The microscope

Hold a dewdrop up to a blade of grass and, for a moment, you have built the oldest microscope there is — a curved surface of water bending light so that the tiny veins of the leaf swell into view. Every microscope since, from a jeweller's loupe to the compound brass instruments in a school lab, does exactly this one trick with light: it bends rays with convex lenses so that a very small object throws a very large angle at your eye. Nothing about the object changes size. What changes is how big an angle it commands on your retina, and a bigger angle is read by the brain as a bigger, closer object. That single idea — angular size, not linear size — is the entire physics of magnification, and it is why the topic sits at the very end of the Ray Optics and Optical Instruments chapter, right after you have mastered the thin lens formula and the lens maker's formula. A simple microscope is nothing more than one convex lens of short focal length, held close to the eye, doing that angle-enlarging trick directly. A compound microscope stacks two such lenses in series: a short-focal-length objective forms a real, already-magnified image of the object, and a second lens, the eyepiece, then re-magnifies that image the way a simple microscope would. Two modest lenses, used in relay, multiply their powers and deliver magnifications neither could reach alone. This article rebuilds both instruments from the lens formula you already know, derives their magnifying-power formulas, and walks through two worked numericals — one exam-standard, one competitive-exam standard — using the same focal lengths a textbook problem would give you, so you can check every intermediate number as you go.

Key takeaways

A microscope magnifies angular size, not actual size — magnifying power $m$ is the ratio of the angle the image subtends at the eye to the angle the object would subtend at the near point $D = 25\ \text{cm}$ , unaided.
A simple microscope is one convex lens of short focal length $f$ : $m = 1 + D/f$ when the image is at the near point, and $m = D/f$ when the image is at infinity (relaxed eye).
A compound microscope uses a short-focal-length objective ( $f_o$ ) to form a real, inverted, enlarged image, which a short-focal-length eyepiece ( $f_e$ ) then re-magnifies as a simple microscope would.
Total magnifying power is a product: $m = m_o \times m_e$ , with the useful tube-length approximation $m \approx \dfrac{L}{f_o}\left(1 + \dfrac{D}{f_e}\right)$ for the near-point setting and $m \approx \dfrac{L}{f_o}\cdot\dfrac{D}{f_e}$ for the infinity setting.
Both $f_o$ and $f_e$ must be small, and the objective must be short-focus, for large magnification — this is exactly why microscope objectives look like tiny, fat, stubby lenses.
Sign convention is not optional in the objective calculation: a wrong sign on $u_o$ silently flips the answer for $v_o$ and every magnification that follows.
Key idea: a microscope multiplies angular size, and the compound microscope reaches far higher magnifying power than a single lens by cascading a short-focus objective's real image into a short-focus eyepiece acting as a simple magnifier, m = m_o × m_e.

The core insight

A microscope enlarges the angle an object subtends at the eye rather than the object itself, and a compound microscope multiplies this effect by first forming a real, magnified image with a short-focal-length objective and then re-magnifying that image with a short-focal-length eyepiece acting as a simple microscope, so total magnifying power is the product $m = m_o \times m_e$ .

Where it fits in the syllabus

This topic closes the "Optical instruments" section at the end of the CBSE (NCERT) Class 12 Ray Optics and Optical Instruments chapter, immediately after spherical mirrors, refraction, and thin lenses — mirror formula, lens formula, lens maker's formula, power of a lens, and lenses in contact. You are expected to know the working principle and ray diagram of the simple microscope (a single convex lens used as a magnifying glass) and the compound microscope (objective plus eyepiece), and to apply their magnifying-power formulas for image formation at the near point and at infinity. Aberrations, scattering, and lenses separated by a distance beyond the microscope's own tube length are outside this chapter's prescribed scope, so none of that appears here.

The physical picture

Simple microscope. Bring a small object closer and closer to your eye and it looks bigger — up to the near point $D$ (taken as $25\ \text{cm}$ for a normal eye), beyond which the eye can no longer focus it comfortably. A convex lens defeats this limit: placed with the object just inside its focal length, it forms a virtual, erect, enlarged image that the eye can view from much closer than $D$ , so the image subtends a far larger angle than the naked object ever could at the near point.

m = 1 + \frac{D}{f} \qquad \text{(image at the near point)}

m = \frac{D}{f} \qquad \text{(image at infinity, relaxed eye)}

where: $D = 25\ \text{cm}$ is the conventional near point of a normal eye, and $f$ is the focal length of the convex lens (positive, since it is convex). Reading these physically: making $f$ smaller — a more strongly curved, "fatter" lens — increases $m$ in both cases, which is why a magnifying glass with a stubby, thick shape magnifies more than a thin, flat one. The image-at-infinity case gives one magnification unit less than the near-point case, because the eye's own accommodation effort at the near point contributes that extra " $+1$ ."

