Refracting & Reflecting Telescopes

Every great observatory is equation (1) attached to an enormous mirror. The Hubble Space Telescope is a $2.4\ \mathrm{m}$ reflector; the ground-based giants — Keck, the VLT, the Extremely Large Telescope under construction — carry mirrors from $10$ to nearly $40\ \mathrm{m}$ across, and every metre of that width is bought for exactly the two reasons this article gives: to gather more light (as the square of the aperture) and to resolve finer detail. The James Webb Space Telescope is a $6.5\ \mathrm{m}$ segmented mirror working in the infrared, chosen because a mirror carries no chromatic aberration across a huge wavelength range. Radio telescopes are the same idea with dishes hundreds of metres wide, and linked across continents they synthesise apertures the size of the Earth to reach resolving powers that imaged a black hole's shadow. Even the humble camera lens borrows the objective's job, and the autofocus in your phone is a tiny telescope-plus-eye system — but the frontier, as ever, belongs to whoever can build the widest aperture and point it at the dark.

Look at a star through nothing but air and its light arrives as a bundle of essentially parallel rays, because it is so far away — and the angle between the rays from two different stars, or the two edges of the Moon, is what your eye reads as their separation and size. The problem with far objects is never that they are small; it is that they subtend a tiny angle. A telescope's entire job is to take that small incoming angle $\alpha$ and present the eye with a larger outgoing angle $\beta$, so the object appears bigger and two close objects pull apart. It does this in two stages, just like the compound microscope, but arranged for parallel incoming light. The objective — a long-focus converging lens — collects the parallel rays from the distant object and brings them to a real, inverted image at its focal plane. Because the object is at infinity, that image is small and sits exactly one focal length $f_o$ behind the objective. The eyepiece — a short-focus lens — is then placed so that this real image lies at its front focus, and it renders the image back into parallel rays, which the relaxed eye focuses at ease. The upshot: parallel rays in at angle $\alpha$, parallel rays out at a larger angle $\beta$, and $M = \beta/\alpha = f_o/f_e$.

Notice the geometry that fixes the tube length. In normal adjustment the objective's real image sits at the objective's focus and at the eyepiece's focus — the two focal points coincide at the common focal plane — so the lenses are separated by $L = f_o + f_e$. This is the exact opposite of the microscope, whose tube length is the image throw $L$, not a sum of focal lengths. And it is why a telescope is long: the objective focal length dominates.

With the final image at infinity (the relaxed-eye setting astronomers use, since a star is comfortably viewed with an unaccommodated eye), the angular magnification is

where: $f_o$ is the objective focal length and $f_e$ the eyepiece focal length. Read it directly: magnification is how many eyepiece focal lengths fit into the objective's. A $f_o = 100\ \mathrm{cm}$ objective with a $f_e = 5\ \mathrm{cm}$ eyepiece gives $M = 100/5 = 20$. The lenses sit

apart — here $L = 100 + 5 = 105\ \mathrm{cm}$, so the instrument is over a metre long. To magnify more you lengthen the objective or shorten the eyepiece; a $f_o = 1.2\ \mathrm{m}$ objective with a $f_e = 6\ \mathrm{cm}$ eyepiece also gives $M = 1.2/0.06 = 20$, in a tube $L = 1.26\ \mathrm{m}$ long. The long objective is why classical refractors are unwieldy giants and why the biggest refractor ever built, at Yerkes, is a $1.0\ \mathrm{m}$ lens on a tube many metres long.

If instead you bring the final image to the near point $D = 25\ \mathrm{cm}$ — squeezing a little extra angle at the cost of accommodating the eye — the eyepiece works as a near-point magnifier of the intermediate image, and the magnification rises to

For the same $f_o = 100$, $f_e = 5\ \mathrm{cm}$: $M = 20 \times (1 + 5/25) = 20 \times 1.2 = 24$, a $20\%$ boost over the normal $20$. The factor $(1 + f_e/D)$ is the telescope's version of the microscope's "$+1$" — the small premium the near-point setting always pays. Astronomers rarely use it (the relaxed setting is easier for long viewing), but boards ask for it, so state which adjustment your number belongs to.

