Tag: option c: imaging

Questions Related to option c: imaging

In an astronomical telescope of refracting type:

  1. Eyepiece has greater focal length

  2. Objective has greater focal length

  3. Objective and eyepiece have equal focal length

  4. Eyepiece has greater aperture than the objective


Correct Option: B
Explanation:

An astronomical telescope of refracting type has objective of greater focal length to have more surface area for incoming light from celestial bodies. 

Four lenses of focal length $+15cm,+20cm,+150cm$ and $+250cm$ are available for making an astronomical telescope. To produce the largest magnification, the focal length of the eyepiece should be

  1. $+15cm$

  2. $+20cm$

  3. $+150cm$

  4. $+250cm$


Correct Option: A
Explanation:

In astronomical telescope, magnification is inversely proportional to focal length of eye piece. Hence, it will be minimum i.e. +15cm as in question.

The objective of a telescope A has diameter 3 times that of the objective of telescope B, How much greater amount of light is gathered by A as compared to B?

  1. $\dfrac { 1 }{ 9 } $

  2. $\dfrac { 1 }{ 3 } $

  3. 3

  4. 9


Correct Option: A

In telescope, ratio of resolving power due to light of $\lambda=400\mathring {A}$ and $\lambda =6000\mathring {A}$ is ______

  1. $4:5$

  2. $3:2$

  3. $2:3$

  4. $5:4$


Correct Option: B

For an electron microscope, which of the following is false ?

  1. It uses magnetic lens to converge electron beam.

  2. Its resolving power is directly proportional to accelerating potential of electron.

  3. Its resolving power is inversely proportional to wavelength of electrons

  4. Magnification attained with the help of of it of the order of $10^6$.


Correct Option: B

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

  1. $16^o$

  2. $50^o$

  3. $10^o$

  4. $24^o$


Correct Option: A

A telescope has an objective of focal length $50 cm$ and an eye-piece of focal length $5 m$ THe least distance of distinct vision is $25 cm$. The telescope is focused for distinct vission a scale $200 cm$ away from the objective.The separation between the two lenses is nearly 

  1. $71 cm$

  2. $61 cm$

  3. $81 cm$

  4. $51 cm$


Correct Option: A

By which instrument we collect the space information:

  1. Convex lens

  2. Microscope

  3. Hubble Telescope

  4. None of these


Correct Option: C
Explanation:

Hubble Telescope is a device used to collect space information.

The focal lengths of the objective and the eyepiece of an astronomical telescope are 20 cm and 5 cm respectively. If the final image is formed at a distance of 30 cm from the eyepiece, find the separation between the lenses required for distinct vision

  1. 32.4 cm

  2. 42.3 cm

  3. 24.3 cm

  4. 30.24 cm


Correct Option: C
Explanation:

$f _{0}=20cm$


$f _{e}=5cm$

$V _{e}=30cm = D$

$L=?$

$L _{D}=f _{0}+ \dfrac{Df _e}{D+f _e}$ $=20+\dfrac{5\times 30}{35}=24.3 cm$

The focal length of the objective of an astronomical telescope is 1 m and it is in normal adjustment.Initially the telescope is focussed to a heavenly body. If the same telescope is to be focussed to an object at a distance of 21 m from the objective,then identify the correct choice

  1. eye piece should be displaced by 2 cm away from the objective

  2. eye piece should be displaced by 2 cm towards the objective

  3. eye piece should be displaced by 5 cm towards the objective

  4. eye piece should be displaced by 5 cm away from the objective


Correct Option: D
Explanation:

We know, $ \dfrac {1}{f} = \dfrac {1}{v} - \dfrac {1} {u} $


We seeing heavenly body,  $u =  \infty$
$v = f = 1 m$

When seeing 21 m far
$u = - 21 m $
$f = 1 m$
$v = \dfrac {f u} {(f+u)} = 21 / 20 = 1.05 m$

So, eye piece need to move $1.05-1 = 0.05 m$ further away from objective

Answer. D) eye piece should be displaced by $5 cm$ away from the objective