Showing posts with label focal-length. Show all posts
Showing posts with label focal-length. Show all posts

Saturday, April 28, 2012

Does the shutter speed and focal length rule of thumb apply to cropped sensor cameras?

Question

So, the rule of thumb for shutter speed is that it shouldn't be less than 1 / focal length. Well, that's straight forward on full frame cameras, but what about cropped sensor cameras? Is it going to be 1 / equivalent focal length? By equivalent focal length, I mean original focal length × crop factor.

My guess is: no, it's just 1 / focal length, cause the lens's focal length didn't physically change.

Asked by Akram Mellice

Answer

According to this Wikipedia article on the secondary effects of crop sensors:

The old rule of thumb that shutter speed should be at least equal to focal length for hand-holding will work equivalently if the actual focal length is multiplied by the FLM [focal length multiplier] first before applying the rule.

So, yes, use the 35mm equivalent focal length as your reference for minimum shutter speed.

Answered by John Rygielski

Wednesday, April 25, 2012

What is the actual distance for taking a photograph with a 4.3-180mm Super Zoom camera lens vs. a 55-200mm DSLR lens?

Question

What will be the true actual distance for both lens in terms of actual meters?

Asked by Jack

Answer

I think I understand what you're getting at here. The puzzle is: the superzoom camera has a big "times zoom" number, whereas the SLR lens much smaller. That's explained in the answers to What does 'how much zoom' mean?, and reading that should make this somewhat more clear. The key point is that the "zoom times" number is just a measure of flexibility, not an indicator of magnification.

What you want is a way to tell how much bigger that telephoto lens will make stuff in your pictures — how far away you can be from your subject and make a decent picture. The answers to What does "angle of view equivalent to that of some lens in 35mm format" mean? will tell you some of that, but it's in detached, mathematical terms of angles and so on.

Fortunately, those angles can be translated into the real world. You can work out the math yourself, but there's also handy online (and smartphone) tools to do it for you so you don't have to remember the formulas. There's a nice simple one at http://www.tawbaware.com/maxlyons/calc.htm#fov_calculator. For this purpose, scroll down to the Dimensional Field of View Calculator.

First, put in 200 for the lens focal length, and 1.5 for the "Focal length multiplier" (which is another word for sensor crop factor) — this matches a Nikon, Sony, or Pentax entry-level DSLR with a 200mm lens. Put in 25 ft for the distance to subject, and press compute. You can see in the results that at that distance, this focal length gives you a frame of about 3 feet wide and 2 feet tall.

Next, change the focal length to 180, and change the "focal length multiplier" (remember, crop factor) to 5.6 to match the superzoom. Leave the distance the same, and hit compute again. Now, the imaginary captured rectangle is much, much smaller: 10.7" across and 7.1" high.

Imagine two prints, both the same size, taken from each picture. The superzoom image would be a much narrower angle, but printed to cover the same area as the wider angle, so there'd be more magnification.

In this case, it's about 3× in each direction — if you imagine dividing the field of view of the 200mm DSLR lens in thirds, and then taking just one third and expanding it, there's your direct comparison.

But wait! Before you run off to buy the superzoom based on this alone, consider that here's where Why doesn't it make sense to compare an entry-level DSLR with a super zoom? might kick in. Because that 55-200mm is physically much bigger, and because it has less of zoom range, it may have less compromise in image quality. Plus, the DSLR has a much larger sensor, which will matter particularly if you're not in full sunlight. For these reasons, it may be that a tiny fragment of the DSLR image cropped and expanded is as good or even better than the full image straight from the superzoom.

It may be the case that either gives perfectly adequate results for what you want to do. In that case, there are other factors: size, price (especially price!), and zoom flexibility vs. system flexibility to consider, and that's what you should base your decision on.

Answered by mattdm

Tuesday, April 24, 2012

What does “the telephoto end” and “the wide angle end” of a zoom lens mean?

Question

In the Canon Rebel T3i Manual, page 59, it's mentioned that "If you have a zoom lens, use the telephoto end to fill the frame with the subject from the waist up." Same note on page 61 and another one for landscape telling to use the "wide angle end".

I have EF-S 18-55 mm IS II lens which is a zoom lens. I've googled "telephoto end or wide-angle end", but all the results were lenses not an end. I don't know if I got it wrong but I thought that "end" would mean something that can be mounted on my zoom lens itself.

