The telescope

Point a small backyard telescope at Saturn and the planet stops being a dot and becomes a world — a pale disc wrapped by a thin, tilted ring, sitting steady in the eyepiece while your naked eye, looking at the same patch of sky a second earlier, saw only a faint star-like point. Nothing about Saturn changed. What changed is the angle it subtends at your eye. The telescope did not make Saturn bigger or closer; it took the tiny angle the planet already carves out in the sky and stretched it into a much larger angle, so that the image on your retina — and therefore the image in your brain — is enormously magnified. This is the one idea the whole chapter has been building toward. A magnifying glass and a compound microscope both take something near and enlarge the angle it subtends by shortening the effective distance to the eye. A telescope faces the opposite problem: the object — a planet, a star cluster, a distant tower — is so far away that its rays arrive at the instrument essentially parallel, no matter how you move it. There is no "distance" left to shorten. Instead, a telescope must gather these parallel rays and re-aim them so that the final image, though it may be no closer in a physical sense, occupies a much larger slice of your field of view. Two designs achieve this. The astronomical (refracting) telescope uses two convex lenses — a large, long-focus objective that captures the parallel light and forms a small real image, and a short-focus eyepiece that magnifies that image the way a simple magnifier would. The reflecting telescope replaces the objective lens with a large concave mirror, sidestepping the practical limits that stop refracting telescopes from growing much beyond a few metres across. Both share the same idea and, strikingly, nearly the same formula for how much they magnify. This section builds that formula from a ray diagram, applies it to real numbers, and explains why every giant observatory on Earth — and in orbit — reflects rather than refracts.

Key takeaways

A telescope magnifies the angle a distant object subtends at the eye — it does not make the object closer or larger in any physical sense.
The astronomical telescope uses a large-focal-length, large-aperture convex lens (the objective) to form a small real image of a distant object, and a short-focal-length convex lens (the eyepiece) to view that image, magnified, the way a simple magnifier works.
In normal adjustment (relaxed eye, final image at infinity), the magnifying power is $m = f_o/f_e$ , and the tube length is $L = f_o + f_e$ .
With the final image brought to the near point $D$ instead, the magnifying power rises slightly to $m = \dfrac{f_o}{f_e}\left(1 + \dfrac{f_e}{D}\right)$ , at the cost of eye strain.
The reflecting telescope swaps the objective lens for a large concave (usually parabolic) mirror; its magnifying power in normal adjustment obeys the same ratio $m = f_o/f_e$ , now with $f_o$ the mirror's focal length.
Large telescopes are built as reflectors because mirrors can be supported from behind (no sagging under their own weight), have no chromatic aberration, and are far cheaper to manufacture defect-free at large diameters than an equally large lens.
Key idea: a telescope's objective forms a small real image of a distant object at its focus, and the eyepiece magnifies that image, so the angular magnification in normal adjustment is simply the ratio of the two focal lengths, m = f_o/f_e.

The core insight

A telescope collects the near-parallel rays from a distant object with a large-focal-length objective, forms a small real image at its focus, and magnifies that image with a short-focal-length eyepiece, giving angular magnification $m = f_o/f_e$ in normal adjustment.

Where it fits in the syllabus

This is the closing topic of the "Optical instruments" section of the CBSE Class 12 chapter Ray Optics and Optical Instruments, coming immediately after the compound microscope. The prescribed scope names the astronomical telescope (both refracting and reflecting forms) and asks for its magnifying power — the syllabus does not require aberration theory, resolving power, or the equivalent-focal-length treatment of lenses separated by a distance, so this section stays strictly at the level of the objective–eyepiece ray picture and the two magnifying-power formulas. Everything here rests on ideas already built earlier in the chapter: the thin lens formula, the sign convention for real and virtual images, and the idea (from the simple magnifier and compound microscope) that angular magnification, not linear size, is what "magnifying power" means for the eye.

