Optical instruments — human eye, simple microscope, compound microscope, telescope: Worked examples

This page is a firing range: we shoot every kind of question the parent topic can ask, and knock each one down with a fully worked example. First we lay out the target board — the scenario matrix — so you can see there are no hidden cells. Then we work each cell, forecasting the answer before computing it.

Before we start, three reminders we will lean on in every example:

Definition The Cartesian sign convention (applies to EVERY example, not just when we mention it)

From Lens equation and sign conventions: light travels left→right; distances measured in the direction light travels are positive, those against it are negative. So a real image on the far side has $v > 0$ ; a real object on the near side has $u < 0$ ; a virtual image on the same side as the object has $v < 0$ . The lens equation is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with a converging lens having $f > 0$ . In the "shortcut" formulas ( $M_{D} = 1 + D / f$ , $D / f$ , $L / f_{o} \cdot D / f_{e}$ , $f_{o} / f_{e}$ ) all quantities are already written as positive magnitudes — the signs were used once to derive them (as we show explicitly in Ex 4) and then dropped. So Examples 1, 2, 3, 5, 6 use only magnitudes; Examples 4 and 7 return to signed distances where the raw lens equation is needed.

Recall The master definition (needed from line one)

$M = \frac{θ _{with instrument}}{θ _{naked eye}}$ — a ratio of angles, because retinal image size is set by the angle a thing subtends, not its real height. See Linear vs angular magnification.

The scenario matrix

Every optical-instrument problem falls into one of these cells. The last column names the worked example that covers it.

#	Cell class	What varies / the twist	Covered by
A	Simple microscope, both eye settings	image at $D$ vs at $\infty$	Ex 1
B	Simple microscope, design backwards	given $M$ , find $f$ (power in dioptres)	Ex 2
C	Compound microscope, relaxed eye	multiply, not add	Ex 3
D	Compound microscope, near-point eye + tube length	eyepiece uses $1 + D / f_{e}$ ; find $L$ from lens eqn	Ex 4
E	Telescope, both eye settings	normal vs near-point adjustment	Ex 5
F	Telescope, word problem (real sky)	angle in, physical size out	Ex 6
G	Degenerate / limiting input	$u \to f$ ( $M \to \infty$ ), $f \to \infty$ ( $P \to 0$ ), object at $D$ , $f_{e} \to f_{o}$	Ex 7
H	Exam twist — spot the wrong formula	telescope disguised as a microscope	Ex 8

The sign convention (stated in the definition box above) applies to every example; where signed distances actually matter — Examples 4 and 7 — we restate it in place.

Example 1 — Simple microscope, both eye settings (Cell A)

Forecast: guess before reading on — will the relaxed setting give a bigger or smaller number? (Hint: comfort usually costs magnification.)

Image at near point. Use $M_{D} = 1 + \frac{D}{f}$ (recall $M_{D}$ = image at $D$ , from the subscript box). $M_{D} = 1 + \frac{25}{5} = 1 + 5 = 6.$ Why this step? Putting the virtual image exactly at $D$ is the closest the eye can focus, so the object can sit closest to the lens → largest angle → largest $M$ .
Relaxed eye. Use $M_{\infty} = \frac{D}{f}$ ( $M_{\infty}$ = image at infinity). $M_{\infty} = \frac{25}{5} = 5.$ Why this step? Object at the focus $\Rightarrow$ rays leave parallel $\Rightarrow$ eye muscles relaxed. No "+1" because the image is at infinity, not at $D$ .
Difference. $M_{D} - M_{\infty} = 6 - 5 = 1.$ Why this step? Always exactly $1$ : $(1 + \frac{D}{f}) - \frac{D}{f} = 1$ . The "+1" is the price of the sharpest possible view.

Verify: Units: $D$ and $f$ both in cm, so $D / f$ is a pure number — magnification is dimensionless ✓. Sanity: $M_{D} > M_{\infty}$ ✓, difference is exactly $1$ ✓.

Example 2 — Design backwards: given $M$ , find $f$ and power (Cell B)

Forecast: A stronger magnifier — will its focal length be longer or shorter than the $5 cm$ lens of Ex 1?

Invert the relaxed formula. $M_{\infty} = \frac{D}{f} \Rightarrow f = \frac{D}{M}$ . $f = \frac{25}{10} = 2.5 cm .$ Why this step? We are given $M$ and want $f$ ; the relaxed formula is the simplest link between them, so we solve it for $f$ .
Convert to power. From Power of a lens and dioptres, $P = \frac{1}{f ( in metres )}$ . $f = 2.5 cm = 0.025 m \Rightarrow P = \frac{1}{0.025} = 40 D .$ Why this step? Power is defined in dioptres, which needs $f$ in metres, so we convert first.

