Optical instruments — human eye, simple microscope, compound microscope, telescope: Question bank

Definition Every symbol on this page, in plain words

Before you touch a single trap, know exactly what each letter means (all lengths in cm).

$θ$ — the visual angle: the angle a bundle of rays from the object makes at your eye. Bigger $θ$ = bigger picture on the retina.
$D$ — the near point distance, $25 cm$ for a normal eye: the closest a normal eye can focus.
$f$ — focal length of a lens ( $f_{o}$ objective, $f_{e}$ eyepiece). Short $f$ = strongly bending.
$M$ — angular magnification: $M = θ_{with instrument} / θ_{naked eye}$ . $M_{D}$ is its value with the image at the near point; $M_{\infty}$ with the image at infinity (relaxed eye).
$m_{o} = v_{o} / u_{o}$ — the objective's linear magnification (image height ÷ object height).
$u_{o}, v_{o}$ — object and image distances for the objective. Sign convention (Lens equation and sign conventions): distances are measured from the lens, positive along the outgoing-light direction. A real object gives $u_{o} < 0$ ; a real image on the far side gives $v_{o} > 0$ , so $m_{o} = v_{o} / u_{o}$ comes out negative — the minus sign is the inverted image, and we quote its magnitude.
$L$ — tube length: the objective–eyepiece separation, $\approx$ the distance between their two foci.
$h$ — a height (of object or of the intermediate image), $α, β$ — angles at the objective and eyepiece.

Formula Where

$M_{D} = 1 + D / f$ and $M_{\infty} = D / f$ actually come from These two appear all over the page, so derive them once. A magnifier of focal length $f$ holds the object at distance $u$ (so its angle is $θ = h / u$ ) and throws a virtual image to a distance $v$ where the eye can focus. The naked-eye reference angle is $θ_{0} = h / D$ , so $M = θ / θ_{0} = D / u$ .

Relaxed eye: send the image to infinity by putting the object at the focus, $u = f$ . Then $M_{\infty} = \frac{D}{f} .$
Image at near point: force the virtual image out to $v = - D$ . The lens equation $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ gives $\frac{1}{u} = - \frac{1}{D} - \frac{1}{f}$ , so $M_{D} = \frac{D}{u} = 1 + \frac{D}{f} .$ Why the "+1"? The image sits at $D$ instead of infinity, which lets the object creep slightly closer than $f$ ; that extra closeness is exactly one unit of $D / u$ — a genuine geometric bonus you pay for with eye strain. The figure below shows the two ray geometries side by side.

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

True or false — justify

A magnifying glass makes an object bigger because it increases the object's real size.

False — nothing physical changes size; the lens increases the angle

θ

the object subtends at your eye, and retinal image size follows that angle (see Linear vs angular magnification).

Angular magnification

M = 1

means the instrument is useless.

True in the magnifying sense —

M = 1

means the same visual angle as the naked eye at the near point, so the retinal image is no bigger.

For a simple microscope, viewing with the image at the near point always gives more magnification than viewing with a relaxed eye.

True —

M_{D} = 1 + D / f

exceeds

M_{\infty} = D / f

by exactly

1

, but the "+1" costs eye strain, so relaxed viewing trades magnification for comfort.

A telescope forms a magnified real image of a star that you could catch on a screen at the eyepiece.

False — in normal adjustment the final rays leave the eyepiece parallel (image at infinity); a telescope magnifies the angle, not the linear size, so there is no finite screen image to catch.

Increasing the objective focal length increases magnification for both a microscope and a telescope.

False — telescope

M = f_{o} / f_{e}

grows with long

f_{o}

, but microscope objective wants short

f_{o}

since

m_{o} \approx L / f_{o}

grows as

f_{o}

shrinks.

The compound microscope's total magnification is the sum of the two lenses' powers.

False — the eyepiece magnifies the already-magnified intermediate image, so effects multiply:

M = m_{o} m_{e}

, exactly like two photocopiers in series.