Asked by Akram Mellice

Answer

'End' in this case is just referring to the end of the zoom range available. So zooming in on an 18-55 lens would take you to the 55mm end of the zoom range, which would be the telephoto end, and the 18 mm end of the zoom range would be the wide angle 'end' (although on this lens the telephoto end isn't very telephoto). So while it may sound like a physical 'end' they're referring to, it's just the ends of the zoom range of the lens they are referring to

Answered by Dreamager

Sunday, April 8, 2012

Why are the area of aperture, focal length, and amount of light specified in these terms?

Question

So I'm looking at the wikipedia article on Aperture, and I'm a bit confused. I understand aperture to be the diameter of the hole that lets light in. In the article, someone states that

The amount of light captured by a lens is proportional to the area of the aperture, equal to:

Area = pi ( focal length /(2 * f number) )^2

But in the f number article, they define f number as

f = focal length / aperture

It then seems trivial to substitute in:

Area = pi ( focal length / (2 * focal length / aperture) )^2

Area = pi ( focal length * aperture / (2 * focal length) )^2

Area = pi ( aperture / 2)^2

Area = pi ( radius )^2

So, my question is: Why did they bother putting something so blatantly obvious into the article? Didn't we already know that the area of the aperture was pi * r^2? Is this just me reading too much into a wikipedia article?

Answer

The focal length and f-number are photographically relevant quantities, so the formula is expressed in terms of those convenient variables. Photographers don't generally know the radius of their lens aperture for every f-stop.

Answered by coneslayer

Sunday, April 1, 2012

Calculating focal length from photograph

Question

I posted this question over on Stack Overflow, but someone suggested this site might be of more help to me.

I'm doing a college project where I'm trying to calculate the focal length of a camera from a photo. The project is part of my final year computing course (that's why I assumed Stack Overflow would be the best place to ask)

I'm using vanishing lines identified by the user to try and calculate the focal length. say you have a cube, the user can select 3 sides to intersect. For each side, 2 lines are drawn to infinity. Where 2 lines intersect = intersection point. This is done for all 3 sides giving 3 intersection points. These points are then used to draw a new triangle and from this triangle to focal length can be calculated as shown in this Research research. But my values are way off. For this image the focal length is known as 18mm but the value I get is 478.634....

enter image description here

Would anyone have any experience with this type of problem?

I have also come across this formula on this website. Does anyone know what

alphaX (fx = (imageWidth)2*tan(alphaX))

would represent.

Any help would be appreciated

UPDATE:.

Ive made a few pics to show how I calculate the focal length (images not to scale or size). Starting with a photo of a cube Cube I extend each one of the 3 sides till they intersect (shown by the green dots). Then using those intersection points I construct a rectangle as shown here Intersecting points. Then I calculate the ortho-centre of the triangle. Then using the formula here I calculate the focal length as shown here enter image description here. In my application the focal length is simply a value (integer) that is calculated from the square root of the length of the intersecting points (IP1) to the centre, multiplied by the length from the centre to the opposite midpoint (MP2)

Asked by Hans Moolman

Answer

(1) By inspection it is "clear" [tm] that a relatively wide angle lens is in use. In a 35mm full frame system a guesstimate far far far closer to 18mm than 480 mm would be arrived at.

(2) Without having got my brain fully around the triple vanishing point method described in a reference, I would think that for vanishing points to be relevant you would need to deal with lines which were parallel in reality but rendered apparently not so by perspective OR perhaps wit lines whose actual interposing angle was known and which were rendered at another apparent angle in the image. However, in the example photo given, the actual angle between the two walls on either side of the island seems to be arbitrary and unknown (except to their builders) and that changing this angle changes the answer you would get. HOWEVER, I may just be wholly missing the point :-).

(3) I am reasonably certain that the formula quoted by the original poster in your reference as

  • fx = (imageWidth)2*tan(alphaX))

Is intended to read

  • fx = (imageWidth) / (2 * tan(alphaX) )

Where:

fx = focal length
imagewidth = sensor width
alphaX = HALF angle of view

This is consistent with the diagram below whose precision of rendition will hopefully be excused or ignored. The above formula then reduces to a simple geometric statement.
I have used slightly different terms to assist and/or confuse the mental process :-)

enter image description here

HALF the angle of view = alpha has tan(alpha) = Y/X = height/distance.
The identical tan is produced by half the image height on the sensor / focal length.
So
(sensor_height/2) / focal_length = height/distance = tan (alpha)
so
(IW/2) / fx = tan_alpha
invert
fx / (Iw/2) = 1/tan_alpha
fx = Iw/(2 x tan_alpha)

Answered by Russell McMahon

Sunday, March 25, 2012

Do bigger focal lengths capture more light?

