Preface: if there is a better-suited forum for this question, please direct me. I figured it'd be better to post into Photography and not Physics as some of you might have already examined this particular issue. Thank you.


I am wondering about the physics side of things. Let's say I position a camera 25cm away from an ideal 4W light bulb (all 4 Watts go into electromagnetic radiation, and we won't differentiate between visible and IR/UV spectrums). Let's assume that the distance is measured from the bulb's filament to the camera's sensor surface. To have somesthing specific, let's say it's Canon 5D Mk4, with a Zeiss Milvus 2/35 lens attached. The flange focal distance is 4.4cm, and the lens' length until the front glass element is about 8.5cm.

My question is: as the light travels from the bulb to the glass, it loses denstiry (pardon my lack of proper terminology). Once it enters the lens system, however, does the process continue in some form? Or is all the light transferred to the imaging circle? I do not consider factors such as vignetting and glass transmission in this case. In the end, I would like to be able to estimate the amount of energy that reaches my sensor by knowing the distance to the light source and its initial brightness, etc.

Below is my initial understanding of the issue; however, it might be 100% wrong in every aspect. You can safely ignore it.

If I wanted to calculate the amount of light that hits the sensor, consulting the inverse square law to a reasonable precision, would I use the distance to the lens' surface, or to the sensor's surface? Or something else, like the focal center of the lens? The difference of ~12cm at this distance is quite significant. I assume that the relevant measure is the lens' entrance pupil. Perhaps it should be the whole frontal glass element?

I cannot measure the entrance pupil, but let us assume its diameter is 1cm. The area would be ~3.14cm^2 (or 0.000314m^2), that should be close enough. The end of the lens is approximately 13cm (0.13m) away from the light source. From that it is easy to calculate that the lens' entrance pupil receives about 0.15% of the total light emitted by the bulb, or 0.006 Watts. Meanwhile, if the lens' tip was positioned full 25cm away from the bulb, it would receive only 0.04%, or 0.0016 Watts. Nearly 4 times less.

But, the tip is positioned 13cm away from the light. My question is, does the light continue to dissipate within the lens? Is the "distance" within the lens the same as its physical length, or is it expanded or compressed somehow? Or is all the energy that enters the Entrance Pupil transferred to the imaging circle? Or, maybe I am wrong about everything from the get-go?

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    \$\begingroup\$ What photographic problem are you trying to solve? Modern cameras render such questions moot, since metering is done through the lens at pretty much the same optical distance from the back of the lens as the sensor is, if it is not done directly using the sensor. Having said that, the answer to most of your sub-questions is, "It depends on the specific lens in question." Not all lenses have equal transmission of the light that falls upon their objective, nor is the entrance pupil located the same distance behind the front element for all lenses. \$\endgroup\$
    – Michael C
    Feb 6, 2019 at 5:02
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    \$\begingroup\$ Refractive surfaces, such as lens elements, can spread light (reducing the field density) or concentrate light (increasing the field density). Most lenses have elements that do both at some point in their optical formula. But in the end you do not need to estimate the amount of light energy reaching your sensor. The sensor's sole purpose is to measure intensities of light. That's it. All you need to do to know how much light energy fell on your sensor is to use the raw data from the sensor combined with the sensor's response characteristics. \$\endgroup\$
    – Michael C
    Feb 6, 2019 at 5:06
  • \$\begingroup\$ vtc b/c 1. Looks like physics homework. 2. You say you want to estimate light energy hitting the sensor, but you also want to ignore nearly every factor that would reduce that energy. That makes your "estimate" pretty much useless. 3. Look up T-stops. \$\endgroup\$
    – xiota
    Feb 6, 2019 at 8:43
  • \$\begingroup\$ Thank you for your comments. I am not trying to solve a "photographic" problem, but I am trying to understand the principles behind. I understand your concern that someone would come here for homework solutions, xiota, but I would be surprised if any school actually gave this as homework. \$\endgroup\$ Feb 6, 2019 at 11:36
  • \$\begingroup\$ In any case, thank you for suggesting T-stops and the workings of the lens. I have mentioned in my post, that for these purposes I would like to ignore the Transmission properties of the lens. I am interested in how light travels within the lens system, whether it acts more akin to optical fibre or to compressed space where non-parralel light still loses intensity over distance \$\endgroup\$ Feb 6, 2019 at 11:38

3 Answers 3


Typically, in radiometry you try to reduce an object to a point source, or a uniform emitter of some primitive geometry (disk, square, etc). We'll say you're far enough away that the light bulb behaves like a point source.

From the inverse square law, you find that there is infinite power when in contact with the bulb and infinitesimal power when extremely far from the bulb. This is not a loss of energy, just the energy spreading out. When you collect some of the power with a lens, there are two cases -- one is to "collimate" it (make it travel in straight lines) in which case the energy density is now constant as a function of distance traveled, or project it to a focus. Your camera lens does the latter. Since you want to ignore secondary effects like transmission, the lens is an ideal collector and projects all of the power to an infinitesimal spot on the sensor with infinite energy density (i.e., W/cm^2 = ∞).

The meat of your question is where is the lens collecting energy exactly, and how big is that thing.

Starting from the top, the front element of your lens is used by the entire field of view. Especially in a wide angle lens, this is not all used by a point source. Consider a fisheye lens, where the front element is huge but if you look through it the bright circle you see is quite small. That bright circle you see is the aperture collecting light from an infinite number of point sources like your light bulb.

