why the front element diameter of a lens must be l...

Because the f-number is set by the entrance pupil, and the front of the lens must be able to “see” that pupil without clipping it.

1) The key definition

The f-number $N$ is defined as

$$N=\frac{f}{D_\text{EP}}$$

where:

• $f$ = (effective) focal length
• $D_\text{EP}$ = entrance pupil diameter (the apparent diameter of the aperture stop as seen from the front of the lens)

So

$$D_\text{EP}=\frac{f}{N}$$

That $\frac{f}{N}$ number is not “front element diameter”—it’s the diameter of the entrance pupil.

2) Why the front must be at least that big (or you lose the f-number)

For an on-axis object at infinity, the lens must pass a bundle of rays whose limiting opening, as seen from the front, is the entrance pupil. If the front clear opening (front element + barrel) is smaller than $D_\text{EP}$, then the front becomes the limiting aperture instead. That reduces the entrance pupil to the front opening:

• effective $D_\text{EP}$ becomes smaller
• therefore $N = f/D_\text{EP}$ becomes larger (the lens becomes “slower”)

So: to actually achieve an f-number $N$, the lens needs a front clear aperture $\ge D_\text{EP} = f/N$.

3) Why it’s usually larger than $f/N$ in real lenses

Even if the on-axis rays fit, real lenses also need to avoid vignetting (clipping) for off-axis field points.

For a field angle $\alpha$, the ray bundle is tilted and shifted at the front. A simple way to see the sizing is:

$$D_\text{front} \;\gtrsim\; D_\text{EP} \;+\; 2\,L\,\tan(\alpha)$$

where $L$ is the distance from the entrance pupil location to the front limiting surface.
So wider field ($\alpha$ bigger) and/or pupil deeper inside the lens ($L$ bigger) forces a larger front diameter.

On top of that, designers also add margin for:

• mechanical mounts and filter threads
• thick elements (you need clear aperture, not just glass diameter)
• pupil magnification effects (the stop’s apparent size can be larger than its physical size because the front group images it)

4) Is it strictly “must be larger”?

Strictly speaking:

• The front clear aperture must be ≥ $f/N$ to achieve that f-number on-axis.
• It is often larger than $f/N$ to cover the field without vignetting and to allow mechanical structure.

A contrived lens with the aperture stop right at the front and a narrow field could have a front clear aperture very close to $f/N$. But most practical camera lenses need more.

Example

A 50 mm $f/1.4$:

$$D_\text{EP} = \frac{50}{1.4} \approx 35.7\text{ mm}$$

So the lens must present ~36 mm entrance pupil from the front. The front element will typically be noticeably bigger (often 45–55 mm OD) to avoid vignetting and fit mechanics.

If you tell me what kind of lens you’re thinking about (phone, DSLR prime, wide-angle, telephoto, zoom), I can relate the “how much bigger than $f/N$” to its field of view and pupil position.

Because “same f-number” does not mean “same entrance-pupil diameter.” It means the ratio of focal length to entrance pupil is the same:

$$N=\frac{f}{D_{\text{EP}}}\quad\Rightarrow\quad D_{\text{EP}}=\frac{f}{N}$$

So if two lenses are both f/2.8:

• 24 mm f/2.8: $D_{\text{EP}} \approx 24/2.8 \approx 8.6$ mm
• 200 mm f/2.8: $D_{\text{EP}} \approx 200/2.8 \approx 71$ mm

The telephoto must have a much larger entrance pupil to be f/2.8, so it will naturally “look” bigger.

Why the wide-angle can still be “fast” with a small-looking pupil

A wide-angle projects a smaller image scale (less magnification) onto the sensor. Even with a smaller entrance pupil, the cone angle at the image plane corresponds to the same f-number, so the sensor irradiance for an extended scene stays tied to $N$, not to the absolute hole size.

Extra wrinkle: wide angles often demagnify the stop (retrofocus designs)

The “hole” you see from the front is the entrance pupil: the aperture stop as imaged by the front lens group. In many wide-angle (retrofocus) lenses, the front group makes the stop appear smaller and deeper inside (pupil magnification < 1), so it can look even smaller than you might expect. Telephotos often do the opposite (pupil magnification ≥ 1), making the pupil look larger.

So the short answer: same f-number → same ratio, not same diameter; telephotos have longer $f$, so their entrance pupils are larger, and lens design can amplify the difference in “apparent” pupil size.

The exit pupil is the image of the aperture stop as seen from the image side (through the rear elements). It’s the “hole” the sensor thinks it’s looking at.