Question

I never understood what the f really stands for in the f-stop values, like f/1.8. Wikipedia explains it in various articles, but I still find it very confusing. What I understand, though, is that it has something to do with focal length.

Should I assume, therefore, that bigger focal lengths capture more light? For example, should 85 mm f/1.8 lens capture a lot more light than for example 24 mm f/1.8?

References to back up my confusion:

The article about aperture goes on to claim:

The amount of light captured by a lens is proportional to the area of the aperture, equal to:

enter image description here

Where f is focal length and N is the f-number.

The "f-number" is mentioned here. But the article about F-number claims:

In optics, the f-number of an optical system expresses the diameter of the entrance pupil in terms of the focal length of the lens; in simpler terms, the f-number is the focal length divided by the "effective" aperture diameter.

This seems very recursive. Why does the aperture article refers to both the focal length and the f-number, when the latter article claims that f-number already carries the property of focal length?

What's going on here?

Asked by RiMMER

Answer

The f-number is in use to express how much light a lens can capture, so the 85mm f/1.8 and 24mm f/1.8 can capture the same amount. Here, f is the focal length, and f/1.8 means that maximum aperture diameter is 47.2mm in first example and 13.3mm in second.

What you have to consider here is that the 85mm lens has a much narrower field of view, therefore it has to gather the same amount of light from a much smaller area - to compensate the narrow view, aperture has to be bigger.

How much the aperture has to be bigger is linearly correlated to focal length. We could say that a 24mm lens with 13.3mm aperture can gather as much light as a 85mm lens with 47.2mm aperture, but talking about the F-number makes this much easier to notice.

Answered by Imre

Wednesday, March 21, 2012

Why is the Canon MP-E 65mm F/2.8 Macro Lens not called a zoom?

Question

The Canon MP-E 65mm works differently than my other macro lenses. Compare:

  • With a standard 35mm macro, I can frame first and then adjust focus which only changes framing very slightly with.
  • With the MP-E 65mm changing focus completely changes framing.

Assuming that the angle of view corresponds to a 65mm lens at 1X magnification, it seems that by 5X, the angle of view is about that of a 325mm lens! In other words, manual focusing (this lens does not autofocus at all) looks like it has the same effect as zooming on a zoom lens with the plane of focus fixed at the closest focus-distance.

So why is this not called a MP-E 65-325mm instead of simply 65mm? More importantly, what does the 65mm focal-length represent? And how can I use that to frame by shots with less guesswork?

Asked by Itai

Answer

It's called a 65mm because that's the combined focal length (light bending ability) of the lens elements. The focusing mechanism on the MPE-65 works like a set of bellows, i.e. it just moves the whole lens back and forth. You can see this as the physical length of the lens barrel changes significantly when focusing. This has the effect of changing both the minimum focus distance and the angle of view, however the focal length of the lens elements remains unchanged.

It's common for prime lenses change field of view slightly on focusing, the effect is just exaggerated with the MPE-65. Lots of rules break down when you get into macro and super-macro photography, focal length, f-stop etc. cease to matter as the formulas for things like depth of field and exposure don't give accurate answers at macro distances. You have to ignore what it says on the lens and use your judgement or trial an error to begin with.

Note it's standard practice to state the focal length of a lens when focussed at infinity (to account for minor variations as mentioned above). As the MPE-65 doesn't focus to infinity I assume they measure it for the 1x setting, so the value is not directly comparable with other lenses.

Answered by Matt Grum

Tuesday, March 13, 2012

What are the technical camera specs for the HTC Wildfire S?

Question

I am trying to do some CGI manipulations on pictures I've taken. In order to fake reality with some 3D object inserted in the picture, I need some information to configure my virtual camera so it can match my real one.

The problem is that I'm using the camera of my Smartphone, which is a HTC Wildfire S. There are no specs concerning the camera in the manual nor on the Internet.

I need to find out the focal length and the width of CCD sensor (and maybe some other things I might be asked by the CGI software later on). Is there a good and simple way to find these?

Asked by Oltarus

Answer

Answered by MikeW

Monday, March 5, 2012

How to test actual focal length?