Compound microscope. A single lens runs out of practical magnification quickly, because pushing $f$ too small makes the lens hard to hold and grind, and the image quality suffers. The fix is to split the job: a short-focus objective, placed very close to the object (just outside its focal point $f_o$ ), forms a real, inverted, already-magnified image. That real image then sits inside the focal length of a second short-focus lens, the eyepiece, which magnifies it further exactly as a simple microscope magnifies an object — producing a final large virtual image that the eye views comfortably.

m = m_o \times m_e

m = \frac{v_o}{u_o}\left(1 + \frac{D}{f_e}\right) \approx \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right) \qquad \text{(final image at near point)}

m = \frac{v_o}{u_o}\cdot\frac{D}{f_e} \approx \frac{L}{f_o}\cdot\frac{D}{f_e} \qquad \text{(final image at infinity)}

where: $v_o, u_o$ are the objective's image and object distances (from the lens formula, with the usual sign convention), $m_o = v_o/u_o$ is the objective's linear magnification, $m_e$ is the eyepiece's angular magnification (the simple-microscope formula applied to the eyepiece alone), and $L$ is the tube length — approximately the separation between the objective's real image and the eyepiece. The approximation $v_o \approx L$ holds only when $f_o$ and $f_e$ are both small compared with $L$ , which is exactly the regime a real compound microscope is built in. Reading the formula physically: total magnification grows the more you shrink both focal lengths and the longer the tube — three separate knobs a microscope designer can turn.

Figure

A ray diagram of a compound microscope showing the object near the objective lens forming a real inverted image which lies inside the focal length of the eyepiece, and the eyepiece forming a large final virtual image far to the left

Derivation AThe total magnifying power of a compound microscope

The goal is to build the compound microscope's magnifying power from two pieces of physics a student already owns — the thin lens formula applied to the objective, and the simple-microscope angular magnification formula applied to the eyepiece — and to show precisely why the two combine as a product rather than a sum, and why the tube-length version $L/f_o$ is only an approximation to the exact $v_o/u_o$ . No new physics is introduced; this is bookkeeping of angles and image distances done carefully.

Step 1. Set up the objective stage. The object of height $h$ sits just outside the objective's focal length $f_o$ , at object distance $u_o$ . The thin lens formula $\dfrac1{v_o} - \dfrac1{u_o} = \dfrac1{f_o}$ gives the real image distance $v_o$ , and the objective's linear (transverse) magnification is $m_o = v_o/u_o$ , the ratio of image height to object height (with sign carrying the inversion information). Because $u_o$ is only slightly larger than $f_o$ , this real image is much taller than the object: $m_o$ is large in magnitude.

Step 2. Treat this real image as the "object" for the eyepiece. The eyepiece is positioned so that the objective's real image, of height $h_o = m_o h$ , falls just inside the eyepiece's focal length $f_e$ — exactly the condition for the eyepiece to act as a simple microscope on it. By the simple-microscope result already derived for a single lens, the eyepiece's angular magnification is $m_e = 1 + D/f_e$ (final image at the near point) or $m_e = D/f_e$ (final image at infinity).

Step 3. Multiply, because the two magnifications act on independent variables and compound in sequence. The objective magnifies height by the factor $m_o$ ; the eyepiece then magnifies the angle subtended by that already-enlarged height by the factor $m_e$ . Since the final angle at the eye scales with the intermediate image's height, and that height already scales with the original object's height by $m_o$ , the two multiply: total magnifying power $m = m_o \times m_e = \dfrac{v_o}{u_o}\left(1 + \dfrac{D}{f_e}\right)$ , or $\dfrac{v_o}{u_o}\cdot\dfrac{D}{f_e}$ for the infinity setting. This is a product, not a sum, because each stage multiplies whatever angle or height it receives from the previous stage — exactly like compounding two percentage increases.