The physical picture

Before any formula, get the geometry straight. Because the object — a star, a planet, a distant building — is so far away, every ray from a single point on it that reaches the objective lens is essentially parallel to every other ray from that point. A convex lens brings a bundle of parallel rays to a single point at its focus. So the objective, of focal length $f_o$ , forms a small, real, inverted image of the distant object exactly at its focal plane. This image is real precisely because the object is effectively at infinity, and it is small precisely because the object is enormously distant — the image height at the focal plane is what will be angularly magnified next. The eyepiece, of much shorter focal length $f_e$ , is then placed so that this small real image lies at (or just inside) its own focus. Used this way, the eyepiece behaves exactly like a simple magnifying glass: it takes the small real image formed by the objective and produces a large virtual image that the eye examines with a much bigger visual angle than the image itself subtended when viewed with the naked eye at that distance.

m = \frac{\text{angle subtended by the final image at the eye}}{\text{angle subtended by the object at the eye, viewed directly}} = \frac{\beta}{\alpha}

where: $\alpha$ is the tiny angle the distant object itself subtends at the unaided eye, and $\beta$ is the (much larger) angle the final image subtends when viewed through the instrument. Magnifying power, exactly as for the microscope, is this ratio of angles — never a ratio of physical sizes, since the object and the final image can be worlds apart in actual size and distance. In normal adjustment, the eyepiece is positioned so the small real image formed by the objective sits exactly at the eyepiece's own focal point. The eyepiece then sends out parallel rays for each point of that image, so the final image is at infinity and the eye, relaxed (its own focal length unstrained, the way it looks at a distant hill), views it in comfort. Because the objective's image lies at the objective's focus $f_o$ from the objective, and the eyepiece's focus is $f_e$ from the eyepiece, and in this arrangement these two foci coincide at the same point, the two lenses sit a fixed distance apart:

L = f_o + f_e

where: $L$ is the separation between the objective and the eyepiece — physically, the length of the telescope tube — and $f_o, f_e$ are the objective's and eyepiece's focal lengths. This is why a telescope built for high magnification with a very long-focus objective is necessarily a long instrument, a genuine mechanical constraint that limits how much magnification a portable telescope can offer.

Figure

A ray diagram of an astronomical telescope in normal adjustment, showing parallel rays from a distant object entering a large objective lens on the left, converging to a small real image at the shared focal point, then diverging through a small eyepiece lens on the right to emerge as a second set of parallel rays for a relaxed eye, with the objective focal length and eyepiece focal length marked along the axis

The reflecting telescope replaces the objective lens with a large concave mirror (in practice ground to a parabolic shape to avoid spherical aberration at the edges). Parallel light from a distant object strikes the mirror and converges toward its focus exactly as it would through a lens; a small secondary mirror (plane or convex) intercepts this converging beam before it reaches the focus and redirects it sideways or back through a hole in the primary, where an eyepiece views the image just as in the refracting design. The magnifying-power reasoning is identical, only $f_o$ is now the focal length of the mirror rather than a lens:

m = \frac{f_o}{f_e}

where: the symbols mean the same thing as before — the ratio of the light-collecting element's focal length to the eyepiece's focal length — but the physical element producing $f_o$ is now a mirror, which is why the reflecting design is preferred whenever the telescope must be very large.

Derivation

Derivation AMagnifying power of an astronomical telescope in normal adjustment

The goal is to show, using nothing beyond the small-angle approximation already used for the simple microscope, that in normal adjustment a telescope's magnifying power reduces to the clean ratio $m = f_o/f_e$ — independent of how far away the object actually is, which is itself a striking and useful fact, since a telescope's magnification does not depend on the distance to the star or planet being viewed.

Step 1. Set up the two angles that define magnifying power. Let $\alpha$ be the angle the distant object subtends at the unaided eye, and let $\beta$ be the angle the final image subtends at the eye when viewed through the telescope in normal adjustment. By definition, $m = \beta/\alpha$ .

Step 2. Express $\alpha$ using the objective's image. Because the object is effectively at infinity, all its rays reach the objective essentially parallel, and the small real image it forms sits at the objective's focus, at a distance $f_o$ from the objective (and, in this normal-adjustment arrangement, also at distance $f_e$ from the eyepiece, since the two foci coincide). The angle $\alpha$ subtended by the distant object at the unaided eye is, to a very good approximation for angles this small, the same as the angle its own image (of height $h$ , say) subtends back at the objective from that image point: $\alpha \approx h/f_o$ , using $\tan\alpha \approx \alpha$ for small angles.

Step 3. Express $\beta$ using the eyepiece's view of the same small image. The eyepiece looks at the very same real image, of height $h$ , but from a much shorter distance $f_e$ (since the image sits at the eyepiece's focus too). By the identical small-angle reasoning, the angle this image subtends at the eyepiece — and hence, since the eye is placed close to the eyepiece, at the eye — is $\beta \approx h/f_e$ .