Question

From Matt Grum's comment to my previous question, I learned that manufacturers may casually "round" actual focal length of a lens to some nice number that gets printed on the box and stored into EXIF. From his answer to the same question, it seems I would need to know actual focal length of a lens to test what aperture is used.

I have also heard that most lens will change focal length when focused very close.

How would I go about testing what focal length my lens is actually using when focused on a given distance? EXIF obviously won't help me here, because data is put there by manufacturer.

Asked by Imre

Answer

There is a mathematical / measurement method to calculate the effective focal length of a lens by measuring its angle of view.

The formula for angle of view is given as
enter image description here

To calculate effective focal length (f), the formula comes down to:
f = d / (2 * tan(α/2)) -> Equation1

Where d represents the size of the sensor in the direction measured. d would be 24 in case you are using a full frame camera.

Let us now have the following setup for measuring α

enter image description here

You have a camera sitting at a height H from the ground and a distance of X from the wall with a scale. Now take a picture and you should be able to read the maximum height the lens can see (this would be H + Y).
Now knowing X and Y, we can calculate half the angle of view (i.e. α/2) using this link (X would be the opposite side and Y the adjacent side)

Now that you have figured out α/2, use it on Equation1 to calculate the effective focal length of the lens.

The value is only accurate as your measurements.

Edit 1:
In reference to mattdm’s question: Are the manufacturer-stated sensor dimensions close enough?
With reference to sensor sizes of camera’s in these links: here and here
We can logically assume that camera makers or at least Canon and Nikon round their sensor sizes 1/10 of an mm. i.e. there is a possibility of +/- 0.05mm error in case they round the sensor size.
Let us consider 3 type of lenses:
1. Wide angle lens (say 13mm, angle of view: 85.4)
2. Normal lens (50mm, angle of view 27.0)
3. Telephoto lens (300mm, angle of view: 4.58)

The effect of a 0.05mm change in sensor size are:
change for wide Angle lens = 0.05 / (2 * tan(85.4/2)) = 0.04613 mm appx.
Which represents an difference of 0.35% (i.e. (0.04613 / 13) * 100 )

change for normal lens= 0.05 / (2 * tan(27/2)) = 0.012 mm appx.
Which represents an difference of 0.024% (i.e. (0.012/ 50) * 100 )

change for telephoto lens= 0.05 / (2 * tan(4.58/2)) = 0.0019 mm appx.
Which represents an difference of 0.0006% (i.e. (0.0019/ 300) * 100 )

We can thus see that with a 13mm wide angle lens and taking a 0.05mm error in manufacturers’ measurement, the change in the focal length is only 0.35%.

I hope that my math is correct.

Edit 2:
In reference to Imre's question about measurements for X & H,
H should be measured from ground to the horizontal center of the sensor.
X is the distance between sensor and the wall.

Answered by Vivek

Wednesday, February 22, 2012

How to compare the focal length of a 50mm prime lens with the default lens of a point and shoot camera?

Question

To imagine the 50mm focal length of a prime lens 1.4F, can I simply zoom the lens of my point and shoot camera up to 50mm?

Will both the focal lengths be same?

Any other way to imagine the focal length 50mm of prime lens 1.4F when what you only have is a point and shoot camera with 70mm max zoom?

Asked by Anisha Kaul

Answer

If your point-and-shoot has the typical 1/2.3" format sensor and you are trying to compare it to a 50mm lens on a cropped-sensor DSLR (in your case, a Nikon, if I recall correctly), then there's a little bit of math involved.

The compact's sensor has a 3:4 aspect ratio. It measures 6.16mm by 4.62mm, with a diagonal of 7.70mm.

The Nikon DX sensor has an aspect ratio of 2:3 and measures 23.6mm by 15.8mm. That would give a diagonal of 28.4mm.

It's normal to compare lenses based on the diagonal of the negative/sensor. I don't like that approach, though, since the aspect ratio is different. It's best to compare images with the same aspect ratio, and in the case of both of these sensor formats, the full length of the short side of the sensor will be used when an image is printed in a 4:5 aspect ratio. That is, an 8x10 picture made with either camera would involve cropping the longer dimension. So it's safe to compare just the shorter dimension of each camera/lens combination.

The 50mm lens on the Nikon is about 3.165 times the length of the shorter side (the height when the camera is held horizontally). That means that in order to get the "same" 8X10 from your compact camera, the lens would have to be set to 3.165 times the shorter side of its sensor, or about 14.6mm.

