I have seen many camera review sites illustrate the effect of using different focal lengths on the same photograph using frames. There are also similar illustrations to compare different sensors (full frame vs APS-H vs APS-C vs micro 4/3). I find this kind of visualization to be very useful in comparing the actual effect of the longer focal length\different sensor size. Of course the ability to do a wider focal length visualization would also be equally useful, but that's not possible.

Is there any ready-made software or plugin that enables this? If not, is there any simple technique to visualize the effect of longer focal lengths on an image (one would of course have to use the zoom information in the image EXIF data to accurately represent this)? All I can think of is to use some trigonometry to do the necessary calculations for the frame sizes for cropping.


If using GIMP (and probably any other image editing software), then you could use the selection tool and set the size of the crop frame to a required size and aspect ratio. This way, you can see the relative sensor size superimposed on your image while letting you panning it around.

Note that this technique is good only for smaller-than-the-actual sensor sizes, unless you "cheat" by stating the original was shot with a bigger sensor than actual.

UPDATE: Unfortunately, using GIMP, you cannot simply set the ratio of the size if the selection box to match the crop factor (or focal lengths ratio). As @Stan Rogers commented, you'll have to set the size in pixels based on the simple focal length ratio. Then you can move the selection box to the desired location and if you like, add a rectangular frame to the image itself, so you can compare several selection sizes.

In order to add the rectangle, use the "Edit" -> "Stroke Selection..." dialog. The default settings will stroke a solid rectangle on your image.

  • The question in this case is how do I determine the required size to crop the image? Say, I have an image taken at 18mm and I want to visualize the 50, 100 & 200 mm crops, then what should I be doing?
    – ab.aditya
    Feb 22 '11 at 8:41
  • 5
    For 50mm, the crop dimensions would be (18/50)*x, where x is the width or the height of the image in pixels.
    – user2719
    Feb 22 '11 at 8:58
  • To the anonymous reader - I just noticed that there is one downvote on this answer. I don't really care about the vote itself, but PLEASE, can you tell WHY do you think there is a problem with the answer? This is generally true for all downvoters out there. To the benefit of the community, please try to explain your vote, and give the poster a chance to correct his potential mistakes.
    – ysap
    Feb 23 '11 at 2:52

Nikon has a simulator of lenses. Try it here. It supports up to 600mm

UPD: As Joanne C noticed:

Bear in mind that the Nikon DX format (APS-C) is actually a little larger than the Canon APS-C format. Not a huge difference, but it's there.

  • 2
    Bear in mind that the Nikon DX format (APS-C) is actually a little larger than the Canon APS-C format. Not a huge difference, but it's there. Nevertheless, I've always thought Nikon's utility is pretty handy and good relative comparison.
    – Joanne C
    Mar 31 '11 at 2:23
  • Nikon's APS-C (DX) has 1.5x factor, Canon's is 1.6x. It's different but almost unnoticeable
    – t3mujin
    Mar 31 '11 at 17:54
  • Canon also has 1.3x crop sensors used in 1D.
    – Imre
    Apr 1 '11 at 17:54

You don't actually need trigonometry — just basic arithmetic. The zoomed-in focal length gives a field of view as if you'd cropped the image by the ratio of the old focal length over the new: that is, if you have an image taken at 50mm, you can see the field of view of a 75mm lens simply by cropping by ⁵⁰⁄₇₅ths — which is ⅔.

This simple relationship is why the "crop factor" (sometimes, unfortunately, called "focal length multiplier") works. If your sensor is ⅔ the width of a full-frame sensor, that's cropping by a factor of 1.5 (the inverse of ⅔). So, you get the field of view of a lens with 1.5× the focal length on full frame — a 50mm lens on APS-C gives you the same field of view as a 75mm lens on full frame.

To put some numbers to it: if your 50mm-focal-length starting point is a 6-megapixel 3000×2000 image, cropping it to 2000×1333 will give you the field of view of a 75mm lens: in pixels, 3000 × 50 ÷ 75 horizontally, and 2000 × 50 ÷ 75 vertically. (A tangent, if you pardon the trig pun: You'll notice that this is quite a big hit in resolution — you loose a number of pixels equal to the crop factor — the ratio between focal lengths — squared. This is why optical zoom is usually preferable to "digital zoom", which is just cropping. And, generally, smaller sensors cram more pixels into the smaller sensor in order to compensate for the crop, which, depending on the level of technology used works to some degree. But that's a whole different discussion.)

