# Tag: Lens

## nikon – Bought “Mint” lens from Japan, but then I see this

I just received a supposedly “mint” Nikkor 28mm f2.8 AIS lens from Japan. It sure does look unused, but when inspecting the rear element I found this stuff:

The smaller spot near the top looks like it’s just beneath the outer glass, while the bigger “spot” (it’s actually two spots) looks like it’s deeper in. And yes, I did try cleaning it, but to no avail. It’s clearly not just smudges on the surface.

It’s not really visible while doing the LED-test, but outdoors lighting makes it very visible from the right angle as seen on the photo.

What do you think it is?

Cheers,

Vincent

## What size lens will give a 2ft by 2 ft view at a range of 2 ft?

The easiest way is put a zoom lens on your camera and figure out what focal length you need empirically.

Since you don’t say what your sensor size or camera model is, you’ll need to do the maths yourself.

Angle of view for your situation is about 2 * arctan(1/2) = 53°.

From a Wikipedia article: angle of view = 2 * arctan(d/(2*f)

Where: d = distance; f = focal length

d, in your case, is 610mm = 2 feet.

Solving for f: focal length = d/(2 * tan({angle of view}/2))

For a full frame sensor and using the 24mm dimension, you’ll need a focal length of 24mm.

For an APS-C sensor, 16mm.

Some caveats:

- distance is measured from the film (sensor) plane to object. Decent cameras have a marking on the body where the film plane is located.
- The equation for finding focal length is for a simple lens. Camera lenses aren’t simple, especially when shooting close up.
- When focusing at closer ranges, the magnification of your lens will change. Unless you calibrate your lens, you’ll be slightly off from the equation.

## Best budget wide-angle FX lens for Nikon F-mount

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What is, in your opinion, the best wide-angle (not fisheye) FX lens? I’d prefer prime one, but it’s not hard criterium for me. I want best possible image quality and price below $500 used. I like sharp and crisp image. Thank you for your advices.

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## optics – How do I calculate f/stop change for teleconverter that increases lens magnification?

The job of the camera lens is to project an image of the outside world onto the surface of film or digital sensor. The image size of object (magnification) is determined by the actual size of the object intertwined with distance from the camera and the focal length of the camera lens used. If you increase the focal length of the camera lens, the projection distance is also increased. This results and image that displays greater magnification. As an example, if you increase the distance screen to projector of a slide or movie projector and re-focus, the image projected on the screen is enlarged.

Peter Barlow, English Mathematician / Optician, invented an achromatic (without color error) supplemental lens that increased the magnification of telescopes in 1833. The Barlow lens design is the one used in modern teleconverters.

Such supplemental lenses increase the versatility of our camera lens. Commonly they double or nearly double the focal length. A 2X teleconverter doubles the focal length granting a 2X focal length increase which results in 2X grater magnification.

This increased magnification comes with a price. Along with the increased image size comes a reduction in the intensity of projected image. To calculate the impact of this magnification gain on image brightness, we square the magnification gain. Thus for a 2X teleconverter the math is 2 X 2 = 4. We find the reciprocal of this reduction factor by annexing 1/ before the number. Thus, a reduction factor of 4 tells us that the amount of light reaching film or image chip is ¼ or 25% of the former.

Now the f-number system we use is based on an incremental change of 2. In other words, each f-number change doubles or halves the exposing energy. Thus, we divide the magnification increase granted by the teleconverter by 2 to find out how many f-stops reduction results. In this case, a 2X doubling of the magnification results in a reduction factor of 2 X 2 = 4. This value, divided by 2 = 2. This tells us that the functioning f-number is 2 f-stops so we open up 2 f-stops. Go left on the below f-number set.

The f-number set:

1 – 1.4 – 2 – 2.8 – 4 – 5.6 – 8 – 11 – 16 – 22

Thus if the f-number is f/8 and we add a 2X teleconverter, the working f-number changes two f-stope to f/4. Also note – the f-number set is its neighbor multiplied going right by the square root of 2 = 1.4.

Let me that that understanding the resulting reduction factor holds for figuring out exposure when adding filters (filter factor). This value is a multiplier used to manipulate exposure time. Thus if the factor is 4, we can multiply the exposure time by this factor to calculate a compensating exposure time.