A long converging lens is not the only way to gather starlight to a focus; a concave mirror does it too, and better. A reflecting telescope replaces the objective lens with a large concave mirror, and this brings three decisive advantages. First, a mirror has no chromatic aberration — it does not disperse light into colours, because reflection, unlike refraction, does not depend on wavelength — so the coloured fringing that plagues cheap refractors simply vanishes. Second, a mirror can be made enormous: it is supported from behind across its whole back, whereas a lens can only be held at its rim and sags under its own weight past about a metre. Third, mirrors are cheaper per unit area. Newton built the first one in 1668, folding the light out to the side with a small flat diagonal; Gregory, Cassegrain and later Herschel gave other fold-geometries, and every major observatory telescope today is a reflector.

And here is the lesson beginners miss. The number stamped on a toy telescope — "$525\times$!" — is almost meaningless, because magnification is cheap (any short eyepiece raises it) and past a point only magnifies the blur. What actually limits an astronomical telescope is its aperture, the diameter of the objective. Aperture governs two things that magnification cannot buy. Light-gathering scales as the aperture squared (area), so doubling the aperture makes faint stars four times brighter and reveals ones the smaller instrument never showed. Resolving power — the finest angle it can split — improves in proportion to aperture, set by the diffraction (Rayleigh) criterion

where: $\lambda$ is the wavelength and $a$ the aperture diameter; $\theta_{\min}$ is the smallest angle (in radians) between two stars the telescope can still separate. This is quoted from wave optics — its proof is the diffraction of the next chapter — but the message is plain. For a $10\ \mathrm{cm}$ aperture in green light ($\lambda = 550\ \mathrm{nm}$), $\theta_{\min} \approx 1.22 \times (550 \times 10^{-9})/0.10 \approx 6.7 \times 10^{-6}\ \mathrm{rad}$, about $1.4$ arcseconds — fine enough to split many double stars, and a limit set by width, not by any eyepiece. This is the exact analogue of the microscope's resolving power, now written as an angle, and it is why observatories are named for the size of their mirrors.

Board level — an astronomical telescope in both adjustments (CBSE and NEET pattern)

An astronomical telescope has an objective of focal length $100\ \mathrm{cm}$ and an eyepiece of focal length $5\ \mathrm{cm}$. Find (a) its angular magnification and tube length in normal adjustment, and (b) its magnification when the final image is formed at the near point of $25\ \mathrm{cm}$.

Step 1. Normal adjustment, magnification: $M = f_o/f_e = 100/5 = \boxed{20}$. The telescope makes a distant object subtend twenty times the angle it would to the naked eye.

Step 2. Normal adjustment, tube length: the two foci coincide, so $L = f_o + f_e = 100 + 5 = \boxed{105\ \mathrm{cm}}$ — the instrument is just over a metre long, dominated by the objective.

Step 3. Near-point adjustment: apply $M = \dfrac{f_o}{f_e}\left(1 + \dfrac{f_e}{D}\right) = \dfrac{100}{5}\left(1 + \dfrac{5}{25}\right) = 20 \times 1.2 = \boxed{24}$.

Step 4. Sense-check: the near-point setting gives more magnification ($24 > 20$) by the factor $1 + f_e/D = 1.2$, the standard $20\%$ premium, at the cost of accommodating the eye. State which adjustment each number belongs to — boards mark the distinction.

Competitive level — design a telescope and estimate its resolving power (JEE-Main and NEET pattern)

An astronomical telescope must give an angular magnification of $20$ in normal adjustment using an objective of focal length $1.2\ \mathrm{m}$. (a) Find the eyepiece focal length and the tube length. (b) If the objective aperture is $100\ \mathrm{mm}$, estimate the smallest angular separation it can resolve in green light ($\lambda = 550\ \mathrm{nm}$).

Step 1. Design the eyepiece from $M = f_o/f_e$: $f_e = f_o/M = 1.2/20 = 0.06\ \mathrm{m} = \boxed{6\ \mathrm{cm}}$. The tube length is $L = f_o + f_e = 1.2 + 0.06 = 1.26\ \mathrm{m} = \boxed{126\ \mathrm{cm}}$.