Entrance vs exit pupil (why we care)

• Entrance pupil → sets the object-space cone (what the scene “sees” looking into the lens). This is what the usual f-number definition uses: $$N=\frac{f}{D_\text{entrance}}$$
• Exit pupil → sets the image-space cone (what angles the rays hit the sensor at) and interacts with vignetting, telecentricity, microlenses, etc.

The two pupil diameters are related by pupil magnification:

$$P = \frac{D_\text{exit}}{D_\text{entrance}}$$

Wide-angle retrofocus designs often have $P<1$ (exit pupil smaller than entrance), and many telephotos have $P\ge 1$ (exit pupil comparable or larger).

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How the exit pupil connects to “how steep the rays are”

What the sensor really “feels” is not just pupil size, but exit pupil size AND where it sits.

If the exit pupil is at a distance $L$ in front of the image plane, then (paraxially) the marginal ray half-angle $u'$ at the image plane is roughly:

$$\tan u' \approx \frac{D_\text{exit}/2}{L}$$

and the image-space f/# is approximately:

$$N_\text{image} \approx \frac{1}{2\tan u'} \approx \frac{L}{D_\text{exit}}$$

So:

• Bigger $D_\text{exit}$ → wider cone (faster)
• Closer exit pupil (smaller $L$) → wider cone (also “faster” at the sensor, and steeper ray angles)

Why wide angles can be “problematic” on digital

Wide angles tend to have:

• large field angles (chief rays already want to come in oblique)
• exit pupil sometimes not very far from the sensor

That makes ray incidence angles steep, which can cause:

• edge shading / “color shift” on some sensors
• microlens inefficiency at edges
• mechanical vignetting sensitivity

This is why some digital-oriented wide angles aim for image-side telecentricity (pushing the exit pupil far away so chief rays are more perpendicular).

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Does exit pupil change the f-number on the barrel?

For “normal” photography focused near infinity, the marked f-number is tied to the entrance pupil and exposure behaves as expected.

But for finite conjugates / macro, the effective f-number at the sensor depends on pupil magnification too. A common general form is:

$$N_\text{eff} \approx N \,\frac{(1+M)}{P}$$

where:

• $M$ = image magnification at the sensor (1.0 at 1:1 macro)
• $P = D_\text{exit}/D_\text{entrance}$

So retrofocus wide angles ($P<1$) can “lose light” faster in close-up than a lens with $P\approx 1$, all else equal.

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Quick intuition

• Entrance pupil answers: “How big is the ‘window’ from the scene’s point of view?”
• Exit pupil answers: “How wide/steep is the cone hitting the sensor, and where is it coming from?”

If you want, tell me two example lenses (like 24mm f/2.8 and 200mm f/2.8) and whether you mean “looks smaller” when viewing from the front, rear, or through the viewfinder—and I’ll map exactly which pupil you’re seeing and why.

If you take two lenses with the same f-number (say 24 mm f/2.8 and 200 mm f/2.8) and look in through the rear, you’re looking at the exit pupil: the aperture stop as imaged by the rear lens groups.

What you’ll see (typical)

24 mm f/2.8 (wide-angle, usually retrofocus)

• The exit pupil usually looks like a small bright disk (the “hole”) floating deeper inside the lens.
• Relative to the mount opening, it often looks quite small.
• It may also appear to “move” (parallax) more as you shift your eye, because it’s deeper.

200 mm f/2.8 (telephoto)

• The exit pupil usually looks like a much larger disk, often closer to the rear of the lens.
• It can look like it fills a big fraction of the mount.

Why they look so different

1. Same f-number ≠ same pupil diameter
   The entrance pupil scales with focal length:

$$D_\text{entrance}=\frac{f}{N}$$

So at f/2.8:

• 24 mm ⇒ $D \approx 8.6$ mm
• 200 mm ⇒ $D \approx 71$ mm

Telephotos must have a much bigger pupil to be the same f-number, so their exit pupil tends to be big too.

2. Exit pupil size depends on “pupil magnification” (design-dependent)
   The exit pupil is the stop magnified or demagnified by the rear optics:

$$P=\frac{D_\text{exit}}{D_\text{entrance}}$$

Wide retrofocus designs often end up with smaller exit pupils (often $P<1$), while many telephotos have $P\approx1$ or $>1$. So the wide can look even smaller from the rear than you’d “naively” expect.

3. Exit pupil position matters for the “look” Even if two exit pupils had similar diameter, the one that’s deeper inside the lens can look smaller because you’re viewing a virtual image at a different apparent distance.

Easy way to observe it correctly

1. Set the lens to wide open (smallest f-number).
2. Remove the lens, point it toward a bright wall/sky.
3. Hold it ~20–40 cm from your eye and look into the rear.
4. The bright circular “window” you see is the exit pupil.
5. Stop down the aperture: that circle shrinks—confirming you’re really seeing the stop image.