Suppose the exposure without filter or teleconverter is 1/400 of a second at f/8. We mount a filter of teleconverter with a factor of 4.

The revised exposure time is 4/1 X 1/400 = 4/400 = 1/100 the revised shutter time @ f/8

Or 1/400 second @ f/4

## optics – Calculating f/stop chance for teleconverter that increases lens size

With a rear-mount converter, it’s always the factor of 2 (unless there’s a gross mismatch between lens and converter), but with your front-mount one, it’s not that simple.

The relevant question is where the light path is effectively limited.

Or, to put it another way: Does all the light collected in your converter’s front lens reach the sensor, or is some part of it blocked by the front opening of your base lens? If part is blocked, then the large converter front lens doesn’t help and is just a waste of material.

As a quick check, you can detach the combo from the camera body and look into it from the rear side, a few centimeters behind the lens, roughly where you’d expect the sensor. Do that with aperture full open. If you can fully see the circular edge of the converter’s front lens, then the 40mm-based calculation is indeed valid (all the light collected on that 40mm circle reaches the sensor). If you don’t, that means that the 30mm opening of the base lens still is the limiting factor, and the 30mm calculation will probably give better results.

And, if you want to do some experiments, compare the exposure times for full-open shots with and without the converter (of course, in a constant-lighting situation). If there’s a factor of 4 between the two times, the classical calculation applies, if it’s a factor of about 2, your 40mm-based calculation is correct.

Having said all that, I bet that you won’t be able to see the converter front lens edge, and that the exposure-time experiment will result in a factor of 4.

## neutral density – Is there an optimal order when using both polarized and ND filters on a lens?

Try it both ways and see what happens. Maybe it is different one way from the other. Now you have two possibilities to create a picture you want. Or maybe it doesn’t make a difference in a way that matters to you.

The most likely way it will make a difference is mechanically.

One way is probably a little easier in terms of making photographs than the other. Maybe it is easier to turn the polarizer if it is first. Maybe it is easier to turn if it is second.

Maybe it is easier to compose and set exposure with one filter on. (or maybe not).

The right way is the way that works for you. Not what someone on the internet says…well except for this of course.

## optics – In what way does the lens mount limit the maximum possible aperture of a lens?

There are two hard limits on how fast a lens can be:

**The first is a thermodynamic limit.** If you could make a lens arbitrarily fast, then you could point it to the sun and use it to heat your sensor (not a good idea). If you then get your sensor *hotter* than the surface of the Sun, you are violating the second law of thermodynamics.

This sets a hard limit at f/0.5, which can be derived from the conservation of etendue. Well, technically it’s more like T/0.5. You *can* make lenses with f-numbers smaller than 0.5, but they will not be as *fast* as their f-numbers suggest: either they will work only at macro distances (with “effective” f-numbers larger than 0.5), or they will be so aberrated as to be useless for photography (like some lenses used to focus laser beams, which can only reliably focus a point at infinity on axis).

**The second limit is the mount.** This limits the angle of the light cone hitting the sensor. Your trick of using a diverging element dos *not* work. You certainly get a wider entrance pupil, but then you have a lens combination which has a *longer* focal length than the initial lens. Actually, your trick is very popular: it’s called a “telephoto” design. Bigger lens, same f-number.

If the lens mount allows for a maximum angle α for the light cone, then the fastest lens you can get will have an f-number equal to

N= 1/(2×sin(α/2))

or, equivalently, *N* = 1/(2×NA), where NA is the numerical aperture. This formula also shows the hard limit at 0.5: sin(α/2) cannot be larger than 1. Oh, BTW, if you try to derive this formula using small-angle approximations, you will get a tangent instead of a sine. Small-angle approximations are not good for very fast lenses: you should use the Abbe sine condition instead.

The same caveat about f-numbers v.s. T-numbers applies to this second limit. You can get a lens with an f-number smaller than 1/(2×sin(α/2)), but it will work as macro-only, and the bellows-corrected f-number will still be larger than the limit.

## Derivation

This section, added on Nov. 26, is intended for the mathematically inclined. Feel free to ignore it, as the relevant results are already stated above.