Step 4. Divide. The unknown image height $h$ , which we never needed to compute, cancels exactly: $m = \dfrac{\beta}{\alpha} = \dfrac{h/f_e}{h/f_o} = \dfrac{f_o}{f_e}$ . The magnifying power in normal adjustment depends only on the two focal lengths, never on the object's actual distance or size — which is exactly why the same telescope gives the same $20\times$ , say, whether pointed at the Moon or at a star thousands of light-years away.

Worked examples

Board level - magnifying power and tube length from given focal lengths

The objective of an astronomical telescope has a focal length of $100\ \text{cm}$ and the eyepiece has a focal length of $5\ \text{cm}$ . Find (a) the magnifying power in normal adjustment, (b) the length of the telescope tube in this setting, and (c) the magnifying power if the final image is instead formed at the near point $D = 25\ \text{cm}$ .

Step 1. Write down the given data. $f_o = 100\ \text{cm}$ , $f_e = 5\ \text{cm}$ , $D = 25\ \text{cm}$ .

Step 2. Magnifying power in normal adjustment, using $m = f_o/f_e$ directly:

m = \frac{100}{5} = 20

The telescope magnifies the angular size of a distant object $20$ times.

Step 3. Tube length in normal adjustment, using $L = f_o + f_e$ :

L = 100 + 5 = 105\ \text{cm}

Step 4. Magnifying power with the final image at the near point, using $m = \dfrac{f_o}{f_e}\left(1+\dfrac{f_e}{D}\right)$ :

m = \frac{100}{5}\left(1 + \frac{5}{25}\right) = 20 \times 1.2 = 24

Answer: $m = 20$ in normal adjustment, tube length $L = 1.05\ \text{m}$ , and $m = 24$ with the final image at the near point. Moving the final image from infinity to the near point buys a modest gain in magnification — $20$ to $24$ , about $20\%$ — at the cost of a strained, actively-accommodating eye rather than a relaxed one, the identical trade-off met with the simple magnifier and the compound microscope.

Competitive level - designing a telescope for a target magnification and tube length

A telescope is to be built with a magnifying power of $20$ in normal adjustment and an overall tube length of exactly $105\ \text{cm}$ . Find the focal lengths $f_o$ and $f_e$ of the objective and eyepiece that must be used.

Step 1. Write the two governing equations as a pair of simultaneous relations in $f_o$ and $f_e$ :

\frac{f_o}{f_e} = 20 \qquad\text{and}\qquad f_o + f_e = 105

Step 2. Use the first equation to eliminate $f_o$ : since $f_o = 20 f_e$ , substitute into the second equation, $20f_e + f_e = 105$ , giving $21 f_e = 105$ .

Step 3. Solve for $f_e$ , then back-substitute for $f_o$ :

f_e = \frac{105}{21} = 5\ \text{cm}, \qquad f_o = 20 \times 5 = 100\ \text{cm}

Step 4. Sanity-check both conditions: $f_o/f_e = 100/5 = 20$ ✓, and $f_o + f_e = 100+5 = 105\ \text{cm}$ ✓. Answer: $f_o = 100\ \text{cm}$ and $f_e = 5\ \text{cm}$ — exactly the pair used in the board-level example above, now reached by working backwards from the design specification rather than forwards from given lenses. Timing note: under ninety seconds. The two governing equations, $m = f_o/f_e$ and $L = f_o + f_e$ , are linear once you substitute one into the other, so this is ordinary simultaneous-equation algebra dressed up in optics language — recognise the pattern (a ratio condition plus a sum condition) and it collapses to one line, $f_e = L/(m+1)$ , $f_o = mf_e$ , which you can even memorise for design-style telescope questions.

Common traps

Swapping which lens is the objective: in a telescope the objective has the larger focal length and the eyepiece the smaller one — the opposite pairing from the compound microscope, where the objective's focal length is the small one. This reversal is one of the single most common marks lost on this topic.
Applying $m = f_o/f_e$ when the question specifies the final image at the near point, or applying the near-point formula when the question specifies normal adjustment (relaxed eye, image at infinity) — always read which condition is being asked before picking the formula.
Treating a bigger objective focal length, by itself, as making the telescope "better" in every sense. A large focal length raises magnifying power, but it is the objective's aperture (diameter), not its focal length, that governs how much light it gathers and how sharply it can resolve close-together objects — brightness and resolution are separate practical concerns from angular magnification.
Assuming the reflecting telescope needs an entirely new formula. It uses the same $m = f_o/f_e$ ratio; only the physical element producing $f_o$ changes from a lens to a mirror.
Forgetting that $L = f_o + f_e$ holds only in normal adjustment — moving the eyepiece to focus the image at the near point instead of infinity changes the lens separation slightly, so the simple sum no longer applies exactly.

