Optical instruments

Hold a magnifying glass over the small print of a newspaper and the letters swell into comfortable, readable shapes. Point a pair of binoculars at a bird on a distant branch and it seems to leap closer. Look through a laboratory microscope at a drop of pond water and a whole invisible world of paramecia and algae swims into view. None of these devices creates light or magic — each is simply a clever arrangement of one or two lenses (or a mirror plus a lens) that bends the light already leaving an object so that the angle it makes at your eye is bigger than it would be if you looked at the object unaided. That bigger angle is what your brain reads as "bigger." This chapter is the payoff of everything learned earlier in Ray Optics: the thin lens formula, magnification, and lens power all quietly do their job inside three everyday instruments — the simple microscope, the compound microscope, and the astronomical telescope. The simple microscope is a single convex lens used close to the eye. The compound microscope adds a second lens to push magnification far higher, at the cost of a shorter working range. The telescope flips the geometry entirely: instead of bringing a small nearby object closer, it collects the nearly-parallel rays from something impossibly far away and re-bends them so a small eyepiece lens can present them at a large, comfortable angle. By the end you should be able to derive and apply the magnifying-power formula for each instrument, know exactly when the "+1" appears and when it vanishes, and understand why every large modern astronomical telescope uses a giant mirror rather than a giant lens as its light-collecting objective.

Key takeaways

A simple microscope is one convex lens held close to a small object placed inside its focal length; the eye sees a magnified, virtual, erect image.
Magnifying power of a simple microscope: $m = 1+\dfrac{D}{f}$ when the image is at the near point, $D=25$ cm, and $m=\dfrac{D}{f}$ when the relaxed eye views the image at infinity.
A compound microscope uses an objective (short $f_o$ , forms a real inverted enlarged image) followed by an eyepiece (acts as a simple microscope on that image); overall magnifying power is the product $m=m_o\times m_e$ .
An astronomical telescope uses an objective of large focal length $f_o$ to collect near-parallel rays from a distant object, and an eyepiece of small focal length $f_e$ ; for the final image at infinity, $m=\dfrac{f_o}{f_e}$ .
A reflecting telescope replaces the objective lens with a large concave mirror, avoiding the manufacturing limits and chromatic problems of very large lenses, while keeping the same $m=f_o/f_e$ formula.
Telescope resolving/light-gathering power depends on the objective's diameter, not on magnifying power, which depends only on the ratio of focal lengths.
Key idea: every optical instrument works by re-bending light so that the visual angle at the eye is larger than the object's unaided angle, and each instrument's magnifying power reduces to a simple ratio or sum of the focal lengths and object/image distances of its lens(es).

The core insight

Optical instruments magnify not by making objects bigger but by increasing the angle they subtend at the eye, and for the simple microscope, compound microscope, and astronomical telescope this angular gain reduces to $m=1+D/f$ , $m=m_o m_e$ , and $m=f_o/f_e$ respectively.

Where it fits in the syllabus

This is the closing topic of Unit IX, "Ray Optics and Optical Instruments," in CBSE Class 12 Physics, and it directly follows the topic on refraction through a prism. It rests entirely on tools built earlier in the chapter — the thin lens formula $1/v-1/u=1/f$ , magnification, and lens power — applied now to two-lens (or lens-plus-mirror) systems. The prescribed scope is the working and magnifying-power formulas for the simple microscope, the compound microscope, and the astronomical telescope in both its refracting (two-lens) and reflecting (mirror objective) forms. Aberrations, the displacement method for focal length, and lens combinations separated by a distance are explicitly outside this year's syllabus, so this article stays strictly at the level of magnifying-power formulas and their direct application, without drifting into instrument-design detail.