A larger telescope objective always gives higher magnification.

False — a wider objective improves light-gathering and resolving power, but magnification depends only on the focal length ratio

f_{o} / f_{e}

, not the aperture diameter.

For a magnifying glass, moving the object closer to the lens always increases magnification without limit.

False — the object must stay inside the focal length;

M_{D} = 1 + D / f

is fixed by

f

and

D

, not by nudging the object, so there is a hard ceiling.

Higher magnification always means you see more of the object at once.

False — magnification and field of view trade off: cranking

M

up narrows the cone of angles the eyepiece accepts, so you see a smaller patch, just enlarged.

Spot the error

"A student computes a telescope's power as

M = D / f_{o} = 25/100

The near-point distance

D

has no business here — the object is at infinity, so the reference angle is the star's real angle

α = h / f_{o}

, giving

M = f_{o} / f_{e}

, not

D / f

"The telescope's

M = f_{o} / f_{e}

is just asserted; there's no reason behind it."

Trace the rays: parallel light at angle

α

focuses to a spot of height

h

in the objective's focal plane, so by small angles (see figure below)

α = h / f_{o}

. That same height sits at the eyepiece focus, leaving as parallel rays at

β = h / f_{e}

. Divide:

M = β / α = (h / f_{e}) / (h / f_{o}) = f_{o} / f_{e}

"To magnify a cell

\times 300

we simply pick one lens with

f

tiny enough that

1 + D / f = 300

A single lens can't reach such power without absurdly short

f

and severe aberrations; that's why the compound microscope splits the job across an objective and eyepiece that multiply.

"The intermediate image in a compound microscope is virtual and erect."

Wrong — the objective forms a real, inverted, enlarged image just inside the eyepiece's focus; the eyepiece then treats that real image as its object.

"Tube length of a compound microscope in normal setting is

f_{o} + f_{e}

That's the telescope formula. For the microscope

L \approx v_{o} + f_{e}

(roughly the separation of the two foci); the object sits just outside

f_{o}

, not at it.

"For a telescope the eyepiece should have a long focal length to see more."

Backwards —

M = f_{o} / f_{e}

increases when

f_{e}

is short; a long eyepiece focal length shrinks the magnification.

"A person's near point is

25 cm

, so every instrument formula uses

D = 25 cm

Only for a normal eye — a hypermetropic eye has a larger near point, changing

D

(see Defects of vision — myopia, hypermetropia); and telescope formulas ignore

D

entirely.

"Since

m_{o} = v_{o} / u_{o}

, making

u_{o}

huge boosts objective magnification."

Large

u_{o}

means the object is far, so the real image barely magnifies; the objective magnifies most when the object sits just outside

f_{o}

(small

∣ u_{o} ∣ \approx f_{o}

), giving

m_{o} \approx L / f_{o}

. Recall

u_{o} < 0

for a real object, so a "huge

u_{o}

" means far away, not close.

"Just make magnification enormous and you'll resolve any tiny detail."

Empty magnification — beyond a limit set by the objective's numerical aperture, you enlarge blur, not detail; for a microscope useful

M

is capped at about

1000 \times

the numerical aperture, past which the diffraction-limited resolution (Resolving power and diffraction limit) refuses to improve.

Why questions

Why does an instrument's magnification use the ratio of angles rather than the ratio of image heights?

Because the size of the picture on your retina is set by the angle a bundle of rays makes at the eye, so two very differently sized objects look "equal" if their visual angles match.

Why is a simple microscope's relaxed-eye power exactly

1

less than its near-point power?

M_{D} - M_{\infty} = (1 + D / f) - (D / f) = 1

; the "+1" is the extra angle you gain by straining to pull the virtual image in from infinity to the near point

D

Why do we want a short focal length for a magnifying lens but a long one for a telescope objective?

