Tag: observing space: telescopes

Questions Related to observing space: telescopes

An astronomical telescope has an eye piece of focal length $5\ cm$. If magnification produced is 14 in normal adjustment, then calculate the length of the telescope.

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

  1. $75\ cm$

  2. $9\ cm$

  3. $50\ cm$

  4. $55\ cm$

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Correct Option: A

A narrow vertical slit of width 2 mm is placed in front of a telescope.This setup is used to observe a car with its head light 1.2 m apart. The diameter of the objective of the telescope is 2 cm and the wavelength of light from the headlights is $ 5000 \mathring { A }  $. The distance of the car when the two headlights of the car are just resolved is:-

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

  1. 1.5 Km

  2. 2.6 Km

  3. 4.8 Km

  4. 3.93 Km

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Correct Option: A

The ratio of resolving power of telescope, when lights of wavelength $4400\overset{o}{A}$ and $5500\overset{o}{A}$ are used, is _________.

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

  1. $16:25$

  2. $4:5$

  3. $9:1$

  4. $5:4$

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Correct Option: D
Explanation:

Resolving power $\infty \dfrac{1}{\lambda}$
$\dfrac{(R.P.) _1}{(R.P.) _2}=\dfrac{\lambda _2}{\lambda _1}=\dfrac{5500}{4400}=\dfrac{5}{4}$.

The diameter of the moon is $3.5\times 10^{3} km$ and its distance from the earth is $3.8\times 10^{5} km$. The diameter of the image of the moon seen by a telescope having focal length of the objective and eye-piece as $400\ cm$ and $10\ cm$ respectively will be

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

  1. $11^{\circ}$

  2. $21^{\circ}$

  3. $31^{\circ}$

  4. $41^{\circ}$

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Correct Option: B

The magnification produced by an Astronomical telescope in normal

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

  1. $f _{0}+f _{e}$

  2. $f _{0}\times f _{e}$

  3. $\dfrac{f _{ 0} }{ f _{e}}$

  4. $\dfrac{f _{e} }{ f _{0}}$

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Correct Option: C

A telescope has a objective of focal length $60\, cm$ and an eye-piece of focal length $5\, cm$. The least distance of distinct vision is $25\, cm$. The telescope is focussed for distinct vision on a scale of $300\, cm$ away. The separation between the objective and the eye-piece is

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

  1. $71\, cm$

  2. $67\, cm$

  3. $83\, cm$

  4. $79\, cm$

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Correct Option: A

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.