Common trapA telescope's magnifying power says nothing about how bright or sharp the final image is

It is tempting to treat magnifying power as the single number that describes a telescope's quality, but it does not. Two telescopes can have identical $f_o/f_e$ and therefore identical magnifying power, yet one can show a crisp, bright image of Saturn's rings while the other shows a dim, barely-resolved blur — because the one with the larger-aperture objective (bigger diameter, not necessarily bigger focal length) collects more light and can separate finer angular detail. This is exactly why professional observatories chase ever-larger mirror diameters rather than ever-longer focal lengths: light-gathering power and resolving power depend on aperture, while magnifying power depends only on the ratio of focal lengths. On an exam, never answer "what limits how much detail we can see through this telescope" with the magnifying-power formula alone.

You should be able to…

Draw the ray diagram of an astronomical telescope in normal adjustment, correctly placing the objective, the shared focal point, and the eyepiece.
Derive and apply $m = f_o/f_e$ for normal adjustment, using the small-angle approximation on the objective's real image.
Apply $m = \dfrac{f_o}{f_e}\left(1+\dfrac{f_e}{D}\right)$ when the final image is formed at the near point instead of infinity, and explain the accompanying eye-strain trade-off.
Calculate the tube length $L = f_o + f_e$ in normal adjustment, and solve simple design problems that go backwards from a target $m$ and $L$ to the required focal lengths.
Explain qualitatively how a reflecting telescope works, and give at least two concrete reasons it is preferred over a refracting telescope for large apertures.
Distinguish magnifying power (a focal-length ratio) from light-gathering power and resolution (governed by the objective's aperture), and know which one each affects.

Ask the tutor: want to see how changing only the eyepiece's focal length trades off magnifying power against tube length, or how the same objective mirror performs in a reflecting telescope versus a refracting one — ask the AI tutor to work through it with you.

Why does a telescope need such a long tube if the object is infinitely far away?

The tube length has nothing to do with the object's distance — it only depends on the two focal lengths of the lenses you chose, through $L = f_o + f_e$ . A long-focus objective is what gives high magnifying power ( $m = f_o/f_e$ ), but that same long focal length also means the small real image forms far from the objective, and the eyepiece has to sit right where that image is. So a telescope built for high magnification is necessarily long, purely as a geometric consequence of where the objective's image happens to form — not because the star being viewed is far away.

If the telescope's image is inverted, how do astronomers make sense of what they see?

The image really is inverted in a simple astronomical telescope, and astronomers generally do not mind — a star or a galaxy looks the same right-side up or upside down, so there is no everyday cue that tells you the orientation is wrong. For observing on Earth (a bird, a ship, a landscape) an upright image obviously matters, which is why terrestrial telescopes and binoculars add extra optics (often prisms) purely to flip the image back the right way round, at the cost of a bit of length and light loss. For pure astronomy the inversion is simply ignored.

Does a bigger magnifying power always mean I can see fainter or more distant objects?

No, and this trips up a lot of students. Magnifying power only stretches the angle an object already subtends — it cannot add light that was never collected in the first place. Whether a faint star or a fine surface feature on Jupiter is visible at all is governed by the objective's aperture (its diameter), which sets how much light is gathered and how much fine angular detail can be resolved. A telescope with huge magnifying power but a small objective aperture will simply show a large, dim, blurry image; astronomers chasing faint galaxies build the biggest-diameter mirrors they can, not the highest-magnification eyepieces they can.

Why is the reflecting telescope's mirror shaped like a parabola instead of a simple sphere?

A simple spherical mirror does not bring all parallel rays to exactly the same focus — rays striking near the edge focus slightly differently from rays near the centre, blurring the image, an effect called spherical aberration. A parabolic mirror is shaped precisely so that every parallel ray, regardless of where it strikes the mirror, converges to the same single focal point. This refinement is beyond what CBSE Class 12 requires you to derive or even name in an exam answer — it is enough to know qualitatively that large research telescopes use precision-ground parabolic mirrors rather than plain spherical ones.