If you tell me the exact two lenses (or at least “DSLR retrofocus wide vs rangefinder wide”), I can predict the usual exit-pupil position too (close to sensor vs pushed forward), which is the big reason some wides cause steep ray angles and edge color shifts on certain sensors.

Yes—within one lens, the entrance pupil and exit pupil can be different sizes. The difference is set by pupil magnification (a design property).

1) The basic relationship

• Entrance pupil: image of the aperture stop seen from the front
• Exit pupil: image of the aperture stop seen from the rear

Their size ratio is the pupil magnification $P$:

$$P \;=\; \frac{D_\text{exit}}{D_\text{entrance}} \qquad\Rightarrow\qquad D_\text{exit} \;=\; P\,D_\text{entrance}$$

If $P=1$: entrance and exit pupils are the same size.
If $P<1$: exit pupil looks smaller than entrance.
If $P>1$: exit pupil looks larger than entrance.

This happens because the front group and rear group “magnify” the stop differently when you view it from each side.

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2) Is this different for wide vs telephoto?

Often, yes (but not always). Typical tendencies:

Wide-angle SLR lenses (retrofocus / inverted telephoto)

• Very commonly have $P<1$
• So the exit pupil tends to be smaller than the entrance pupil, and it may appear deeper inside the lens.

That’s one reason wide angles often have more sensitivity to sensor microlenses/edge shading: the image-side ray geometry is strongly shaped by a small/near exit pupil.

Telephoto designs

• Often have $P \approx 1$ or $P>1$
• So the exit pupil is similar or larger than the entrance pupil.

(But lens families vary: some telephotos are designed more telecentric on the image side, which can shift the exit pupil position a lot.)

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3) What you’ll see if you compare one lens front vs rear

Set the lens wide open (smallest f-number), point it at a bright surface, and look:

• From the front: you see the entrance pupil (apparent aperture).
• From the rear: you see the exit pupil.

If the rear “hole” looks noticeably smaller than the front “hole,” that lens has $P<1$. If it looks bigger, $P>1$.

If you tell me what mounts/lenses you mean (e.g., “24mm f/2.8 SLR” and “200mm f/2.8”), I can describe the expected direction (exit smaller/larger) more concretely for those designs.

Because the front element is sized mostly by the entrance pupil, while the rear element is sized mostly by the field coverage + the steepness of the converging cone near the sensor. In compact wide-angles those two constraints can be very different, so you often end up with small front / big rear.

1) Wide angle ⇒ entrance pupil can be small

For a given f-number $N$,

$$D_{\text{entrance}}=\frac{f}{N}$$

A wide angle has a small focal length $f$, so even at the same f/# its entrance pupil is small. Example at f/2.8:

• 24 mm: $D_{\text{entrance}}\approx 8.6$ mm
• 200 mm: $D_{\text{entrance}}\approx 71$ mm

So the front doesn’t need to be huge just to meet the f/#.

2) But wide angle ⇒ rays are very oblique at the image side

Wide angle also means large field angle. For off-axis image points, the whole cone of rays is tilted and laterally shifted as it approaches the sensor.

A simple “clear aperture” sizing idea for some surface a distance $z$ in front of the image plane is:

• shift term (chief ray): roughly grows with field angle
• cone radius term (marginal ray): roughly $\approx \dfrac{z}{2N_{\text{image}}}$

So the needed diameter is roughly:

$$D_{\text{needed}} \;\gtrsim\; 2(\text{chief-ray height at that surface} \;+\; \text{cone radius})$$

Near the rear of a wide-angle, the chief ray can be far off-axis (big shift) and the cone is still fairly fat—so rear elements must be large to avoid mechanical vignetting.

Telephotos have much smaller field angles, so those off-axis shifts are modest; the rear doesn’t need to balloon the same way.

3) “Thickness limited” makes the rear constraint worse

If overall length is limited, designers often can’t “spread out” the geometry. That tends to push the exit pupil closer to the sensor (or forces strong power near the rear). When the exit pupil is close, the cone has to open quickly:

Paraxially,

$$N_{\text{image}} \approx \frac{L}{D_{\text{exit}}}$$

where $L$ is distance from exit pupil to image plane.

Smaller $L$ (compact lens) ⇒ to keep the same f/# you need a larger $D_{\text{exit}}$ (or you accept more vignetting / worse ray angles). A larger exit pupil typically means larger rear optics.

4) Wide-angle families that show “small front / big rear”

This look is especially common in compact wide-angle designs (e.g., more symmetric/rangefinder-style or some mirrorless wides) where the rear group sits close to the sensor and must pass very oblique bundles. You’ll often see a big rear element (sometimes even protruding) because that’s the only place you can buy the needed clear aperture without making the lens much longer.