If your compact has a different-sized sensor, the math will still hold. Find out what format of sensor it uses, then multiply the shorter side of the sensor by 3.165 to find out how far out the lens needs to be zoomed to approximate the 50mm lens. Do note, though, that the field of view for our hypothetical 8X10 print is the only thing the two cameras will have in common.

Or, if you want to do it without the math, my rather standard 8-1/2" (about 21.5cm) tall head will completely, but just, fill the frame of a Nikon DX in "landscape" orientation from the top of my bald pate to the bottom of my chin using a 50mm lens from almost exactly three feet away. So a volunteer or a ruler three feet (90cm) away or a mirror 18 inches (45cm) away will be enough to show you how far to zoom.

Answered by Stan Rogers

Monday, February 13, 2012

If I have an 85mm lens for portraits, is a 50mm useful as well?

Question

I recently got my Canon EF 85mm f1.8, after lot of thought over if I should buy Canon EF 85mm f1.8 or Canon EF 50mm f1.4, I decided on 85mm so that I can shoot portraits of people even from a distance, though I might have difficulty in closeup shots where space might be less!

Now I am tempted to buy 50mm too, but of course don't have budget right now for f1.4. So here are questions in my mind:

  1. Should I go for 50mm f1.8 now and be ok with that, or it's better to wait few months and at some point go for 50mm f1.4?

  2. While distance between object and camera with 85mm f1.8 is a known factor, is there anything else which would limit for portraits with 85mm? Should I completely give up thoughts on 50mm, or rather buy it?

Asked by vishal.biyani

Answer

I have the 50mm 1.8 - it takes great pictures but it's not a pro lens - the auto focus is slow and have a tendency to hunt a bit, the manual focus ring is tiny and the build quality is ... well, don't expect much (but it is cheap and optically wonderful).

I've heard the 85mm 1.8 is a great lens and if I had it I would go for something around 30mm (Canon or 3rd party) for a second lens, not the 50mm.

The 50mm is like an 85mm on full frame, it's great for head shots and can take half body shots if you have the space, something in the 28-35 range is like 50mm full frame good for full body and small groups.

I have the 50mm 1.8 and the 18-135mm zoom, some time ago I checked the EXIF data on all my pictures and most of the pictures I take with the 18-135 (by a really large margin) are in the 30-35 range, remember, the "normal lens" for a Canon APS-C is around 30mm (if my math is right it's 27mm) not 50mm.

Answered by Nir

Sunday, February 12, 2012

Should I start with an 85mm or 50mm prime lens for portraits with an APS-C DSLR?

Question

I recently got my Canon EF85mm f1.8, after lot of thought over if I should buy Canon EF 85mm f1.8 or Canon EF 50mm f1.4. I decided on 85mm so that I can shoot portraits of people even from a distance, though I might have difficulty in closeup shots where space might be less!

Now I am tempted to buy 50mm too, but of course don't have budget right now for f1.4 :) So here are questions in my mind

1) Should I go for 50mm f1.8 now and be ok with that, or it's better to wait few months and at some point go for 50mm f1.4?

2) While distance between object and camera with 85mm f1.8 is a known factor, is there anything else which would limit for portraits with 85mm? Should I completely give up thoughts on 50mm, or rather buy it?

Asked by vishal.biyani

Answer

I have the 50mm 1.8 - it takes great pictures but it's not a pro lens - the auto focus is slow and have a tendency to hunt a bit, the manual focus ring is tiny and the build quality is ... well, don't expect much (but it is cheap and optically wonderful).

I've heard the 85mm 1.8 is a great lens and if I had it I would go for something around 30mm (Canon or 3rd party) for a second lens, not the 50mm.

The 50mm is like an 85mm on full frame, it's great for head shots and can take half body shots if you have the space, something in the 28-35 range is like 50mm full frame good for full body and small groups.

I have the 50mm 1.8 and the 18-135mm zoom, some time ago I checked the EXIF data on all my pictures and most of the pictures I take with the 18-135 (by a really large margin) are in the 30-35 range, remember, the "normal lens" for a Canon APS-C is around 30mm (if my math is right it's 27mm) not 50mm.

Answered by Nir

Thursday, January 5, 2012

Why does my aperture setting change as I zoom the lens on my Nikon D90?