Verify: Plug back: $D / f = 25/2.5 = 10 = M$ ✓. Shorter $f$ than Ex 1's $5 cm$ (as forecast — stronger lens is fatter/shorter $f$ ) ✓. Power positive because it's a converging lens ✓.

Example 3 — Compound microscope, relaxed eye (Cell C)

Forecast: Guess the order of magnitude — tens? hundreds? And decide now: do the two stages add or multiply?

Objective (linear) magnification. $m_{o} \approx \frac{L}{f _{o}}$ . $m_{o} = \frac{20}{1} = 20.$ Why this step? The object sits just outside $f_{o}$ , so $u_{o} \approx f_{o}$ and $v_{o} \approx L$ , giving $m_{o} = v_{o} / u_{o} \approx L / f_{o}$ .
Eyepiece magnification (relaxed). $m_{e} = \frac{D}{f _{e}}$ . $m_{e} = \frac{25}{2.5} = 10.$ Why this step? The eyepiece is just a simple microscope acting on the intermediate image, relaxed setting → use $D / f_{e}$ .
Combine — multiply. $M = m_{o} m_{e} = 20 \times 10 = 200.$ Why this step? The eyepiece magnifies the already magnified intermediate image, so magnifications compound like two photocopies (×2 then ×3 = ×6), never add.

Verify: $20 \times 10 = 200$ ; the wrong "add" answer would be $30$ . A ×2-then-×3 copier gives ×6 not ×5, confirming multiplication ✓.

Example 4 — Compound microscope, near-point eye + find tube length (Cell D)

Figure — Optical instruments — human eye, simple microscope, compound microscope, telescope

Forecast: Near-point eyepiece uses $1 + D / f_{e}$ instead of $D / f_{e}$ — so is this $M$ bigger or smaller than the relaxed version would give?

Read the figure above first. It lays the tube out on the axis: the objective sits at position $0$ , the real intermediate image forms at $v_{o} = 12 cm$ to its right (orange arrow), and the eyepiece sits a little further right. The eyepiece treats that intermediate image as its own object at distance $∣ u_{e} ∣$ to its left, and throws a virtual final image all the way out to $D = 25 cm$ (plum, dashed, far left). The bracket at the bottom shows why $L = v_{o} + ∣ u_{e} ∣$ : the tube must physically span from the objective to the eyepiece, i.e. objective→image ( $v_{o}$ ) plus image→eyepiece ( $∣ u_{e} ∣$ ).

Objective, exact lens equation. $\frac{1}{v _{o}} - \frac{1}{u _{o}} = \frac{1}{f _{o}}$ with $v_{o} = + 12$ , $f_{o} = + 2$ . $\frac{1}{u _{o}} = \frac{1}{v _{o}} - \frac{1}{f _{o}} = \frac{1}{12} - \frac{1}{2} = \frac{1 - 6}{12} = - \frac{5}{12} \Rightarrow u_{o} = - 2.4 cm .$ Why this step? We need the true object distance to get the exact $m_{o}$ ; don't approximate when a value is handed to us. The negative $u_{o}$ just confirms the object is on the incoming-light side, as the convention demands.
Objective magnification. $m_{o} = \frac{v _{o}}{u _{o}} = \frac{12}{2.4} = 5.$ Why this step? Linear magnification of the real intermediate image is $v_{o} / u_{o}$ ; we take magnitude for its size factor.
Eyepiece, near-point. $m_{e} = 1 + \frac{D}{f _{e}} = 1 + \frac{25}{5} = 6.$ Why this step? Final image at $D$ (not $\infty$ ) → use the "+1" form, exactly as the simple microscope at the near point.
Total. $M = m_{o} m_{e} = 5 \times 6 = 30.$ Why this step? Same multiply rule as Ex 3.
Tube length — geometric argument tied to the figure. In the figure, the eyepiece's object is the intermediate image; its virtual final image is on the same side as that object (both to the left of the eyepiece), so by the sign convention $v_{e} = - 25$ , with $f_{e} = + 5$ . Solve the lens equation for the eyepiece object distance $u_{e}$ : $\frac{1}{v _{e}} - \frac{1}{u _{e}} = \frac{1}{f _{e}} \Rightarrow \frac{1}{u _{e}} = \frac{1}{v _{e}} - \frac{1}{f _{e}} = - \frac{1}{25} - \frac{1}{5} = - \frac{6}{25} \Rightarrow u_{e} = - \frac{25}{6} \approx - 4.17 cm .$ The negative sign says the object (the intermediate image) is $4.17 cm$ to the left of the eyepiece — exactly the gap drawn in the figure. The tube must cover objective→image and image→eyepiece: $L = objective \to image v_{o} + image \to eyepiece ∣ u_{e} ∣ = 12 + 4.17 = 16.17 cm .$ Why this step? We add the two physical gaps because the objective and eyepiece are separated by exactly that total distance; the image sits in between them.