Step 4. Replace $v_o$ by the tube length $L$ only as an approximation. The objective and eyepiece are mounted a fixed distance $L$ apart (the "tube length"), and $v_o$ , the distance from the objective to its real image, is close to $L$ whenever both $f_o$ and $f_e$ are small compared with $L$ — which is the design regime of every real compound microscope, since short focal lengths are exactly what large magnification demands. Under that condition $v_o \approx L$ , and since $u_o \approx f_o$ (the object sits almost at the objective's focal point), the ratio $v_o/u_o \approx L/f_o$ , giving the familiar working formulas $m \approx \dfrac{L}{f_o}\left(1 + \dfrac{D}{f_e}\right)$ and $m \approx \dfrac{L}{f_o}\cdot\dfrac{D}{f_e}$ . For an exact numerical answer, always compute $v_o$ and $u_o$ from the lens formula rather than trusting $L$ alone.

Worked examples

Board level - simple microscope magnifying power

A convex lens of focal length $5\ \text{cm}$ is used as a simple microscope by a person whose near point is at $25\ \text{cm}$ . Find the magnifying power when (a) the image is formed at the near point, and (b) the image is formed at infinity.

Step 1. Write down the given data: $f = 5\ \text{cm}$ , $D = 25\ \text{cm}$ . Since both formulas need only these two numbers, no lens-formula calculation is required here — this is a direct substitution problem.

Step 2. Apply the near-point formula. $m = 1 + \dfrac{D}{f} = 1 + \dfrac{25}{5} = 1 + 5 = 6$ .

Step 3. Apply the infinity formula. $m = \dfrac{D}{f} = \dfrac{25}{5} = 5$ .

Step 4. Interpret the two answers together. The magnifying power is $6\times$ when the eye strains to focus the image right at the near point, and $5\times$ — slightly lower, but comfortable, since the eye is relaxed and focused at infinity. This one-unit gap ( $1 + D/f$ versus $D/f$ ) is universal for every simple magnifier and is worth memorising as a pattern, not just a formula.

Competitive level - compound microscope, exact vs approximate

A compound microscope has an objective of focal length $f_o = 1.0\ \text{cm}$ and an eyepiece of focal length $f_e = 5.0\ \text{cm}$ . An object is placed $1.2\ \text{cm}$ from the objective. The final image is formed at the near point $D = 25\ \text{cm}$ . Find (a) the image distance for the objective, (b) the linear magnification of the objective, (c) the total magnifying power, and (d) compare with the tube-length approximation if the objective and eyepiece are $24.0\ \text{cm}$ apart.

Step 1. Apply the lens formula to the objective with correct signs. Using $\dfrac1{v_o} - \dfrac1{u_o} = \dfrac1{f_o}$ and the convention that a real object gives $u_o = -1.2\ \text{cm}$ (measured against the direction of incident light) with $f_o = 1.0\ \text{cm}$ :

\frac{1}{v_o} = \frac{1}{f_o} + \frac{1}{u_o} = \frac{1}{1.0} + \frac{1}{-1.2} = 1.0 - 0.833 = 0.167

v_o = \frac{1}{0.167} \approx 6.0\ \text{cm}

Step 2. Find the objective's linear magnification. $m_o = \dfrac{v_o}{u_o} = \dfrac{6.0}{-1.2} = -5.0$ . The magnitude, $5.0$ , is what enters the magnifying power calculation; the negative sign only confirms the objective's real image is inverted, as it must be.

Step 3. Find the eyepiece's angular magnification and combine. With the final image at the near point, $m_e = 1 + \dfrac{D}{f_e} = 1 + \dfrac{25}{5.0} = 6$ . Total magnifying power: $m = |m_o| \times m_e = 5.0 \times 6 = \mathbf{30}$ .

Step 4. Now check the tube-length shortcut. If the objective and eyepiece are separated by $L = 24.0\ \text{cm}$ , the approximate formula gives $m \approx \dfrac{L}{f_o}\left(1 + \dfrac{D}{f_e}\right) = \dfrac{24.0}{1.0}\times 6 = 144$ — wildly different from the exact answer of $30$ . The mismatch is a warning, not an error: the approximation $v_o \approx L$ is only valid when $v_o$ itself is close to $L$ , and here the exact $v_o = 6.0\ \text{cm}$ is nowhere near $L = 24.0\ \text{cm}$ , so the shortcut cannot be trusted for this particular geometry. Use $L/f_o$ only as a quick estimate or when the problem explicitly says $f_o, f_e \ll L$ ; for an exact numerical, always compute $v_o$ from the lens formula as in Steps 1–3. Timing note: in a competitive exam, compute the objective's $v_o$ first — it takes one line — before reaching for any tube-length shortcut; plugging $L$ in blindly is the single most common way to lose full marks on this topic under time pressure.