The physical picture

Every optical instrument in this topic solves the same underlying problem: the unaided eye can only resolve two points as separate if they subtend a large enough angle at the eye, and it can only focus comfortably on objects between its near point ( $D=25$ cm, the standard value used throughout this syllabus) and infinity. An object that is either too small or too far away subtends too small an angle for its detail to register. Optical instruments do not change the object; they change the angle the light finally makes at the eye. For a simple microscope, a convex lens is placed so the object sits just inside its focus, producing a virtual, erect, enlarged image that the eye can then view comfortably. The magnifying power compares the angle subtended by this image to the angle the object itself would subtend if placed at the near point without any lens:

m = 1+\frac{D}{f}

where: $D=25$ cm is the least distance of distinct vision (assumed value) and $f$ is the focal length of the lens, used when the final image is formed exactly at the near point. If instead the eye is relaxed and views the image at infinity (object placed exactly at the focus),

m = \frac{D}{f}

Physically, the "+1" case squeezes out slightly more magnification because the image is pulled in closer than infinity, but it forces the eye to strain to focus at 25 cm; the infinity case is one unit of magnification lower but restful for long viewing. For a compound microscope, two convex lenses share the work. The objective, of short focal length $f_o$ , sits close to the object and forms a real, inverted, already-magnified image; this image then becomes the "object" for the eyepiece, of focal length $f_e$ , which works exactly like a simple microscope and magnifies that intermediate image again. Because the two magnifications act in sequence on the same light path, the overall magnifying power multiplies:

m = m_o\times m_e = \frac{v_o}{u_o}\times\left(1+\frac{D}{f_e}\right)

where: $v_o,u_o$ are the image and object distances for the objective lens (so $v_o/u_o$ is the objective's linear magnification), and the eyepiece term is the familiar simple-microscope formula with the near-point image case, since a compound microscope's final image is conventionally taken at the near point. For an astronomical telescope, the geometry is the opposite of a microscope: the object is so far away that rays entering the objective are essentially parallel, and the objective (large focal length $f_o$ ) brings them to a real image close to its own focal point. The eyepiece (small focal length $f_e$ ) is placed to view this image the way a simple microscope views a nearby object. For the common case of the final image formed at infinity (relaxed eye, both lenses' foci coinciding),

m = \frac{f_o}{f_e}

where: $f_o$ is the objective's focal length and $f_e$ the eyepiece's; a long objective and a short eyepiece together give high magnification. A reflecting telescope uses the identical formula with $f_o$ now the focal length of a large concave (in practice, parabolic) mirror acting as the objective, with a small eyepiece lens still used to view the image the mirror forms.

Figure

Three optical instruments compared side by side, a simple microscope showing one convex lens with a small object inside its focus producing a large virtual image, a compound microscope showing a short focal length objective forming a real inverted image that the eyepiece further magnifies, and an astronomical telescope showing near parallel rays from a distant object focused by a long focal length objective and viewed through a short focal length eyepiece

Derivation AMagnifying power of an astronomical telescope with the image at infinity

The goal is to obtain the magnifying power $m=f_o/f_e$ of a two-lens refracting astronomical telescope for the common relaxed-eye case, using only the small-angle approximation and the geometry of two thin lenses sharing a focal point, with no calculus and no aberration theory. Magnifying power is defined as the ratio of the angle $\beta$ subtended at the eye by the final image to the angle $\alpha$ that the distant object would itself subtend if viewed directly with the naked eye.

Step 1. Set up the objective. Because the object is effectively at infinity, all rays from any one point on it arrive at the objective lens as a bundle of parallel rays inclined at the small angle $\alpha$ to the principal axis. Such a parallel bundle is brought to a real, inverted image at the objective's focal plane, at a distance $f_o$ from the objective, and the image height there is $h = f_o\tan\alpha \approx f_o\alpha$ for small angles.

Step 2. Position the eyepiece for a relaxed eye. For the final image to form at infinity (so the eye views it without straining to focus), this intermediate real image at height $h$ must itself lie exactly at the focal point of the eyepiece. Since the objective's image is at distance $f_o$ from the objective and must coincide with the eyepiece's focus at distance $f_e$ from the eyepiece, the two lenses are placed a distance $f_o+f_e$ apart, with their foci coinciding at that one point.