That aperture is probably virtual (meaning you can't touch it) and is called the Entrance Pupil (EP). It's the image of the aperture stop seen through all of the lenses in front of it. The EP collects light, so to figure out how much your lens collects you need to know its size and location.

If your object is very far away, the difference between the location of the EP and the location of the image plane is probably a small "correction." Its size can also be calculated as simply the ratio of the focal length and F/#. So for your 35/2 example, it is 17.5 mm in diameter. This is inexact because the F/# on the barrel is a marketing figure and it could really be anywhere from F/1.9 to F/2.3 or so without the vendor violating the CIPA requirements, which conform to ISO 517. The F/# is also only for an object at infinity, and the working F/# is larger. This difference is also known in the photography community as the bellows effect.

You can also measure the EP size and location pretty easily. Put the lens in front of a piece of paper and using a macro lens, take a picture of its aperture. Without adjusting focus, then take a picture of a ruler. Figure out the plate scale (mm/px) and then take the diameter of the aperture in px and calculate its diameter in mm. If the magnification is known (and it is known to very high precision -- the ruler and pixels size in microns let you calculate this) you can work out the total track length of the imaging system. That lets you calculate distances from the sensor of the camera, which is a reasonably well located datum.

  • \$\begingroup\$ Thank you very much. This is a fantastic answer and tells me what I needed to know. I would like to clarify your answer so I don't have any doubts: 1. So is it the front element's area, or the Entrance Pupil's area that determines the amount of energy collected? 2. While usually the difference is too small to matter, to be pedantic, which is the distance at which the energy stops being spread and stars being focused? Is it effectively the front glass element, or the entrance pupil's virtual location? \$\endgroup\$ Feb 6, 2019 at 15:58
  • \$\begingroup\$ 1. EP. 2. The location ("distance") is at the EP. The purpose of the pupils of an optical system is precisely to make these first-order calculations simple. If you want to get into the weeds there is no single place -- each element will contribute to the bending of rays towards the focus. The front element has no special significance. \$\endgroup\$ Feb 6, 2019 at 18:09

Light from a point source obeys the law of the inverse square. Applying this principle, you can calculate the radiant energy that plays on the surface of the lens. Now the camera lens acts like a funnel in that it gathers light. The amount of light energy that traverses the lens is governed by several additional factors. The working surface area (aperture) of the lens is foremost. We are talking about the surface area (capture area) available. The larger the surface area of the aperture, the more light gathered.

Now the job of the lens is to project an image of the outside world on the flat surface of film or digital image sensor. As this light traverses the lens, light energy is lost due to the fact that the various elements of the lens are not flawlessly transparent. Additionally, a camera lens is an array consisting of multiple lens elements. Such a scheme is necessary to mitigate the seven optical aberrations that plague. Each element has two polished surfaces that reflect light. Lens coating mitigates the light loss; nevertheless, 2% or more will be lost per element.

The brightness of the projected image at the focal plane is massively affected by the distance, lens-to-image (focal length). Each doubling of the focal length results in a 2X change in size of the projected image. Each 2X change magnification results in a 4X change in the surface area of the image. The delta (change) has a magnitude of 4X. In other worlds doubling the focal length induces a 4X change in image brightness.

The image is brightest at the center. As you inspect image brightness off axis, you will discover a falloff called a vignette. The vignette has two major contributors. At the center of the image, the lens aperture is a circular opening. An off axis view reveals that the aperture appears elliptical and thus has less surface area -- thus it passes less light.

The lens yields an image that consists of a myriad of tiny points of light. At the center of this image, these points are circular. At the boundaries, these points are ellipses. The light energy conveyed from these ellipses is abridged due to the fact they arrive at an oblique angle.

There are more attenuating factors than mentioned. All this makes your method to account for the energy received at the image plane more daunting.

  • \$\begingroup\$ Thank you Alan. You've provided a detailed explanation, I have learned new nuance, and I hope other readers might, too. However, as I mentioned in my original post, I am more or less away of the vignetting and transmission properties of the lenses. These factors are tested by either the manufacturers, or companies like DxOMark, and can be accounted for. What I am interested, is what distance I should consider when estimating the light falling onto the lens, and what surface. Does the inverse square law still apply within the lens? You mention the pupil's position, is that relevant too? \$\endgroup\$ Feb 6, 2019 at 11:42

I think the distance from a light source to front lens element will accurately follow the law of the inverse square provided the light emanates from a point. If the light rays originate from a broad source or if the incoming light rays are not omnidirectional, the accuracy of the calculation diminishes.

As the light transverse the front lens element group, refraction power negates the law of the inverse square. This is because the actual diameter vs. the apparent diameter, of the aperture stop, differs.

This can be quite useful in the case of a zoom lens. The zoom lens undergoes a gain or loss as to image brightness, with the focal length change. To mitigate, the front lens group changes its spacing to the aperture stop in ratio with the zoom. The result is; the apparent diameter of the stop changes in accordance to the zoom. This action preserves image brightness through most if not all of the zoom (constant aperture).

The fact is, for all lenses, the apparent diameter of the aperture stop differs from actual based on the power of group forward of the iris diaphragm. The bottom line, one rays transverse the front element, the inverse square is negated.


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