Common traps

Confusing the "image at near point" formula ( $1 + D/f$ ) with the "image at infinity" formula ( $D/f$ ) — always check which viewing condition the question specifies before substituting.
Forgetting the sign convention when applying $\dfrac1v - \dfrac1u = \dfrac1f$ to the objective; a wrong sign for $u_o$ flips the sign and magnitude of $v_o$ , and every magnification computed from it.
Treating tube length $L$ as exactly equal to $v_o$ — $L \approx v_o$ is only a good approximation when $f_o, f_e \ll L$ ; for an exact numerical, always compute $v_o$ from the lens formula, as the competitive-level example above shows.
Assuming higher magnifying power always means a clearer, more detailed image — magnifying power measures angular size only; it says nothing about resolution (the ability to separate two close points), which depends on wavelength and aperture, not on $m$ .
Mixing up the objective and the eyepiece in a formula — the objective is always the short-focus lens closest to the object; using $f_e$ where $f_o$ belongs (or vice versa) silently produces a plausible-looking but wrong number.

You should be able to…

State the working principle of a simple microscope and sketch its ray diagram for image formation at the near point and at infinity.
Recall and correctly apply $m = 1 + D/f$ and $m = D/f$ for a simple microscope, choosing the right one from the stated viewing condition.
Describe the construction of a compound microscope (objective plus eyepiece) and explain, in words, the role each lens plays.
Apply the thin lens formula with correct signs to the objective to find $v_o$ and $m_o$ , then combine it with the eyepiece's angular magnification to get the total magnifying power $m = m_o \times m_e$ .
Distinguish the "final image at near point" and "final image at infinity" compound-microscope formulas, and know when the tube-length approximation $L/f_o$ is trustworthy versus when it must be replaced by an exact $v_o/u_o$ calculation.
Explain why magnifying power and resolving power are different physical quantities, and why a large $m$ does not by itself guarantee a sharp, detailed image.

Ask the tutor: want to try a compound microscope numerical with the tube length actually close to $v_o$ , so the approximate and exact formulas agree — or work through why the eyepiece formula uses the simple-microscope result?

Why does a compound microscope give so much more magnification than just using one strong lens?

Because the two lenses multiply their effects rather than adding them. The objective first turns a tiny object into an already-enlarged real image — its linear magnification $m_o$ can easily be $5\times$ or more just from being placed close to its own focal point. The eyepiece then treats that enlarged image exactly as a simple microscope treats an ordinary object, adding its own factor $m_e$ of typically $5\times$ to $10\times$ . Multiplying two moderate factors, say $5 \times 6 = 30$ , easily beats what a single lens could deliver, because a single lens strong enough to give $30\times$ directly would need an impractically short focal length and would suffer badly from image defects.

Is the final image in a compound microscope real or virtual, and does that matter for the formula?

The final image, as seen by the eye through the eyepiece, is virtual — the eyepiece is being used exactly as a simple magnifying glass, and a simple magnifying glass always produces a virtual, erect (relative to the intermediate image) image. Only the objective's image, midway through the instrument, is real. This matters because the simple-microscope formulas $m = 1 + D/f_e$ and $m = D/f_e$ , which assume a virtual image formed by a single convex lens, are exactly the formulas you must apply to the eyepiece stage — not the mirror-style real-image formulas from earlier in the chapter.

Why is D always taken as 25 cm in these formulas, and does the answer change for someone with a different near point?

$D = 25\ \text{cm}$ is simply the conventional near point of a "normal" young eye, adopted so that everyone's magnifying-power numbers are comparable on the same scale; it is not a universal constant of nature like $c$ or $h$ . If a question explicitly gives a different near point — say $20\ \text{cm}$ for a particular observer — you must substitute that value in place of $25\ \text{cm}$ in every formula. A person with a smaller near point (who can focus closer without help) gets a numerically different, but equally valid, magnifying power from the same lens.

If I make fₒ and fₑ both extremely tiny, can I get unlimited magnifying power?

Not in practice, even though the algebra of $m \approx (L/f_o)(1 + D/f_e)$ suggests you can push $m$ as high as you like by shrinking $f_o$ and $f_e$ toward zero. Grinding a lens to an extremely short focal length makes it thick, hard to manufacture accurately, and prone to the aberrations this chapter deliberately keeps out of scope; beyond a point the image gets dimmer and less sharp even as the formula's $m$ keeps climbing. Real microscopes trade off magnifying power against image quality, and that trade-off — resolving power, aperture, aberrations — belongs to a more advanced optics course than this one.