Verify: $m_{o}$ : $∣12/ (- 2.4) ∣ = 5$ ✓. Check lens eqn for objective: $1/12 - 1/ (- 2.4) = 1/12 + 5/12 = 6/12 = 1/2 = 1/ f_{o}$ ✓. $m_{e} = 1 + 5 = 6$ ✓. $M = 30$ ✓. $∣ u_{e} ∣ = 25/6 \approx 4.17$ ✓, $L \approx 16.17 cm$ ✓.

Example 5 — Telescope, both eye settings (Cell E)

Forecast: Which adjustment gives the bigger $M$ — relaxed (normal) or near-point? (Telescopes behave the same direction as microscopes here.)

Read the figure first. In the picture above, orange rays enter from the far-left at the small real star-angle $α$ measured against the optical axis. They cross the objective (teal lens) and meet at the ==image height $h$ == in the common focal plane (dashed line). The teal rays then leave the eyepiece (plum lens) parallel again, but now tilted at the much fatter angle $β$ . The two shaded angles $α$ (orange, at the objective) and $β$ (teal, at the eyepiece exit) are exactly the two angles in $M = β / α$ — the figure is the derivation.

Normal adjustment. Reading the two right triangles in the figure, the same image height $h$ subtends $α \approx h / f_{o}$ at the objective focus and $β \approx h / f_{e}$ at the eyepiece focus. So $M = \frac{β}{α} = \frac{h / f _{e}}{h / f _{o}} = \frac{f _{o}}{f _{e}} = \frac{100}{2} = 50.$ Why this step? The $h$ cancels — the star's real size never enters, only angles. This is why a telescope quotes angular magnification.
Tube length (normal). $L = f_{o} + f_{e} = 100 + 2 = 102 cm .$ Why this step? In normal adjustment the objective's focus and the eyepiece's focus coincide (the dashed line in the figure), so the tube is the sum of the two focal lengths.
Near-point adjustment — sketch the algebra. Now the eyepiece is pulled in so its virtual output image sits at $D$ , not $\infty$ . The eyepiece then acts as a simple microscope on the image of height $h$ , so its angular magnification becomes the near-point form $m_{e} = 1 + \frac{D}{f _{e}}$ applied to the angle. Building $M = β / α$ from scratch: $α \approx \frac{h}{f _{o}}, β = \frac{h}{f _{e}} (1 + \frac{f _{e}}{D}) (eyepiece near-point angle) .$ Divide: $M = \frac{β}{α} = \frac{f _{o}}{f _{e}} (1 + \frac{f _{e}}{D}) .$ Now plug numbers: $M = 50 (1 + \frac{2}{25}) = 50 \times \frac{27}{25} = 54.$ Why this step? Straining the eye to focus the eyepiece output at $D$ (instead of $\infty$ ) squeezes out the extra factor $(1 + \frac{f _{e}}{D})$ — the same "you pay comfort for magnification" idea as the microscope.

Verify: $100/2 = 50$ ✓; $L = 102$ ✓; $50 \times 27/25 = 1350/25 = 54$ ✓. Near-point $>$ normal, as forecast ✓.

Example 6 — Telescope word problem: real sky (Cell F)

Forecast: A whole $0. 5^{\circ}$ blown up 50×... is the answer a few degrees, or tens of degrees (i.e. filling much of your view)?

Apply the definition of $M$ . $M = \frac{β}{α} \Rightarrow β = M α .$ Why this step? $M$ is the ratio of apparent-to-real angle — the whole point of the instrument. Note we work directly with the angles $α, β$ ; the image height $h$ from Ex 5 is not needed here because it cancelled out of $M$ already.
Plug numbers. $β = 50 \times 0. 5^{\circ} = 2 5^{\circ} .$ Why this step? Direct substitution; both angles in degrees, so no conversion needed.

Verify: $50 \times 0.5 = 25$ ✓. Sanity: $2 5^{\circ}$ is a big chunk of the sky (your fist at arm's length is ~ $1 0^{\circ}$ ), matching the "Moon looks huge in a telescope" experience ✓. Units: $M$ dimensionless × degrees = degrees ✓.

Example 7 — Degenerate & limiting inputs (Cell G)

Forecast: Guess each limit: does magnification blow up, vanish, or settle at 1?

