## 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

Best budget wide-angle FX lens for Nikon F-mount – Photography Stack Exchange

## 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

I’m trying to calculate the f/stop change of attaching a x2 teleconverter to a 80mm f/2.8 lens. In this instance, the teleconverter attaches to the end of the lens, and has a larger opening than the original 80mm lens.

The original lens would have a lens diameter 80mm/2.8 which is about 30mm.

the new lens seems to be about 40mm, and it’s a x2 teleconverter, so the new f/stop should be 160mm / 40mm = 4.

All the documentation i found suggested that this is wrong and since it’s a x2 teleconverter the f/stop should also be doubled.

where’s my mistake here?

## 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 dS 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 dS.

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 dS.

(source: edgar-bonet.org)

The etendue of the light beam illuminating dS is

dG = 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

dG = dS ∫ 2π cosθ sinθ dθ
= dS ∫ π d(sin2θ)
= dS π sin2(α/2)

The illuminance at the image plane is then

I = L dG / dS = L π sin2(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 sin2(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 dS on the image plane is conjugated with an infinitesimal portion of the object having a solid-angular size dΩ = dS/f2.

Given that the area of the entrance pupil is πp2/4, the etendue can be computed on the object side as

dG = dΩ π p2 / 4
= dS π p2 / (4 f2)
= dS π / (4 N2)

And thus, the speed of the lens is

speed = π / (4 N2)

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

For how long would I need to use a lens hood during the day and in how many shots – Photography Stack Exchange

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

I noticed that in the front of my lens (`CANON ZOOM LENS EF-S 18-55mm 1:3.5-5.6 IS II`) where the it’s name is printed, that it contains `1:3.5-5.6` which is the widest aperture you can set at `18mm` and `55mm`.

1. What does `1:` mean?
2. Why on the internet it is mentioned as `CANON ZOOM LENS EF-S 18-55mm f/3.5-5.6 IS II` (with `f/`)?

## focal length – How to determine via a math formula when a lens is wide angle or macro

Lets say I have a device with 3 separate lenses. One will be the main reference lens, and then the two others may be a wide angle lens or a macro lens, or also even a replica of the main reference lens and not be different at all.

How can I calculate if any of the 2 extra lenses is a wide angle or a macro lens?

The way I am doing it now is very rough and probably totally wrong.
I use the next formula to calculate the field of view for each lens taking into account that I have the sensor size with and the focal length for all of them:

``````field_of_view = 2 * arc_tangent(sensor_size_width / (2 * focal_length))
``````

Then I compare the obtained field of view against the reference lens one, and if I find that such field of view is around 10% larger then I assume is wide angle, and if is 10% less smaller I assume is a macro lens.

Probably my way to do it is totally wrong, so any tips or corrections will be so much appreciated.

## canon – FD lens aperture is slow

I bought a Canon AE-1 Program on eBay along with a 50mm f/1.4. The camera works well with other lenses, but the 50mm’s aperture seems slow. Moving the bayonet (I think that’s what it’s called) takes much more force than on my other FD lenses.

As a result, the time between pushing the shutter button and the shutter releasing is sometimes hours. However, if I push the shutter button while holding the depth of field preview button, the shutter releases immediately. Similarly, if I push the shutter button and then push the depth of field preview button, the shutter releases. The depth of field preview button is very hard to push. Sometimes the shutter works as it should.

Do I need to lubricate it? How can I fix my lens?

FD 50mm Lens Problem