Question

I am new to the photography field and this something which I am not able to understand. I am playing around with my Nikon D90 along with its kit lens (18-105mm).

While reading about Depth of preview, the author asks to do some basic exercise to better understand DOF, like this:

  • set your aperture to smallest number f/2.8, f/3.5, f/4 with 70mm or longer lens.

When I tried to set aperture to f/3.5, and tried to change focal length, my camera is setting aperture automatically in every every possible mode (which I know of). It is changing aperture in the following fashion:

  1. 18-24 ----> 3.4 to 4
  2. 18-35 ----> 4.5
  3. 18-50 ----> 5
  4. 18-105 ---> 5.6

But if i set my aperture to 5.6 or higher, it does not change when I change the focal length of my camera. I know I am not doing some basic thing right, but I'm still not sure why this is happening. Might some one help me to understand this?

Answer

It is happening because you have a variable aperture zoom lens. The solution is to get a quality lens, otherwise you have to live with the limitations which are actually marked on the barrel of your lens.

It says 18-105mm 1 : 3.5 - 5.6G which means your maximum aperture is F/3.5 at the widest focal-length (18mm) and F/5.6 at the longest (105mm). It changes in increments between that. So, if you are set to F/5.6 then you can zoom with whole focal-length without aperture changing. If you set your aperture to F/3.5 then after a short increase in focal-length, the lens has to diminish its aperture.

Saturday, December 31, 2011

Why doesn't the background become blurred when I change the aperture on my Fujifilm S4000?

Question

I am sorry if my question is too naive, but I am a complete beginner in photography, and I think I still need to understand well what aperture is. I have a Fujifilm S4000 camera, and I know I can't do with it everything that a professional camera does.

I wanted the background of some pictures to be blurred by changing the aperture, but I don't get the results I expected. I think it also has to do with the focal length, am I right? I took two pictures to show you what I mean.

Picture 1 Aperture: f/4 Exposure: 1/2 s Focal length: 7 mm

Picture 2 Aperture: f/10 Exposure: 2.6 s Focal length: 7 mm

I would like to know what settings I can use to get the background to be blurred. I know that with smaller aperture values, the area around the object in focus will be more blurred, but when I change from f/4 to f/10, it doesn't change much.

Answer

There is NO different at ALL because the physical aperture has not changed.

The Fuji Finepix S4000 simulates a small aperture using an ND filter. When you stop-down, the ND filter slides into the optical path. The Aperture written i the EXIF is adjusted to reflect the transmittance of the ND filter but since the size of the opening has not changed, Depth-Of-Field does not change either.

Not only do such cameras have very small sensors and extensive depth-of-field but they are also extremely prone to diffraction. If the aperture was actually stopped down to F/10, images would get quite blurry.

Why doesn't the background become blurred when I change the aperture?

Question

I am sorry if my question is too naive, but I am a complete beginner in photography, and I think I still need to understand well what aperture is. I have a Fujifilm S4000 camera, and I know I can't do with it everything that a professional camera does.

I wanted the background of some pictures to be blurred by changing the aperture, but I don't get the results I expected. I think it also has to do with the focal length, am I right? I took two pictures to show you what I mean.

Picture 1 Aperture: f/4 Exposure: 1/2 s Focal length: 7 mm

Picture 2 Aperture: f/10 Exposure: 2.6 s Focal length: 7 mm

I would like to know what settings I can use to get the background to be blurred. I know that with smaller aperture values, the area around the object in focus will be more blurred, but when I change from f/4 to f/10, it doesn't change much.

Answer

There is NO different at ALL because the physical aperture has not changed.

The Fuji Finepix S4000 simulates a small aperture using an ND filter. When you stop-down, the ND filter slides into the optical path. The Aperture written i the EXIF is adjusted to reflect the transmittance of the ND filter but since the size of the opening has not changed, Depth-Of-Field does not change either.

Not only do such cameras have very small sensors and extensive depth-of-field but they are also extremely prone to diffraction. If the aperture was actually stopped down to F/10, images would get quite blurry.

Thursday, December 22, 2011

What is the meaning of single focal length?

Question

This is a single focal length Nikon lens: http://www.nikon.co.in/products.php?categoryid=1014

What does that mean?

What are its disadvantages as compared to multiple focal length lenses?

Answer

A single focal length, or "prime", lens has a fixed focal length, as opposed to a zoom lens which has a range of focal lengths. So a prime lens will be a 35mm lens, or a 50mm lens, or a 200mm lens. A zoom lens will cover a range, say 70-200mm.