5) Extra modern reason: digital sensors

Many sensors dislike extremely oblique rays (microlenses, color shifts, corner shading). Some wide-angle designs respond by manipulating the exit pupil / ray angles, which often increases the rear group size or complexity.

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If you tell me the lens type (phone module, mirrorless wide prime, SLR retrofocus wide, etc.) I can map which of these is dominating—because an SLR retrofocus wide often ends up with a large front too, while compact mirrorless/rangefinder-style wides are the classic “small front / big rear” look.

Not quite — in that approximation it’s the opposite.

If you define the image-space f/# from the cone angle hitting the sensor,

$$N_{\text{image}} \approx \frac{L}{D_{\text{exit}}}$$

where

• $D_{\text{exit}}$ = exit pupil diameter (aperture stop as imaged from the rear)
• $L$ = distance from the exit pupil plane to the image plane (can be a virtual location)

What happens if $L$ gets shorter?

Holding $D_{\text{exit}}$ fixed:

• smaller $L$ → smaller $N_{\text{image}}$ (a smaller f-number, i.e., a “faster” cone)
• larger $L$ → larger $N_{\text{image}}$ (a larger f-number, i.e., “slower”)

Example: if $D_{\text{exit}}=20\,\text{mm}$

• $L=40\,\text{mm}\Rightarrow N_{\text{image}}\approx 40/20=2$
• $L=20\,\text{mm}\Rightarrow N_{\text{image}}\approx 20/20=1$

So shortening $L$ makes the cone steeper, not slower.

Where your intuition does become true

If by “rear diameter” you mean the physical rear element / mount opening, that’s not necessarily $D_{\text{exit}}$.

• If the rear element is too small, it clips the ray bundle for some field angles.
• That clipping effectively makes the usable exit pupil smaller than the designed $D_{\text{exit}}$.
• Smaller effective $D_{\text{exit}}$ → larger effective f-number (slower) and/or vignetting.

So:

• Shorter $L$ alone (with the same true exit pupil) → smaller $N_{\text{image}}$.
• Shorter $L$ with an undersized rear opening (clipping) → effective $D_{\text{exit}}$ shrinks → effective f-number can increase.

If you want, tell me whether you’re thinking on-axis only or edge-of-frame behavior — the “rear is limiting, so it gets slower” effect is mostly an off-axis/vignetting story.

You’re right to challenge that — with

$$N_{\text{image}} \approx \frac{L}{D_{\text{exit}}},$$

if you hold $N_{\text{image}}$ fixed, then

$$D_{\text{exit}} \approx \frac{L}{N_{\text{image}}}.$$

So smaller $L$ would actually require a smaller $D_{\text{exit}}$ to keep the same image-space f/#. I misspoke earlier if it sounded like “smaller $L$ ⇒ larger $D_{\text{exit}}$” for keeping the same f/# — it’s the opposite.

What is true (and what I meant)

A compact wide lens having a small $L$ means the sensor is “close” (in an optical sense) to the exit pupil. For a given $D_{\text{exit}}$:

• smaller $L$ ⇒ larger cone angle at the sensor ⇒ smaller $N_{\text{image}}$ (faster cone) ⇒ steeper ray angles (often “worse ray angles” for sensors).

To reduce those steep angles (i.e., make $N_{\text{image}}$ larger / more telecentric), you’d want:

• larger $L$ (push exit pupil farther away), or
• smaller effective $D_{\text{exit}}$ (stop down / accept clipping), which makes it slower.

So the “trade” is: compactness pushes you toward steeper image-side angles unless you accept losing light (higher effective f/#, often via vignetting/stop-down).

Why wide compact lenses still end up with a big rear element

Even if the pupil could be small, the rear glass diameter is often set by field coverage, not by $D_{\text{exit}}$.

For off-axis image points, the whole cone is shifted sideways near the sensor. The rear element has to clear:

• the chief-ray height (roughly on the order of the sensor half-diagonal), plus
• the cone radius.

That can demand a large rear clear aperture even if the entrance pupil (and even the exit pupil) isn’t huge.

Bottom line:

• Keeping the same image-space f/# with smaller $L$ would call for smaller $D_{\text{exit}}$.
• But avoiding vignetting across a wide field often forces a large rear element anyway.
• And compactness tends to make ray angles steeper unless you accept a slower effective f/# (often at the edges).

If you tell me the format (full-frame, APS-C, phone sensor) and the approximate focal length/f-number, I can plug in rough geometry to show why the rear element has to be as large as it is.