Step 2. Now the resolving power, a different physical property entirely — set by aperture, not magnification. Use the Rayleigh criterion $\theta_{\min} = 1.22\,\lambda/a$ with $a = 100\ \mathrm{mm} = 0.10\ \mathrm{m}$.

Step 3. Substitute: $\theta_{\min} = 1.22 \times \dfrac{550 \times 10^{-9}}{0.10} = 6.71 \times 10^{-6}\ \mathrm{rad}$. Converting to arcseconds (multiply by $206265$), that is about $\boxed{1.4\ \mathrm{arcseconds}}$ — the finest double star this aperture can split.

Step 4. Read the two answers together. The magnification ($20$) came from the focal lengths; the resolving power ($1.4''$) came from the aperture. They are independent — you could double the magnification with a shorter eyepiece and the resolving power would not change one bit. That independence is the whole point of the aperture warning above. Timing note: about ninety seconds. The design part is one division; the resolving part is one substitution into a quoted formula. The only trap is trying to get resolving power from the magnification — they come from different numbers, so keep aperture and focal length in separate boxes.

Believing a bigger number of times is a better telescope. Magnification is cheap and, past the aperture's resolving limit, empty. Aperture is the figure of merit: light-gathering as aperture squared, resolving power in proportion to aperture. Quote the diameter, not just the "$\times$".

The long-versus-short objective mix-up. A telescope's magnification is $f_o/f_e$, so it wants a long objective and a short eyepiece — the opposite of the microscope. Carrying the microscope's short-objective instinct here is a classic slip.

Tube length errors. In normal adjustment $L = f_o + f_e$ (the two foci coincide). Do not write $f_o - f_e$, and do not import the microscope's tube-length-as-image-throw rule. Near-point adjustment shifts the eyepiece slightly, but boards take $L = f_o + f_e$ unless told otherwise.

Forgetting the near-point premium. Normal adjustment is $f_o/f_e$; the near-point setting multiplies by $(1 + f_e/D)$. Using the wrong one changes $20$ into $24$. Read which adjustment the question specifies.

Board scope. Telescope magnification is examined by CBSE, ICSE, JEE-Main, JEE-Advanced and NEET. It is not in AP Physics 2 (no optical-instrument magnification in the CED), and IB and A-Level carry no geometric optics of instruments. If your board is one of those, this is background reading. The Indian boards additionally like the reflecting-telescope advantages as a short-answer question.

Texts. NCERT Physics Class 12, Ch. 9 ("Ray Optics and Optical Instruments"), the refracting and reflecting telescopes and their magnifying powers · H. C. Verma, Concepts of Physics Vol. 1, "Optical Instruments" · D. C. Pandey, Optics and Modern Physics, the telescope problem set for JEE and NEET.

Provenance. Standard optics. Recompute: normal adjustment $M = 100/5 = 20$, tube length $100 + 5 = 105\ \mathrm{cm}$; the $f_o = 1.2\ \mathrm{m}$, $f_e = 6\ \mathrm{cm}$ case $M = 1.2/0.06 = 20$, $L = 126\ \mathrm{cm}$; near-point $M = (100/5)(1 + 5/25) = 24$, premium factor $1.2$; the design $f_e = 1.2/20 = 0.06\ \mathrm{m}$; light-gathering scaling (aperture doubled → $2^2 = 4\times$); the Rayleigh resolving angle $1.22 \times 550\ \mathrm{nm}/0.10\ \mathrm{m} = 6.71 \times 10^{-6}\ \mathrm{rad} \approx 1.4$ arcseconds.

Prerequisites. Simple and Compound Microscopes — the two-lens, visual-angle idea this article turns outward, and the resolving-power limit now written as an angle. The thin-lens and mirror formulas were established earlier in this chapter and are quoted here. Leads to. Instrument Magnification Numericals — the drill that runs every microscope, telescope and eye-correction case together.

Go deeper (UG). The diffraction theory behind the Rayleigh criterion, aperture synthesis, aberration correction and the full optics of astronomical instruments are developed in Optical Instruments in the university optics track, and the wave physics of resolution opens in the next chapter of this subject. Nothing here depends on it; the magnifications are pure ray optics.