Microscopes

Hold a drop of pond water under a school microscope and paramecia swim into view — creatures far too small for the naked eye to resolve. Every microscope, however elaborate, is doing one simple thing: using a converging lens (or two) to make the angle an object subtends at your eye bigger than it could ever be unaided, so the retina's image grows and detail becomes visible. A single strong convex lens held close to the eye is a simple microscope; put a second lens in front of it to re-image the object first, and you get a compound microscope — the workhorse of every school and research lab. ISC Class XII asks you to derive the magnifying power of both instruments, for the final image at the near point and at infinity, and to know qualitatively what limits how much detail a compound microscope can actually show. Nothing here goes beyond the thin-lens formula you already know; the physics is entirely about angles, not size.

The core insight

A simple microscope enlarges the angle subtended by a nearby object using one convex lens ( $M=1+D/f$ or $D/f$ ), while a compound microscope stacks an objective and an eyepiece so the objective's real magnified image is re-magnified angularly by the eyepiece ( $M=m_o\times M_e$ ), with resolving power set by wavelength and the objective's numerical aperture, not by magnifying power.

The core idea

Why angular size, not linear size, is what matters. The eye judges "bigness" by the angle an object subtends on the retina, not by its actual length. Moving an object closer to the eye already increases this angle — up to the near point $D$ (about 25 cm, the closest distance of comfortable focus), beyond which the eye can no longer accommodate. A lens lets you defeat this limit: it forms a virtual image far enough away to view comfortably while the image still subtends a much larger angle than the object ever could at $D$ . Magnifying power $M$ is therefore always defined as a ratio of two angles — the angle $\beta$ subtended by the final image (through the instrument) to the angle $\alpha$ subtended by the object when placed at the near point without any instrument — using the small-angle approximation $\tan\theta \approx \theta$ :

M=\frac{\beta}{\alpha}

Simple microscope — object inside the focal length. A single convex lens of short focal length $f$ is held so the object sits between the lens and its focus, giving a virtual, erect, enlarged image. Two standard viewing conditions are examined:

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

M=\frac{D}{f}\qquad\text{(final image at infinity, object at the focus, relaxed eye)}

where: $D$ is the least distance of distinct vision (25 cm by convention), and $f$ is the focal length of the lens. Notice the near-point value always exceeds the infinity value by exactly 1 — this is the single most tested relationship in this topic.

Compound microscope — two lenses in series. A short-focal-length objective ( $f_o$ ) faces the object and forms a real, inverted, enlarged intermediate image just inside the focal length of a longer-focal-length eyepiece ( $f_e$ ), which then works as a simple magnifier on that intermediate image. Because the two stages multiply, not add, the total magnifying power is

M=m_o\times M_e

where $m_o=v_o/u_o$ is the objective's linear magnification and $M_e$ is the eyepiece's angular magnification (either $1+D/f_e$ or $D/f_e$ , exactly as for a simple microscope). Writing this out for the near-point case,

M=\frac{v_o}{u_o}\left(1+\frac{D}{f_e}\right)

where: $v_o$ and $u_o$ are the image and object distances for the objective lens, using the usual sign convention. When the object sits just outside the objective's focus so that $u_o\approx f_o$ , the objective's image distance is well approximated by the tube length $L$ (the separation between the two lenses), giving the commonly quoted approximate forms

M\approx\frac{L}{f_o}\left(1+\frac{D}{f_e}\right)\qquad\text{(near point)}\qquad\qquad M\approx\frac{L}{f_o}\times\frac{D}{f_e}\qquad\text{(infinity)}

The design requirement $f_o\ll f_e$ is what makes a compound microscope so much more powerful than a simple one: the objective contributes a large linear magnification while the eyepiece contributes further angular magnification on top of it.

Figure

Ray diagram of a compound microscope showing the objective forming a real inverted intermediate image which the eyepiece then magnifies into a larger virtual image for the eye

Resolving power (qualitative only). Magnifying power tells you how big the image looks; resolving power tells you whether two very close points are actually seen as separate rather than blurring into one. For a compound microscope this ability depends on the wavelength of the light used and the numerical aperture of the objective — shorter wavelength or a larger aperture gives finer resolution. This point is examinable as a statement only; no formula-based derivation is expected at this level.

Derivation AMagnifying power of a compound microscope

Derive the expression for the total magnifying power of a compound microscope with the final image formed at the near point, starting from the definition of angular magnification and the objective's linear magnification, and show why the two contributions multiply rather than add.

Step 1. The intermediate image formed by the objective, of height $h'$ , acts as the "object" for the eyepiece. Its linear magnification relative to the true object (height $h$ ) is the objective's own magnification, $m_o=h'/h=v_o/u_o$ , obtained directly from the thin-lens relation applied to the objective alone.

Step 2. The eyepiece then works exactly like a simple microscope on this intermediate image: it forms a final virtual image at the near point $D$ , subtending angle $\beta$ at the eye, with angular magnification $M_e=\beta/\theta'=1+D/f_e$ , where $\theta'$ is the angle the intermediate image would subtend from the near point without the eyepiece.

