However, there's people who prefer carrying more lenses of different focal lengths and, possibly, better optical quality. I'm one of them. I'm not implying that the 18-200mm is a bad lens: it isn't. I simply made other choices.
Be it as it may, the point of this post is the following: you're framing a picture with your camera and you realize you need a longer lens to get the shot you've got in mind. The question is: Which focal length do I need?
More often than not, you're not carrying many lenses and the answer may be simple: the one you've got. Sometimes the estimate is easy, sometimes it is not. Imagine you're framing the shot with a 35mm and you deem it insufficient. Would a 100mm do the job? Or a 200mm? And what about a 300mm?
Other times you cannot even decide by trial and error: you haven't got any more lenses and you want to buy one. But you want to be sure beforehand.
You've probably experienced such a doubt. A common situation amongst novices with a telephoto lens is shooting the moon. You take your brand new 200mm lens (or whatever) and you frame the moon only to realize that your lens is way too short to get the amazing picture of the moon you wanted to take.
What focal length do I need?
TradeoffsBefore getting deeper into the subject, let me tell you about some tradeoffs. They're obvious, but many people seem to miss them anyway:
- Zoom with your feet.
Zoom With Your FeetThe first advice I can give you is: zoom with your feet. This is the best tradeoff since it's got many advantages. Focal length is just one of the parameters and, sometimes, long lenses may prevent you from doing what a good photographer should do: walk around your subject, look for a good or an unusual point of view, change perspective and so on. The lens is a tool, it's up to you deciding how to use it.
There are times, however, when you can't get closer to your subject. There's no way to get closer to the moon on foot. There are ways to get closer to a leopard in the wild, but you aren't going to do it either. Other times you don't want a bird to see you and fly away. In this case, you do need a longer lens.
How long? Read on.
CropNobody likes cropping: you've spent money on a good sensor and you want to squeeze any possible detail from it. I agree. However, it's better to crop a shot rather than lose it anyway. How much crop you can afford depends on what's the purpose of the image. If you just need a decent print, you can calculate beforehand how much crop you're willing to trade in for a better composition.
A 16 megapixel sensor, for example, gives an image almost 5000 pixels wide and more than 3000 pixel high (landscape frame). If you're going to print such image at 300 dpi, you're going to get approximately a 16" x 10" picture (42 cm x 27 cm). It's big, isn't it? If you crop it to half of its size, you're left with enough resolution to get a 8" x 5" picture (21 cm x 13.5 cm). Big enough for many uses.
Obviously, details aren't going to be as sharp as on the original image but it can be enough.
Look at this picture of the moon, for example:
|200mm - Cropped to 2500x1600 pixels|
It was taken using a 200mm lens and cropped down to approximately 2500x1600 pixels (scaled down further for inclusion in this blog).
It's not state of the art but still, some details of the craters are still visible.
However, this image, as whichever image you see in your viewfinder, can help you estimate the focal length you need to get the picture you'd like.
Angle of ViewHere's some theory to answer this question. The angle of view can be calculated easily using the sensor (or film) size, the focal length of the lens and the distance to the subject. Since we're talking about telephoto lenses, we'll use a simpler model that simplifies a bit the maths and the resulting formula assuming that you're focusing at infinity (or sufficiently far away). This assumption holds in this case: we're talking about getting closer, hence it's safe to assume that we're pretty far away.
The maths tells us that the angle of view a of a given lens/sensor pair is:
where d is the sensor size and f is the focal length. The angle you're measuring (horizontal, vertical or diagonal) depends on how you measure the sensor size (horizontally, vertically or diagonally). As far as it concerns our estimation and the rule of thumb, it doesn't really matter.
Estimating the Required Focal LengthThe previous formula tells us everything, doesn't it? Probably no, in fact, I admit it. We're photographers, not mathematicians. But a very useful rule of thumb comes directly from the properties of the previous equation. Let's plot it from 18mm to 600mm, using a sensor size of 23mm (the sensor size, as I told you, doesn't matter very much since it's effect is just stretching or expanding the function graph):
|Angle of view from 18mm to 600mm|
Let's ignore the y dimension and just focus on its shape. What can we infer? The shortest the focal length you're using (the closer you are to the origin), the steeper the curve. This is something you probably have observed when you zoom in with you telephoto lens: at the beginning, the effect of getting closer is more pronounced, and it slows down as you open your lens. Look how the angle gets smaller and smaller when zooming further at bigger focal length (arrows indicate the size of the angle of view reduction in focal length increments of 100mm, except the first one that is calculated from f=50mm approximately):
|d(Angle of View)/d(focal length)|
Ok, we knew that. What's next? This observation: look at the angle of view delta when scaling the focal length with a fixed multiplier. Let's use 2: the difference between 50, 100, 200 and 400 is very similar. Why? Once more, is a mathematical property of the arctan function. The plot of the arctan function is the following:
|arctan(x), with x in [-2,2]|
In the angle of view equation, f it's in the denominator: this means that when f increases, the arctan argument actually decreases. Look at the plot: if you consider an interval sufficiently near the origin, it very much resembles a line.
That's the trick in our rule of thumb! We're going to approximate the arctan function near the origin with the line y = x. Can we do it? Sure. Take a sensor size of d = 23mm. We're not interested in wider lenses, but even if we were, d/(2f) with f = 18mm gives 0.6. Pretty much into the linear zone of the function anyway. Here's the rule:
We can approximately affirm that, for sufficiently long lenses, multiplying the focal length by a factor of c reduces the angle of view by the same factor.
This is very handy, since you need to take no measurement. Just look into the viewfinder: want a moon twice as big? You need a focal length twice as big.
An ExampleTake a look at the previous picture of the moon:
|200mm - Cropped to 2500x1600 pixels (half the original width)|
The moon size is approximately a quarter of the picture height. If I wanted to fill the frame, our rule of thumb tells us that I'd need approximately a 200mm · 4 = 800mm lens.
But the previous image is a crop: the original image is twice the size. That gives 1600mm, if I wanted to fill that frame.
Big bucks for the moon picture.
Next time you plan to buy a longer lens, frame a picture with your telephoto and use this rule.
P.S.: Plots were generated using the excellent Wolfram Alpha web application.