Here I assume that we use a lossless lens (i.e. it conserves luminance) to focus the light of an object of uniform luminance *L* into an image plane. The lens is surrounded by air (index 1), and we look at the light falling on an infinitesimal area d*S* about, and perpendicular to, the optical axis. This light lies inside a cone of opening α. We want to compute the illuminance delivered by the lens on d*S*.

In the figure below, the marginal rays, in green, define the light cone with opening α, while the chief rays, in red, define the target area d*S*.

_{(source: edgar-bonet.org)}

The etendue of the light beam illuminating d*S* is

d

G= dS∫ cosθ dω

where dω is an infinitesimal solid angle, and the integral is over θ ∈ (0, α/2). The integral can be computed as

d

G= dS∫ 2π cosθ sinθ dθ

= dS∫ π d(sin^{2}θ)

= dSπ sin^{2}(α/2)

The illuminance at the image plane is then

I=LdG/ dS=Lπ sin^{2}(α/2)

We may now define the “speed” of the lens as its ability to provide image-plane illuminance for a given object luminance, i.e.

speed =

I/L= dG/ dS= π sin^{2}(α/2)

It is worth noting that this result is quite general, as it does not rely on any assumptions about the imaging qualities of the lens, whether it is focused, aberrated, its optical formula, focal length, f-number, subject distance, etc.

Now I add some extra assumptions that are useful for having a meaningful notion of f-number: I assume that this is a good imaging lens of focal length *f*, f-number *N* and entrance pupil diameter *p* = *f*/*N*. The object is at infinity and the image plane is the focal plane. Then, the infinitesimal area d*S* on the image plane is conjugated with an infinitesimal portion of the object having a solid-angular size dΩ = d*S*/*f* ^{2}.

Given that the area of the entrance pupil is π*p* ^{2}/4, the etendue can be computed on the object side as

d

G= dΩ πp^{2}/ 4

= dS πp^{2}/ (4f^{2})

= dS π / (4N^{2})

And thus, the speed of the lens is

speed = π / (4

N^{2})

Equating this with the speed computed on the image side yields

N= 1 / (2 sin(α/2))

I should insist here on the fact that the last assumptions I made (the lens is a proper imaging lens focused at infinity) are only needed for relating the speed to the f-number. They are *not* needed for relating the speed to sin(α/2). Thus, there is *always* a hard limit on how *fast* a lens can be, whereas the f-number is only limited *insofar as* it is a meaningful way of measuring the lens’ speed.

## For how long would I need to use a lens hood during the day and in how many shots

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## Question

I was wondering if it is worth it carrying a lens hood around. I read a few articles (1, 2, 3) and now I know what is a lens hood and how it works, but I do not know for how long would I need to have one on my lens during the day and how often would it make a difference in my photos.

## Equipment

Camera: `Canon EOS 2000D`

Lens: `Canon Zoom Lens EF-S 18-55mm 1:3.5-5.6 IS II`

Polarizing Filter (if needed): `K&F Concept C-Series HMC CPL 58mm`

1

The main reason to use lens hood is to stop stray light to come onto your lens and make flare (and decrease the contrast). Also hood can play the role of bumper when lens hit some object.

I personally put it on the lenses very long time and keep it there. The only rare occasions when I remove it is when I change the filter on the lens. But after change I put it on back. Of course this increase a bit the size of the lens but this do not bother me.

I think this is largely up to the individual photographer and situation. Personally, if the lens came with a hood, I pretty much just leave it on, because it does provide a little extra protection for the front element. There are exceptions, though, like if it would interfere with positioning (e.g. with a macro lens shooting extremely close to something – most non-macro lenses have a minimum focusing distance that is much farther out), or if the hood would cast a shadow across the image due to strong directional lighting, or if I artistically decide I actually want to encourage lens flare or other effects that a hood is designed to minimize.

##

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## What does 1: in Canon Zoom Lens EF-S 18-55mm 1:3.5-5.6 IS II mean?

It means exactly the same as `f/`

. `1:`

is just an older syntax for denoting a ratio. You might also occasionally see `1/`

. Since the f-stop is essentially a ratio of focal length to entrance pupil diameter (the apparent size of the aperture when looking into the front of the lens), one of these notations is used to indicate, in this case, that at the 18mm end, your apparent entrance pupil size is `18/3.5 = 5.14mm`

, and at the 55mm end it’s `55/5.6 = 9.82mm`

.