You can use simple (non-trig) geometry to demonstrate this.

You'll need a ruler with millimeter markings, and a blank sheet of paper. I could make some graphics showing all this, but I really strongly believe that it's an exercise which works better if you actually go through it in on actual physical paper. So, if you will humor me and work along....

  1. Along the very bottom edge of the paper, centered in the middle, draw a horizontal line 24mm long. This represents an APS-C sensor.

  2. Measure 50mm up from the very center of that line, and put a dot. This represents the gathering of light within an idealized 50mm lens. (Imagine it as a pinhole camera, if you like.)

  3. Now, draw a line from the left edge of the sensor through the "lens" dot, and continue up through to the top of the paper. Do the same from the right edge, giving you an X shape with the lens point at the center of the X. The top cone of the X represents the horizontal field of view of a 50mm lens on your APS-C sensor.

  4. You can measure the angle with a protractor, if you happen to have one — it should be about 27°. And you can measure the horizontal field of view in millimeters a given distance away from your "camera", by measuring across the cone at the top of the X. (At 10cm away from the idealized lens dot, it should be about 4.8cm.)

  5. Now, measure up 75mm from the middle of your "sensor" line and put another dot, representing an idealized 75mm lens.

  6. Draw an X from the sensor edges through this dot as well. If you measure this angle, it should be about 18.2 degrees, and again, 10cm up from the lens dot, if you measure across, it should be about 3.2cm.

  7. And hey presto: 4.8mm × ⁵⁰⁄₇₅ = 3.2mm. (Of course, your lines are not at the exact same distance from the sensor itself, since you're measuring from the dot representing the lens in order to get the math to come out so nicely. But here we're working with unusually close focusing distances — when you're talking about a subject at normal distances the difference is negligible.)

  8. So anyway, you can then extend your sensor to be 36mm across instead of 24mm — changing it from APS-C to full-frame. Now, draw lines from that new larger sensor through the existing 75mm "lens" point.

  9. Even without measuring, you should be able to see that the angle of view with the larger sensor through the 75mm lens is the same as that with the smaller sensor through the 50mm lens. So there's the "crop factor" equivalence right in front of you. Cool, huh?

Note that this only covers angle of view. Perspective won't change, because you're standing in the same place, but depth of field (and the distribution of the depth of field) will. And of course actual real-world different lenses will have different properties (like distortion) which aren't modeled by this.

But in terms of field of view, that's all there is to it. Nothing beyond middle-school math required.

  • don't you mean "focal point" instead of "aperture"?
    – ysap
    Feb 23 '11 at 1:33
  • @ysap: it has been a ridiculously long time since I studied optics in, um, high school. But I think that a photographic lens doesn't properly have a focal point but rather a focal plane — at the image sensor. The point I'm talking about is (as I look this up quickly) a "nodal point". For the purpose of the simple drawing, I'm imagining a pinhole camera, where the "lens" actually is an aperture.
    – mattdm
    Feb 23 '11 at 1:51
  • @ysap: I've removed "aperture", since that could be misleading. Open to more suggestions on making it better. Thanks.
    – mattdm
    Feb 23 '11 at 1:58

Photo.net have a review of the Tamron SP AF200-500MM F/5-6.3 Di LD (IF) which handily covers the zoom range you discuss. Sample images are reproduced below available by following the link.

  • How are these images licensed?
    – Evan Krall
    Mar 30 '11 at 5:05
  • you'd have to contact photo.net. I assume they made them, so they own the copyright on them.
    – jwenting
    Mar 30 '11 at 6:05
  • It says "© Copyright 2004 Bob Atkins (www.bobatkins.com)" under the review
    – Imre
    Mar 30 '11 at 6:06
  • Does this mean they need to be removed? Sorry, I'm not familiar with the photo.se policy on such things.
    – fmark
    Mar 30 '11 at 6:52
  • The "Legal" page says: "You agree that all Subscriber Content that You contribute to the Network will be licensed under the Creative Commons Attribution Share Alike license." Since you don't have the copyrights for images, you may not give out licenses for them. As I understand, you should either ask Bob Atkins to grant the license, upload your own images, or remove the images and be satisfied with the link to photo.net review.
    – Imre
    Mar 30 '11 at 7:36

This wont be an accurate visualization as it only crops the image. It doesn't show the perspective difference between lenses (WA distortion, image compression,...). I don't think any consumer software could do that. Maybe NASA has something.

