12. Class 12th Physics | Optical Instruments | Refracting Telescope | Ashish Arora


let us now discuss about refracting telescope
this is also called astronomical telescope. because this telescope is used. to provide
the angular magnification for heavenly object or far away located distance object. now in
reflecting telescope here we can write. it provides. angular magnification for. distant
objects. and uses 2 lenses. objective and eyepiece for this purpose. and similar to
compound microscope objective is the one which is closer to the object that is toward object
side and eyepiece is one which is closer to the observers eye, only difference in this
and. microscope lenses are here objective is of quite large aperrture. this is because
as object is located far away the intensity of light rays coming from object. reduces
to very low value. so objective is taken to be of large aperrture so that as many light
rays enters into the object if image intensity will be high. and as object is far away all
rays can be considered as paraxial. so relative spherical aberration will be less. if we just
have a look on optical setup of a telescope. here again. we use 2 lenses, and we’ll analyze
the study of image at 2 location 1 is at near point other is at far point for the observer.
this objective having focal length f-o and this eyepiece having focal length f-e. now
if we consider from far away points parallel rays are incident. onto. the objective. and
say if this angle is alpha which can be taken as the, angular size of object, and if it
is producing an image intermediate image somewhere here. and say it is of height h certainly
the final image due to objective will be obtained in its focal plane or the image distance from,
the objective lens would be equal to the focal length of the, objective lens. and in this
situation this intermediate image will act as an object for eyepiece and light rays will.
incident on the eyepiece and finally, these light rays. will be. falling into the observers
eye and final image will be seen by the observer. so in this situation say if, this is the location
where the image is obtained, this is final image h prime, then here we can say due to
these 2 diverging rays final image will be seen by the, observer at point which is at
a distance near point, of the observer from eyepiece lens as eye is considered to be very
close to the eyepiece. then in this situation we can say if this distance. of intermediate
image is u-e. from the. eyepiece lens, then, this angle beta we can, consider as the angular
size of image produced. so here we can write angular size, of object. in this situation
we write it as alpha. and the value of alpha we can write as h over f-o. and if we calculate
the angular size, of image. this must be beta and the value of beta we can calculate from
this triangle and it can be given as h over u-e where u is the distance of intermediate
image from eyepiece lens. so here, magnifying power. of telescope. if we calculate, then
magnifying power of telescope we can write as, angular size of image by angular size
of object this beta by alpha. so if we substitute the value of beta and alpha here h gets cancelled
out and the value we are getting is f-o by u-e. so you must be very careful here. that
u-e we can easily obtained by, using lens formula for eyepiece lens. say if we. using.
lens formula. for, eyepiece. here we can say the object distance is minus u-e the image
distance is minus d. and focal length can be taken as plus f-e. if we use the values
in lens formula 1 by v minus 1 by u is 1 by f. here you can see we are getting is. minus
1 by d plus 1 by u-e is equal to 1 by f-e. on simplifying you are getting the value of
u-e is equal to. d f-e over d plus f-e. so if we substitute it over here. so here we
can write for image produced. at near point. the magnifying power of refracting type telescope
which can be written as m-p for image production at point d will be. f-o by, here u-e i can
write as d f-e. multiplied by this is d plus f-e. on simplifying here i am getting m-p-d,
is equal to. f-o by f-e we multiplied by this d can be taken as this 1 plus f-e by d. and
as the final image produces inverted we can take this as negative. and this is the magnifying
power of refracting type telescope when final image is produced at near point of the human
eye. and if we consider the situation for image. produced at far point. then here we
know well for this image to be produced at infinity. this u-e must be equal to f-e. so
in this situation if u-e is equal to f-e we can see the value of beta will change to h
over f-e, so the only difference in this situation we are getting is directly the magnifying
power of this telescope for image produced at infinity will be equal to, this directly
written as f-o by. f-e and again it is negative because we are observing an inverted image.
so these are 2 magnifying powers. for final image produce at near point or at far point
in case of refracting type telescope.

You may also like...

19 Responses

  1. Anurag Pradhan says:

    i cannot understand standard reference for angular magnification … o-o
    0
    I——–I

  2. karthik kanna says:

    Respected Sir
    why are we calculating image at far point .
    Does it means that the image is formed at very large distance.

  3. avinash soni says:

    sir how intensity is directly proportional to (aperture)^2. ??

  4. deepak desai says:

    sir while substituting values in lens formulae u have taken u=-ue v=-d but why didn't u take f=-fe?????

  5. ARCHI BINDRA says:

    sir can we consider fo is eq to image dist of objective lens??

  6. akarshkumar srit says:

    sir why we need to put minus in the final magnification in compound microscope final image is inverted but we didn't put minus there

  7. teenrap j says:

    Thanks sir

  8. TheHubtech says:

    Sir why don't we take -ve sign in numericals but in this derivation you said that the image is inverted hence we take it to be -ve? Why the same does not apply for numericals?

  9. rekha gupta says:

    Truly best explanation ever seen

  10. Shlok Saxena says:

    thanks a lot sir

  11. Shilu Kumari says:

    Plz thoda bda likho so that we can easily understand

  12. Gaurav Arora says:

    Sir pls reply sir in first case while calculating magnification you used m=beta /alpha but sir this is valid only when angles are measured for object and image if they are kept at same separation from eye but here object is at infinity and image a tD so how we used this formula

  13. Samkit Jain says:

    Kahan se yeh aadmi Allen ka itna bada teacher ban gaya….kuch samajh nhi aa raha

  14. Samarthya Varghese says:

    Is fo>fe?

  15. Anand Kesar says:

    Your great teacher,

  16. sab ka bhai says:

    Thanks sir you are awesome

  17. Devender nainwal says:

    Seriously sir u are like Richard feynman

  18. Ayan Jain says:

    Sir, does large aperture implies large focal length.

  19. Soumik Gupta says:

    I think the magnifying power of the telesccole (beta/alpha) will be -fo/ue

Leave a Reply

Your email address will not be published. Required fields are marked *