm$
  4. $84cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
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]
$=80+4\times 20+4=164cm$

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

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.