Tag: space exploration and forms of light

Questions Related to space exploration and forms of light

The diameter of the lens of a telescope is 1.22 m., the wavelength of light is $5000{A^0}$ the resolution power of the telescope is 

  1. $2 \times {10^5}$

  2. $2 \times {10^6}$

  3. $2 \times {10^2}$

  4. $2 \times {10^4}$


Correct Option: B

If an object subtend angle of $2^0$ at eye when seen through telescope having objective and eyepiece of focal length $f _0=60cm$ and $f _e=5cm$ respectively than angle subtend by image at eye piece will be

  1. $16^0$

  2. $50^0$

  3. $24^0$

  4. $10^0$


Correct Option: C

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.

  1. $75\ cm$

  2. $9\ cm$

  3. $50\ cm$

  4. $55\ cm$


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:-

  1. 1.5 Km

  2. 2.6 Km

  3. 4.8 Km

  4. 3.93 Km


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 _________.

  1. $16:25$

  2. $4:5$

  3. $9:1$

  4. $5:4$


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

  1. $11^{\circ}$

  2. $21^{\circ}$

  3. $31^{\circ}$

  4. $41^{\circ}$


Correct Option: B

The magnification produced by an Astronomical telescope in normal

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

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

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

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


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

  1. $71\, cm$

  2. $67\, cm$

  3. $83\, cm$

  4. $79\, cm$


Correct Option: A

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. $f _0 + f _e$


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

  1. $164cm$

  2. $124cm$

  3. $100cm$

  4. $84cm$


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]
$=80+4\times 20+4=164cm$