  • 4
    "Telephoto compression" is merely a function of subject distance, not something that is inherent in the lens. As such, cropping will work as-is, out of the box. The corner stretching of rectilinear wide-angles is a lens effect, but should not be an issue for this application.
    – Staale S
    Feb 22 '11 at 17:28
  • There are plenty of consumer programs that can counteract or create wide-angle distortion.
    – Evan Krall
    Feb 23 '11 at 4:52
  • 1
    I think not. Take your favorite zoom lens. Take a shot of someone at it's widest setting, then zoom out and move so that the person fills the frame the same as they did on the previous shot. Is it the same photo? Not hardly. This is the perspective difference that mearly cropping a photo doesn't achieve.
    – Al Graham
    Feb 23 '11 at 15:51
  • Upon thinking on this a bit more, I suspect we're comparing apples and oranges. I'm talking about taking the same shot of a subject at different focal lengths, whereas you seem to be talking about zooming in on a subject (not the same shot).
    – Al Graham
    Feb 23 '11 at 16:06
  • 1
    @Al Graham - Perspective is controlled by one, and only one factor. This is the geometrical relationship between the photographer and the scene. In simple terms, it is all about where the camera is positioned in the scene (I suspect you know and agree with that from your 2nd comment). Any difference in the image of an object taken with different focal lengths is purely due to different lens distortion at the different lengths. In your 1st comment, you emphasized the "fills the frame" for the 2nd shot. You may want to emphasize that for the 1st shot too.
    – ysap
    Feb 24 '11 at 8:12

Tamron has made a focal length comparision page.

Or, you can crop your own images made with a shorter focal length to see how a picture made with a longer lens would look like. You should crop both height and width by the multiplier the other focal length is longer. For example, in order to preview how a picture made with 500mm would look like, you have to divide width and height of an image taken with 200mm by 500/200=2.5 and crop an area with calculated dimensions, preferrably from center of the original image to avoid "shift lens" effect. The resulting image shows you the field of view the 500mm would have. In a cropped image, the effective f-number is also multiplied the crop factor, so if you started with an image taken at 200mm f/4, depth of field in the preview will be similar to 500mm f/10.

  • Minolta used to include such a thing in their lens catalogue. A single photo taken at each focal length between their smallest and largest lens, from the same position and in the same light.
    – jwenting
    Mar 30 '11 at 6:06
  • I remember that, going back to the '70s. Mount Rushmore was what they used then, starting with a circular fisheye that sort of let you know you were in a park of some sort with a bit of a hill at the end of a path, and ending with the 1600mm f/16 reflex lens looking at a part of Teddy Roosevelt's eyeglass frame. It really put things in perspective.
    – user2719
    Mar 31 '11 at 21:14

Canon also has a visualization tool - http://www.usa.canon.com/app/html/EFLenses101/focal_length.html. This one is a step visualization tool unlike the continuous ones provided by Nikon & Tamron. It covers a longer range (15-1200mm) than the other two though, while Nikon has support for different body & lens types.

  • I compared the angles with Nikon's tool mentioned by @Genius; it seems that Canon's comparison is based on full frame.
    – Imre
    Apr 1 '11 at 17:51

For a quick and dirty approach, back in the film days, you could hold an empty slide mount at the focal length from your eye.

Since a lot of fixed lens cameras quote focal length as the 35mm equivalent, that should still work today.

If you can't find a slide mount, you can always cut a 24 x 36mm hole out of a piece of card - or whatever size your sensor is (if working with a lens that quotes the actual focal length - or a suitable scaled 24x36 hole if working with 35mm equivalent focal lengths and a sensor with a different aspect ratio.

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