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### Course: Optics (Essentials) - Class 12th > Unit 3

Lesson 6: How telescopes work# Solved example: magnifying power of telescope

We will calculate the magnifying power of a telescope in normal adjustment, given the focal length of its objective and eyepiece. Created by Mahesh Shenoy.

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- why do we get the magnification positive?(1 vote)
- Thank you ...very helpful !no question.(0 votes)
- Hey is there a way to calculate the limiting magnitude of a telescope from it's magnification?(0 votes)
- Hey! you talked about the normal adjustment between00:43and1:15you have talked about normal adjustment or relaxed eye state. can you also solve some questions about strained eye state?(0 votes)

## Video transcript

a telescope has objective of focal in two meters and an eyepiece of focal length 10 centimeters find the magnifying power this is the short form for magnifying power in normal adjustment so what's given to us what's given to us is that we have a telescope which is kept in normal adjustment mode we'll see what that is in a while and the data is we've been given the focal length of the objective and we've also been given the focal length of the eyepiece so based on this we need to figure out the magnifying power of our telescope the first thing is let's quickly look at what aha what's the principle of a telescope let's quickly recall that and understand what this normal adjustment is so in the telescope a large objective lens focuses the beam of light from infinity to its principal focus forming a tiny image over here it sort of brings the object close to us and then we use an eyepiece which is just a magnifying glass a convex lens and then we go very close to it so to examine that object now normal adjustment more just means that the rays of light hitting our eyes are parallel to each other that means our eyes are in the relaxed state in order for that to happen we need to make sure that the the focal that the that the image formed due to the objective is right at the principle focus of the eyepiece so that the rays of light after refraction become parallel to each other so we are now in the normal it just bent more so we know this focal length we also know this focal length they're given to us we need to figure out the magnification how do we define magnification for any optic instrument we usually define it as the angle that is subtended to our eyes with the instrument - without the instrument we take that ratio so with the instrument can you see the angles of training now is Theta - it's clear right that down so with the instrument the angle subtended by this object notice is Thea - and if we hadn't used our instrument we haven't used our telescope then the angle subtended would have been all directly this angle isn't it if you directly use your eyes then directly these rays would be falling on our eyes and at the angles obtained by that object whatever that object would be that which is just here or not so this would be our magnification and this is what we need to figure out this is the magnifying power so I want you to try and pause the video and see if you can figure out what theta - and theta not are from this diagram and then maybe we can use the data and solve that problem just just give it a try all right let's see theta naught or Tila - can be figured by this triangle by using small-angle approximations remember these are very tiny angles I have exaggerated that in the figure but these are very small angles so we can use tan theta - which is same as T - it's the opposite side that's the height of the image divided by the edges inside which is the focal length of the eyepiece and what is Theta not wealthy or not from here it might be difficult to calculate but that same theta naught is over here as well and so we can use this triangle to figure out what theta naught is and what would that be well that would be again the height of the image divided by the edges inside that is the focal length of the objective and so if these cancel we end up with the focal length of the objective divided by the focal length of the eyepiece and that's it that is the expression for magnification so any telescope problems are asked to us in normal adjustment more I usually like to do it this way I don't have to remember what that magnification formula is if you just remember the principle we can derive it on the spot so now we can just go ahead and plug in so what will we get so focal length of the objective is given to us as 2 meters so that's 2 meters divided by the focal length of the IPS that's given as 10 centimeters can you be careful with the unit's 10 centimeters well we can convert this into centimeters to meters is 200 centimeters and this is 10 centimeters and now this cancels and we end up with 20 so the magnification we're getting is 20 and that's the answer this means that by using the telescope we can see that object 20 times bigger than what we would have seen without the telescope and also in some questions they asked you what should be the distance between the objective and the eyepiece we must maintain a fixed distance and we can figure that distance out the distance is just the focal length of the objective plus the focal length of the eyepiece can you see that and so if that was even then that was asked what is the distance between the objective and the eyepiece or we just add them so that would be 2 meters plus 10 centimeters so you add then I was about 210 centimeter said about 2.1 meters so this would be a pretty pretty long pretty long telescope will be a huge telescope to get this much 9if occasion