Short

f

lets you place a near object very close (huge

D / f

angle boost); a telescope's object is at infinity, and a long

f_{o}

makes its focal-plane image larger so the eyepiece angle

h / f_{e}

is bigger. ::: In short:

D / f

f_{o} / f_{e}

pull the focal lengths in opposite directions.

Why can't you focus an object closer than your near point even though it subtends a bigger angle?

Your eye lens can only shorten its focal length so far (accommodation has a limit); past the near point the image lands behind the retina and blurs, so a bigger angle you can't sharpen is useless.

Why does a compound microscope invert the image while a magnifying glass keeps it erect?

The magnifying glass forms a virtual, erect image (object inside

f

); the compound objective forms a real, inverted image, and the eyepiece re-magnifies without re-inverting, so the net view is inverted.

Why does making the objective aperture larger help even though it doesn't change

M

A wider aperture collects more light and reduces the diffraction limit, so you can see finer, dimmer detail — resolution, not magnification, is what improves.

Why does the size of your eye's pupil cap how bright a telescopic image can be?

The exit beam from the eyepiece must fit through the pupil (

\approx 2

–

7 mm

); if the exit pupil is wider, the overspill is wasted, so a telescope can't make things brighter per unit area than a well-lit naked-eye view — it only spreads light over a bigger angle.

Why does a reflecting telescope avoid one problem that plagues a long refractor?

A curved mirror reflects all colours to the same focus (no chromatic aberration) and needs no huge, sagging glass lens, so large-aperture telescopes are almost always reflectors.

Edge cases

What is a simple microscope's magnification when the object is placed exactly at the focus?

The virtual image goes to infinity (relaxed eye), giving

M = D / f

— the minimum usable magnification, not the maximum.

What happens to the magnifying power as

f \to 0

for a fixed

D

M = 1 + D / f \to \infty

mathematically, but in practice tiny

f

means severe aberrations and an impossibly close object, so the ideal formula breaks down long before infinity.

What magnification does an instrument give for an object placed exactly at the near point with no lens?

M = 1

by definition — that's the reference case

θ_{0} = h / D

every magnifying power is divided by, so "no instrument" gives unity.

For a telescope viewing a nearby object (not at infinity), does

M = f_{o} / f_{e}

still hold exactly?

No — the derivation assumes parallel incoming rays (

α = h / f_{o}

); for a finite-distance object the objective image no longer sits exactly in its focal plane, so the simple ratio is only approximate.

What is the "angular size" of a star through a real telescope, and why isn't it exactly zero?

A star is a point, but diffraction spreads it into an Airy disk of finite angular radius

\approx 1.22 λ / D_{aperture}

(Resolving power and diffraction limit); magnifying just enlarges this blur disk, which is why raw magnification never sharpens a star.

Do the neat formulas hold across the whole field of view, or only near the axis?

Only near the axis (paraxial) — off-axis you meet field curvature (the sharp image lies on a curved surface, so edges blur when the centre is focused) and aberrations (coma, astigmatism) that no single ideal number captures.

When does chasing more magnification make the image worse rather than better?

Once you pass the diffraction/aberration limit: extra

M

magnifies the blur and dims the field (light spread over more area), so there is an optimum

M

beyond which detail and brightness both degrade.

Recall One-line self-test before you leave

Cover each and answer with a reason: Microscope wants short $f$ , telescope wants long $f_{o}$ — why the split? ::: Near object needs a close placement ( $D / f$ ); infinite object needs a big focal-plane image ( $f_{o} / f_{e}$ ). Why multiply, not add, for the compound microscope? ::: The eyepiece magnifies the objective's output, so gains compound. Why does a telescope formula never contain $D$ ? ::: Its object is at infinity, so the reference angle is the real angle $α = h / f_{o}$ , not $h / D$ . Where does the "+1" in $M_{D} = 1 + D / f$ come from? ::: Pulling the virtual image from infinity to the near point $D$ lets the object sit slightly closer, worth one extra unit of $D / u$ .