Step 3. Find the angle subtended after the eyepiece. Rays from the top of this intermediate image, height $h$ , pass through the eyepiece's focus and emerge parallel again (since the image sits exactly at the eyepiece's focal plane), making an angle $\beta$ with the axis given by $\tan\beta \approx \beta = h/f_e$ , again using the small-angle approximation for the short focal length $f_e$ .

Step 4. Take the ratio. Magnifying power is $m=\beta/\alpha$ . Substituting $h=f_o\alpha$ from Step 1 into $\beta=h/f_e$ gives $\beta = (f_o\alpha)/f_e$ , so

m=\frac{\beta}{\alpha}=\frac{f_o\alpha/f_e}{\alpha}=\frac{f_o}{f_e}

The object's own angle $\alpha$ cancels out entirely, which is exactly why the formula depends only on the two focal lengths and not on how far away the object actually is — true for any sufficiently distant object, whether it is the Moon or a star.

Worked examples

Board level - Magnifying power of a simple magnifying glass

Question: A magnifying glass has a focal length of 5 cm. Find its magnifying power (a) when the final image is formed at the near point ( $D=25$ cm), and (b) when the final image is formed at infinity.

Step 1. Identify the formula for the near-point case. With $D=25$ cm and $f=5$ cm,

m=1+\frac{D}{f}=1+\frac{25}{5}=1+5=6

Step 2. Identify the formula for the infinity case. Here the "+1" is dropped because the image, not just the object, is effectively at infinity:

m=\frac{D}{f}=\frac{25}{5}=5

Step 3. Compare and interpret. The near-point case gives the larger magnifying power, m = 6, while the relaxed-eye case gives m = 5. The difference of exactly one unit is the "+1" term itself; despite the lower number, viewing with the image at infinity is more comfortable over long periods since the eye muscles stay relaxed rather than straining to focus at 25 cm.

Competitive level - Redesigning a telescope for higher magnification

Question: An astronomical (refracting) telescope has an objective lens of focal length 100 cm and an eyepiece of focal length 5 cm. Find its magnifying power for the final image at infinity, and find what new eyepiece focal length would be needed to double the magnifying power without changing the objective. Also state the length of the tube (distance between the lenses) in both cases.

Step 1. Compute the original magnifying power using $m=f_o/f_e$ :

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

Step 2. Compute the original tube length. For the final image at infinity, the lenses are separated by $f_o+f_e = 100+5=105$ cm.

Step 3. Find the new eyepiece focal length for double the magnification, keeping $f_o=100$ cm fixed. We need $m'=40=100/f_e'$ , so $f_e'=100/40=2.5$ cm.

Step 4. Compute the new tube length and compare. The new separation is $f_o+f_e'=100+2.5=102.5$ cm — only slightly shorter, because the change is entirely in the tiny eyepiece, not the long objective. The magnifying power doubles to 40 by halving the eyepiece focal length from 5 cm to 2.5 cm, while the tube length barely changes, showing that the eyepiece is the "fine control" on magnification and the objective the dominant length. Timing note: this is a two-line, two-formula problem once you recognise $m=f_o/f_e$ and $L=f_o+f_e$ — budget under 2 minutes; do not waste time re-deriving the formula in an exam, just quote and substitute.

Common traps

Using $m=1+D/f$ for the simple microscope even when the final image is meant to be at infinity — the "+1" belongs only to the near-point case; drop it for the relaxed-eye case.
Forgetting that the compound microscope's overall magnification is a product, $m_o\times m_e$ , not a sum, of the objective's linear magnification and the eyepiece's angular magnification.
Assuming a telescope's magnifying power depends on the objective's diameter — magnifying power depends only on the ratio of focal lengths $f_o/f_e$ ; diameter instead controls light-gathering power and resolution, a separate consideration that is why large-diameter mirrors, not just long focal length, matter for real astronomical telescopes.
Mixing up which lens is the objective and which is the eyepiece in the two instruments — in both, the objective forms a real intermediate image and the eyepiece magnifies it further, but the object distance relative to focal length differs sharply: microscope objective sits very close to a small nearby object just outside its focus, while a telescope objective effectively receives an object at infinity so its image forms essentially at its own focal point.
Forgetting to convert focal lengths and $D$ to the same length unit before substituting, especially when a question mixes centimetres and metres.