Prime lenses are generally sharper, less expensive, lighter weight and have larger maximum apertures, which is useful in low light. This is because these lenses can have a much simpler design, since the lens only has to work at the one focal length, and can be optimised for it.

Zoom lenses have the advantage that you can use one lens and cover a range of focal lengths, rather than owning several prime lenses and having to swap lenses all the time. You can also zoom in or out to compose a shot. With a prime lens you'd have to move closer to, or farther away from, the subject ("zoom with your feet"), or switch lenses.

Wednesday, December 21, 2011

What does it really mean that telephoto lenses “flatten” scenes?

Question

This is another perspective distortion question. I understand that perspective distortion is primarily a function of capture and viewing distance, but I am still working on fully understanding the causes of the "flattening" of scenes that is credited to telephoto lenses. This seems to refer to the fact that our ability to discern the differences in size attributable to differences in distance from the lens for subjects beyond a certain distance from the lens. If this is correct, then is it just that with the longer distance from the subject when using a telephoto lens, this compression happens closer to the foreground, rather than in the mid- or background with a wide lens? Is it fair to say that the compression effect occurs at a certain distance from the photographer regardless of lens, but the field of view influences how close to the foreground this effect takes place? Can you share any information to help people like me understand this?

Answer

The flattening or compression effect is not caused by a particular kind of lens, it applies to all lens in the same way. Actually, this property of lenses applies to our own eyes as well. The factor that affects flattening is the distance from the camera to the subjects.

Consider the following exercise:

Place two friends 1 meter away from each other. Place yourself in line with them, so that you are 1 meter away from one of them, and 2 meters away from the other. Move a bit to the side, so that you can see both. From this position, you can easily estimate what the distance between your two friends is. Now walk 100 meters back along the imaginary line determined by your two friends. Again, move a bit to the side so that you can see both. Can you really tell now how far away from each other they are? Will you see your friends in a significantly different way if one of them moves an additional meter apart from the other? You won't, because from 100 meters away, a difference of a meter is not significant.

The thing is, our eyes have a fixed field of view, so in addition to flattening, we experience a scale reduction as we move farther away from the subjects. With a camera you can magnify the effect by using a long lens. But the compression will be exactly the same no matter what lens you use.

There are a few good examples on the Wikipedia page on Perspective Distortion.

From that page, I like this animation of a cube that goes from extreme extension to extreme compression, when changing the focal length and distance to camera:


Image copyright 2007 SharkD, licensed CC-BY-SA 3.0

Thursday, December 8, 2011

How do I get the 'equivalent' focal length for a DSLR lens for my camera?

Question

On my Nikon D3000, how do I figure the focal length equivalence? In other words, if I want a lens that is just like a 50mm lens on a 35mm film SLR, I need to multiply by a factor. What is the factor, and how is it derived?

Also, when I buy a new lens for it, and the box says 70mm-300mm, I assume that's the true focal length of the camera and not the equivalent, so I have to apply the factor. Is that correct?

Answer

What is the factor, and how is it derived?

The factor (also commonly called a 'crop factor') is a measurement of how much larger a full-frame image sensor is than the sensor in your camera. People say the D3000's crop factor is 1.5 because a full-frame sensor is 1.5x larger than the D3000 sensor.

A full-frame sensor is 36mm x 24 mm. The sensor in the Nikon D3000 is 23.6mm x 15.8mm.

Comparing the sensor sizes along each dimension:

36mm / 23.6mm = 1.52 (approximately)

24mm / 15.8mm = 1.52 (approximately)

You can see how the full-frame sensor is around 1.5x the size of the D3000 sensor, hence the 1.5x crop factor.

Monday, December 5, 2011

What is it called when an object appears to be the same size with the eye and through the camera viewfinder,?

Question

I have a Canon 18-135 lens. I am looking into the camera viewfinder with my right eye and not through the viewfinder with my left. I start from 18mm and keep zooming in till I see a particular object as the same size with both the eyes.

I read the focal length to be just above 50 mm (say, 55mm). What is special about this 55mm with reference to the object I was focussing on?

Answer

There's a specification on (d)SLR bodies called viewfinder magnification; this refers to how large an object appears in the viewfinder when a 50mm lens is mounted and focused at infinity.

