Telescopes

Transcript

0:00Introduction

Hello everyone! Let me start with a question. The telescopes at big observatories have huge objectives, the large lens or mirror at the front. So a bigger objective must magnify more, right? Hold that thought; we will test it later in this video. Today we learn how a telescope works, how much it magnifies, and how long its tube is. Questions on telescopes come up often in JEE, so let's begin.

0:27Astronomical telescope: objective and eyepiece

So, here is the astronomical, or refracting, telescope. In the figure, the two dark blue double arrows are lenses, and the orange lines are rays from a far away object, like a star, so they arrive almost parallel. The lens on the left is the objective, with a long focal length, f o, that is f with a small o below, and a large aperture. Okay? Now tell me this. Parallel rays from a distant object enter the objective. Where do they meet? They meet at the focus of the objective, yes, exactly! So the objective forms a real image in its focal plane: the small brown arrow in the middle of the figure. Clear? Next, the lens on the right is the eyepiece, with a short focal length, f e, with a small e below. It magnifies that real image. With me? Now, magnifying power is measured with angles, not sizes. Alpha, the Greek letter near the objective, is the angle the object makes at the unaided eye. Beta, the Greek letter near the eyepiece, is the angle the final image makes. So capital M, the magnifying power, equals beta over alpha, which is also written on the right of the figure. Right? Now look at the brown arrow once more. Is the final image erect, or is it inverted? It is inverted, very good! The brown arrow points down, and the eyepiece keeps it that way. So the final image is inverted. Got it?

2:01Magnifying power and length

Now the formulas. First, normal adjustment: the final image is at infinity, so the rays leave the eyepiece parallel, as in our figure. Okay? Then M equals f o over f e, the same result you see on the right of the figure. So for a big magnifying power, we want a long objective focal length and a short eyepiece focal length. Clear? Now look at the figure. The first image sits at the focus of the objective. For the rays to leave the eyepiece parallel, where must that image sit with respect to the eyepiece? At the focus of the eyepiece, yes, exactly! So the two focal lengths sit end to end, as the black double arrows under the figure show, f o and then f e. So the length of the telescope is f o plus f e. Okay? Now the second setting: the final image at D, the near point, which is twenty five centimetres for a normal eye; the c m means centimetres. With me? Then capital M equals f o over f e, times one plus f e over D, in brackets. That bracket is more than one, so the magnifying power is a little more than in normal adjustment. Clear? And the length? Now the first image sits a little inside the focus of the eyepiece, at a distance of f e times D, over f e plus D, from the eyepiece. So the length is f o, plus f e times D over f e plus D, a little shorter than f o plus f e. Got it?

3:34Other telescopes, and a bigger objective

Now, other telescopes. For an erect image, the terrestrial telescope adds an erecting lens, of focal length f, between the objective and the eyepiece. It works with its object and image at two f, so the length is f o, plus four times f, plus f e. Okay? The Galilean telescope uses a concave eyepiece, and also gives an erect image. The two vertical bars mean the modulus, so mod f e is f e without its minus sign. Here M equals f o over mod f e, and the length is f o minus mod f e. Clear? Then the reflecting telescope: its objective is a concave mirror, with a parabolic shape. With me? Why a mirror? It has no chromatic aberration and only a little spherical aberration, and a large mirror is lighter and easier to support than a large lens. Right? Now, back to our opening question. We make the objective bigger, with the same focal lengths. Does the telescope magnify more? No, it does not magnify more! M equals f o over f e has no diameter in it. Instead, a larger objective collects more light, proportional to the diameter squared; that is the proportional sign, and the small two up means squared. It also gives better resolution. So the image is brighter and clearer, with the same magnifying power.

5:01Example 1: fo 100 cm, fe 5 cm in normal adjustment

Okay? Now example one, an easy one. This telescope has an objective of focal length one hundred centimetres, and an eyepiece of focal length five centimetres. Find its magnifying power and length in normal adjustment. Pause the video and try it first. Done? Okay, let's check. Normal adjustment, so M equals f o over f e: one hundred over five gives twenty. Clear? And the length is f o plus f e: one hundred plus five gives one hundred and five centimetres. That's it, just two lines! Right?

5:40Example 2: final image at 25 cm, M and tube length

Now example two, a little harder. Same telescope, but now the final image is at twenty five centimetres. Find the magnifying power and the length. Pause the video and try it first. Done? Okay, let's check. The final image is at the near point, so we use the second formula: twenty, times one plus five over twenty five. With me? Five over twenty five is zero point two, so the bracket is one point two. Twenty times one point two gives twenty four, a bit more than twenty, as expected. Okay? Now the length. Same two lenses, so it is still one hundred and five centimetres, right? No, no, careful, that is the trap! The first image is no longer at the focus of the eyepiece. We use our sign convention: distances are measured from the lens, positive along the incident light, and negative against it. The final image is virtual, on the side the light comes from, so v e, its image distance, is minus twenty five centimetres, and the convex eyepiece has f e equal to plus five. Clear? Now rearrange the lens formula, which holds for rays close to the axis, for u e, the object distance for the eyepiece: one over u e equals one over v e, minus one over f e. That is minus one over twenty five, minus one fifth, which gives minus six over twenty five. Right? So the modulus of u e is twenty five over six, about four point one seven centimetres; the minus sign just puts the first image on the incoming side. And see, this matches our formula: five times twenty five, over thirty. Got it? So the length is f o plus the modulus of u e: one hundred plus four point one seven gives one hundred and four point two centimetres. So never reuse f o plus f e when the final image is at the near point. Okay?

7:41Quick revision: M = fo/fe, tube length, bigger objective

So, quick revision. In normal adjustment, M equals f o over f e, the length is f o plus f e, and the final image is inverted. With the final image at D, multiply by one plus f e over D, and the tube gets slightly shorter. And a bigger objective gives more light and better resolution, but the same magnifying power. Now try the practice problems on this page. All the best!