(a) $f \to \infty$ . $M_{\infty} = \frac{D}{f} \to \frac{25}{\infty} = 0$ ; and $M_{D} = 1 + \frac{D}{f} \to 1.$ Why this step? A flat lens ( $P = 0$ , see Power of a lens and dioptres) bends nothing, so it can't enlarge any angle. The near-point form floors at $M_{D} = 1$ — literally "no help beyond the naked eye at $D$ ."
(b) No lens, object at $D$ . With instrument angle $= θ_{0} = h / D$ = naked-eye angle, so $M = \frac{θ _{0}}{θ _{0}} = 1.$ Why this step? Consistency check: "no instrument" must give $M = 1$ by definition. It agrees with the $f \to \infty$ limit of $M_{D}$ in (a) ✓.
(c) $f_{e} \to f_{o}$ . $M = \frac{f _{o}}{f _{e}} \to \frac{f _{o}}{f _{o}} = 1.$ Why this step? Equal focal lengths turn out no angular gain — the eyepiece spreads the angle by exactly the amount the objective shrank it. To magnify you need $f_{o} > f_{e}$ .
(d) Object at the focus, $u \to f$ (the classic $M \to \infty$ edge). Use the signed lens equation $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with a real object $u = - f$ (object at focal point, incoming side): $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{f} + \frac{1}{- f} = 0 \Rightarrow v \to \infty.$ The linear magnification $m = \frac{v}{u} \to \frac{\infty}{f} \to \infty.$ Why this step? When the object sits exactly at the focus, the image races off to infinity, so its size (and the linear magnification) blows up. In practice you can't view it — the image is infinitely far and infinitely large — which is why the usable magnifier is placed with the object just inside $f$ , giving the finite $M_{D} = 1 + D / f$ of Ex 1. This is the boundary case that the finite formulas approach.

Verify: (a) $lim_{f \to \infty} 25/ f = 0$ , $lim 1 + 25/ f = 1$ ✓. (b) any $h, D$ give ratio $1$ ✓. (c) $f_{o} / f_{o} = 1$ ✓. (d) at $u = - f$ , $1/ v = 0 \Rightarrow v = \infty$ , so $m \to \infty$ ✓. All limits are finite or diverge sensibly — no scenario left uncovered.

Example 8 — Exam twist: spot the wrong formula (Cell H)

Forecast: Should a telescope with an $80 cm$ objective magnify only ×6? Trust your gut here.

Identify the error. The student used the eyepiece-as-magnifier formula $D / f_{e}$ . That treats the object as sitting at the near point $D$ . Why this step? Choosing the reference angle is the whole game — you must first ask "where is the object?" A telescope's object is at infinity, not at $D$ .
Correct reference angle. The star's real angle is $α = h / f_{o}$ , not $h / D$ . So use $M = \frac{f _{o}}{f _{e}}$ . $M = \frac{80}{4} = 20.$ Why this step? With the object at infinity the naked-eye angle is the actual $α$ , giving $M = β / α = f_{o} / f_{e}$ .
Why the mistake feels right. Both instruments have an eyepiece acting as a simple magnifier, so $D / f_{e}$ looks familiar — but it only applies when the thing being viewed by the naked eye sits at $D$ , which is false for a star.

Verify: Correct $M = 80/4 = 20$ ; the wrong value $25/4 = 6.25$ is ~3× too small ✓. Sanity: big objective ⇒ big magnification, so ×20 ≫ ×6.25 is the right direction ✓.

Recall One-line self-test

Cover the answers. Which cell uses "+1"? ::: Near-point settings — simple ( $1 + D / f$ ) and compound eyepiece; telescope instead multiplies $f_{o} / f_{e}$ by $(1 + f_{e} / D)$ . Telescope object is at ::: infinity, so reference angle is $α = h / f_{o}$ , not $h / D$ . As $f_{e} \to f_{o}$ a telescope's $M \to$ ::: $1$ (no magnification). As the object reaches the focus ( $u \to f$ ) a simple lens's linear $m \to$ ::: $\infty$ (image races to infinity). No-instrument magnification is always ::: exactly $1$ . Full telescope near-point formula? ::: $M = \frac{f _{o}}{f _{e}} (1 + \frac{f _{e}}{D})$ — the $f_{o} / f_{e}$ times the bracket, not just the bracket.

Connections

Built on Optical Instruments — Eye, Microscope, Telescope
Uses Lens equation and sign conventions (Ex 4, Ex 7)
Uses Power of a lens and dioptres (Ex 2, Ex 7)
Compare with Linear vs angular magnification (Ex 3 vs Ex 5)
Further reading: Resolving power and diffraction limit, Defects of vision — myopia, hypermetropia, Reflecting telescope (Cassegrain) vs refracting