Common trapLarge lenses are not simply scaled-up small lenses, which is why big telescopes use mirrors

It is tempting to think a bigger telescope just needs a bigger, longer-focal-length version of the same objective lens used in a small refractor, since $m=f_o/f_e$ rewards a long $f_o$ . In practice this runs into hard physical limits that a simple ray diagram hides: a large glass lens is heavy, sags and distorts under its own weight (it can only be supported at its rim, unlike a mirror which can be supported across its whole back), is extremely difficult and expensive to cast and polish to precision over a large diameter, and — because refraction bends different colours by slightly different amounts — introduces chromatic blurring at the edges. A mirror, by contrast, reflects all colours by exactly the same law regardless of wavelength, can be supported from behind across its entire surface, and can be built far larger in practice. This is precisely why every major modern astronomical telescope is a reflector with a large concave mirror as its objective, even though the magnifying-power formula $m=f_o/f_e$ looks identical either way.

You should be able to…

State and apply the magnifying-power formulas for a simple microscope, for both the near-point case ( $m=1+D/f$ ) and the infinity case ( $m=D/f$ ).
State and apply the magnifying-power formula for a compound microscope, $m=m_o\times m_e$ , correctly identifying the objective and eyepiece roles.
Derive and apply the magnifying-power formula for an astronomical telescope, $m=f_o/f_e$ , for the final image at infinity, and compute the tube length $f_o+f_e$ .
Explain, in physical terms, why large modern astronomical telescopes use a concave mirror rather than a lens as the objective.
Distinguish magnifying power (depends on focal-length ratio) from light-gathering/resolving power (depends on objective diameter).

Ask the tutor: Want a side-by-side walkthrough comparing how the microscope and telescope each use their objective and eyepiece differently, or help setting up a magnifying-power numerical step by step?

Why does a shorter eyepiece focal length give more magnification, when a shorter objective focal length gives less?

The two lenses play opposite roles in the formula $m=f_o/f_e$ . A longer objective focal length $f_o$ spreads the intermediate real image over a larger height for a given incoming angle, which is why it sits on top. A shorter eyepiece focal length $f_e$ means that same image height is viewed from very close up, producing a larger final angle at the eye, which is why it sits on the bottom — smaller $f_e$ means larger $m$ . Physically, you are trading a wide, gentle bend at the big front lens for a tight, sharp bend at the tiny lens right next to your eye.

Is the image in a telescope or microscope always upside down?

For a single simple microscope the final image is virtual and erect, same way up as the object. For a compound microscope and for a basic two-lens astronomical telescope, the objective forms a real, inverted intermediate image, and the eyepiece (acting like a simple microscope on that inverted image) does not flip it back — so the final image is inverted relative to the object. For astronomical viewing this rarely matters, since there is no "up" in space, but terrestrial telescopes and binoculars add extra erecting lenses or prisms specifically to flip the image back the right way up for comfortable everyday use; that erecting system is beyond this syllabus.

Why is the near point taken as exactly 25 cm and not some other number?

$D=25$ cm is a conventional, rounded average value for the closest distance at which a normal healthy human eye can focus comfortably without strain, used uniformly across this syllabus so that magnifying-power numbers are comparable between students and problems. Real near points vary somewhat from person to person and increase with age, but for every calculation in this chapter you should simply use $D=25$ cm unless a question explicitly states a different value.

Do I need to memorise where the object and image actually sit inside a compound microscope, or just the final formula?

For CBSE Class 12, you need the final magnifying-power formula $m=m_o\times m_e=(v_o/u_o)\times(1+D/f_e)$ and the qualitative picture of the objective forming a real inverted image that becomes the eyepiece's object — not a full ray-by-ray derivation with sign conventions for every case. Know which lens plays which role, know that the two magnifications multiply, and be able to substitute numbers into the formula; a detailed multi-step ray-tracing derivation of this particular formula is not required at this level.