Step 3. The total magnifying power is defined as the ratio of the final angle $\beta$ to the angle $\alpha$ the original object would subtend at the near point unaided. Since $\theta'/\alpha$ equals the intermediate image's linear enlargement of the object, that ratio is exactly $m_o$ ; combining with Step 2 gives $M=(\beta/\theta')\times(\theta'/\alpha)=M_e\times m_o$ , i.e. the two magnifications multiply because they act on the light in series, one stage feeding its enlarged angle into the next.

Step 4. Substituting the explicit forms gives the final result, $M=\dfrac{v_o}{u_o}\left(1+\dfrac{D}{f_e}\right)$ , and replacing $v_o$ with the tube length $L$ under the usual approximation $u_o\approx f_o$ yields the commonly used working formula $M\approx\dfrac{L}{f_o}\left(1+\dfrac{D}{f_e}\right)$ .

Worked example

A compound microscope has an objective of focal length 1 cm and an eyepiece of focal length 5 cm, separated by 20 cm. An object is placed 1.2 cm from the objective. Find the magnifying power for the final image formed (a) at the near point ( $D=25\ \text{cm}$ ) and (b) at infinity.

Step 1. Find $v_o$ from the thin-lens formula for the objective, with $f_o=1\ \text{cm}$ and $u_o=-1.2\ \text{cm}$ : $\dfrac{1}{v_o}=\dfrac{1}{f_o}+\dfrac{1}{u_o}=1-0.833=0.167$ , so $v_o\approx6\ \text{cm}$ .

Step 2. Objective's linear magnification: $m_o=v_o/u_o=6/(-1.2)=-5$ , magnitude 5 (the negative sign confirms the intermediate image is inverted; only the magnitude is used in the magnifying-power formula).

Step 3. Eyepiece's angular magnification at the near point: $M_e=1+D/f_e=1+25/5=6$ . Total magnifying power at the near point: $M=|m_o|\times M_e=5\times6=\mathbf{30}$ .

Step 4. For the final image at infinity, the eyepiece term simplifies to $M_e=D/f_e=25/5=5$ instead of 6. Total magnifying power at infinity: $M=|m_o|\times M_e=5\times5=\mathbf{25}$ .

Answer: The compound microscope's magnifying power is 30 for the final image at the near point and 25 for the final image at infinity — consistent with the simple-microscope rule that the near-point value exceeds the infinity value, here by a factor of 5 (the objective's own magnification) applied to the "+1" difference.

Common traps

Using $D/f$ instead of $1+D/f$ for a simple microscope (or for the eyepiece stage of a compound microscope) when the final image is at the near point — the "+1" vanishes only for the infinity case.
Adding $m_o$ and $M_e$ instead of multiplying them; the two stages act in series on the light path, so the total magnifying power is always a product, $M=m_o\times M_e$ .
Approximating $v_o\approx L$ (tube length) without checking that $u_o\approx f_o$ first; for exact numericals always compute $v_o$ from the thin-lens formula as in the worked example, rather than jumping straight to the tube-length shortcut.
Confusing magnifying power with resolving power — a lens can magnify an image as much as you like, but if the objective's aperture and the light's wavelength do not resolve two close points, magnifying the blur only produces a bigger blur, not more detail.

Ask the tutor: Want a walkthrough of the full ray-diagram derivation for the compound microscope, or a reverse problem where you're given the magnifying power and asked to find the tube length?

Why does the "+1" appear for the near-point case but disappear for the infinity case

The "+1" comes from adding the angle due to the image itself to the angle already present from viewing along the lens axis; when the object sits exactly at the focus, the emergent rays are parallel and this axial contribution vanishes, leaving just $D/f$ . At the near point the object is very slightly inside the focus, so both contributions survive and add, giving $1+D/f$ . It is a direct algebraic consequence of the lens formula, not an arbitrary rule to memorize.

Do I need to memorize both the exact and the tube-length-approximate compound microscope formulas

Yes, but understand which to use when. If a numerical gives you $u_o$ and $f_o$ explicitly (as in the worked example), always compute $v_o$ from the lens formula and use the exact $M=(v_o/u_o)(1+D/f_e)$ . If the problem only gives you the tube length $L$ and expects a quick estimate, the approximate $M\approx(L/f_o)(1+D/f_e)$ is acceptable and is what most textbook derivations present first.

Why can a compound microscope magnify a specimen a lot but still not show two very close bacteria as separate

Magnifying power and resolving power answer different questions. Magnifying power only enlarges the angle the image subtends, which you control almost arbitrarily by choosing lenses. Resolving power is limited by diffraction at the objective's aperture and by the wavelength of light, so no amount of extra magnification beyond the resolution limit reveals new detail — you just get a bigger, blurrier image, which is why oil-immersion objectives and shorter-wavelength illumination are used in real microscopes.

Microscopes

The core idea

Worked example

Common traps

Before this

Where this leads