On mid-range DSLRs, which typically have around 0.95x magnification, an object will appear to be life-sized at 52.6mm. With entry-level DSLRs, you might have around 0.8x magnification, so you'd have to zoom in to 62.5mm to get a 1:1 magnification. I'm betting your camera has somewhere around 0.9 to 0.95x magnification.

This has very little to do with the object, except that the front of the camera is slightly closer to the object. With far-away subjects (landscapes, etc), the distance between your eyes and the front of your camera won't matter much, but close subjects will appear larger to the camera than to your eye even with a 1:1 magnification.

Thursday, December 1, 2011

What mm lens for distance shots?

Question

If I am looking to take landscape and wildlife shots and some of those shots being from a decent distance what mm lens would I need? I am not sure what else I should be adding here, I am just looking for decent quality above all else.

Answer

The mm of a lens is the focal length. To take close-up shots of things far away, you need a "long" focal length, which means high mm.

An entry or mid-level Canon camera has an "APS-C" sized sensor. The sensor size is really what determines what the meaning of the focal length is — focal length and sensor size together give you angle of view. (See my visual exercise for understanding that.)

To get another visual idea of what all this means, look at the answers to How can I visualize or simulate the effect of different focal lengths?, particularly the one that points to Nikon's web-based lens simulator. As mentioned, the numbers are slightly different from Nikon to Canon (choose Nikon's "DX" format to be closest to an entry-level Canon), but close enough that it really doesn't matter for basic purposes. Drag the slider at the bottom to see the effect of a focal length ranging from 10mm to an extreme 600mm.

After playing with that, come back here to read some explanation of what all this means:

Canon's APS-C is slightly smaller than that from Nikon, Sony, or Pentax. This isn't a big deal, but for those systems the numbers below should be a few percent larger for the same meaning. All of the definitions are rough categories, though, so it's close enough that it really doesn't matter.

Anyway, on Canon's APS-C, a focal length of about 27mm is considered normal — that means when printed at normal size, you get a perspective that seems natural and about what you might expect your memory of the scene to be.

Focal lengths in somewhere in that neighborhood are also considered normal: probably up to 40mm or so, and down to about 22mm.

Once you get smaller than 22mm, you're into wide angle. Wide angle won't be terribly flattering for portraits (it will cause features to be exaggerated if your subject is close enough to fill the frame), but the more-normal end of the wide range is pretty common for street photography. And, wide angle is very common for the landscape shots you mention.

Down below 15mm, it's "ultra-wide", where perspective distortion starts to be a key feature of the way the lens sees. This can be used for landscapes, but you have to be very aware of the distortion and the field of view of the lens — see this question on ultra-wide composition.

On the other side, there's the telephoto range. Telephoto has a technical meaning, but generally we use it to just mean a lens with a relatively long focal length (high mm). The range from around 50mm to 85mm is considered short telephoto or portrait — the latter because, well, it's a good range for flattering perspective for portraits, and it's not so narrow that you have to stand hundreds of feet from your subject.

Beyond that is the real telephoto range. This is generally what people want for action and wildlife shots, because they enable you to get right into the action without actually being close.

Consumer-priced telephoto zooms usually go to 200mm, 250mm, or 300mm. Lower-cost lenses in this range tend to have less ability to let in light — they are "slow lenses". The fancier, "faster" lenses in this range used by professionals and enthusiasts for sports and wildlife cost one to two thousand dollars. And for fast lenses beyond 300mm, you could pay several times that, even.

But that's probably not what you're looking for right now. It might be — one advantage of just getting the nicer equipment to start is that you can forget about buying stuff for a while and just go out and enjoy while you grown into your gear, rather than growing out of it soon and wasting money trading up. So if you know a 70-200mm f/2.8 pro lens is right for you, it's not a mistake to just start there.

However, it's likely that you'll feel most comfortable starting out with a standard two-lens zoom kit. This isn't a bundle as sold by those shady camera stores: it's something from the camera makers. Almost all entry-level SLRs come with a 18-55mm f/3.5-5.6 zoom (note that the numbers go from wide angle at 18mm up through normal to portrait length at 55mm — this is a very versatile range, which is why it's the default). Many are also offered in two-lens kits with a 50-200mm telephoto zoom — or else, a 50-200mm lens is available separately cheaply. Sometimes, there's "upgrade" options like a 55-300mm zoom, also usually of lower-end features and quality for a similarly-reasonable price. You won't go wrong buying a setup like this from any of the top handful of brands.