# What is the actual significance of the focal point of a lens?

I know how to find the focal length using the lens formulas, and I know where the focal point is on a diagram.

I've read a few definitions for the focal point such as (Cbakken.net)

Concave lens

Converging light rays striking a concave lens but headed towards a point on the other side can be bent until they emerge parallel to the axis. The point that causes this to happen is called the focal point.

and

The focal point of a concave lens is the point where light rays parallel to the axis seem to diverge from after passing through the lens. The distance from the lens to this point is called the focal length of the lens.

Convex lens:

The focal point of a convex lens is the point where light rays parallel to the axis are brought to a point. The distance from the lens to this point is called the focal length of the lens. and

Diverging light rays striking a convex lens can be bent until they emerge parallel to the axis. The point where this happens is called the focal point.

But what hasn't clicked is why is the focal length significant or useful? If I had a lens, why would I care where the focal point is? Is it simply the point where the imagine (like in a telescope) goes into focus and becomes sharp?

Also, for a convex lens. If I have my eye at the focal point, would the image appear tiny? Since all the light rays converge there?

• But what hasn't clicked is why is the focal length significant or useful? If I had a lens, why would I care where the focal point is Well, you would care if, to take a silly example, the focal length of your camera or binoculars, or reading glasses, was twice the normal distance. Also, the longer the focal length, the narrower the angle of view and the higher the magnification. – user108787 Oct 3 '16 at 22:13
• Thanks for the response @CountTo10. Sorry to be a pain by what effect would it have if the binoculars had a focal length twice the normal distance? Does that mean I would have to have binoculars twice as long to get a clear image? – K-Feldspar Oct 3 '16 at 22:33
• Yep, roughly speaking. here's a list of focal lengths and their uses.... less than 21 mm =Extreme Wide Angle 21-35 mm=Wide Angle (Landscape) 35-70 mm =Normal (Street & Documentary) 70-135 mm =Medium Telephoto (Portraiture) 135-300+ mm= Telephoto. I am no expert, and there's a lot more to it, but I am pretty sure that is the basic idea. Look at the focal length of any refractor telescope and how long the tube is. techradar.com/how-to/photography-video-capture/cameras/… – user108787 Oct 3 '16 at 22:54
• As a kid didn't you ever burn ants (or wood) with a magnifying glass on a sunny day. Your target has to be at the focal point. – M. Enns Oct 4 '16 at 1:33

The focal point is where the light from a parallel bundle of rays comes to focus, this location is at one focal length from the lens. This focal length $f$ is important relative to other properties of a lens.

1. It determines the image location using the equation $$\frac{n'}{z'} = \frac{z}{n} + \frac{1}{f}$$ where $z'$ is the distance from the lens to the image, $z$ is the distance from the lens to the object plane (and is typically a negative quantity if the object is to the left of the lens), $f$ is the focal length of a lens. Therefore if the object is really far away (more than several focal lengths), the image will be where the focal point is. $$z' \approx f$$ Think about this because many times the object is far enough away that this approximation is valid. In fact the muscles in your eyeballs which control focus are most relaxed with viewing things that are far away.

2. The image location determines your image height $h'$. The farther your image from the lens, then the larger your image will be. So the combination of $z$, $z'$, and $f$ directly controls yours image height.

To answer your last question: For objects that are small and far away the rays will be approximately collimated (relative to you eyeball), so yes it will be difficult to see them because they are tiny compared to the rods and cones (the pixels of your visual system). But objects that are not small and not far away will have not have a collimate bundle of rays, and therefore a finite image size.

1. Given the focal length $f$ and the diameter of the lens $d$ you can calculate the smallest point that is resolvable. In short the shorter the focal length relative and larger the diameter of the lens the better the detail in the object will show up in your image. This is from the well-known Rayleigh criterion $$\Delta_x = \frac{0.61 \lambda}{ n \sin(\theta)}$$ Where $\Delta_x$ is the smallest distances where the image of two object points will have and still be distinguishable and $\theta$ is half-angle the lens diameter makes with the optical axis as seen from the image and is proportional to the ratio $f/d$.

There are many other reasons why focal length is important. The short answer is: focal length combined with other properties of your lens tells you everything you need to know about image distance, magnification, how much detail you can see (aka resolution), etc...

The focal length tell us how much the light rays will be bent. Think of the light rays as a paper cone, just like the one you get when you buy a snow cone. The mouth of the cone is the size of the lens, and the point of the cone is the focal point. The length of the cone is the focal length. Now picture a cone where the point is very close to the mouth. It would be a very steep short cone. The light rays would come in at very steep angles, and after they crossed the focal point they would spread out quickly. In fact, they would make another cone leaving the focal point that matches the cone that entered the focal point. The rays would continue on spreading out wider and wider. Now picture a cone that is very long, say three feet, a novelty snow cone, notice how these rays come into focus at a much smaller angle, and after they pass the focal point they will form another long cone on there way out. Well, there you have it, the focal length tells us how long the snow cone is. I hope your not offended by my simple examples and language in this answer. It's how I think of focal length. With a little imagination you can see how different cone sizes would be applicable to different applications.

• Not at all offended. Simple examples and analogies like that are wonderful for someone without a strong physics background like myself :) – K-Feldspar Oct 5 '16 at 22:04

Look at the image above where the three laws of optics are shown:

1. Beam passing through the optical centre passes untouched.
2. Beam passing parallel to the optical axis leaves through (projective) focus.
3. Beam passing through (objective) focus leaves parallel to the optical axis.

Using these three laws you can construct how an object will be projected.

It is easy to see that the focal length $f$ and the object position $S_1$ describes:

1. Magnification $h_2/h_1$,
2. Position of the projection $S_2$.

When thinking about the effect of lenses used in binoculars, one must take eye lens into account. And here it is quite tricky.

• I edited in the photo to your post, because people sometimes don't like to go off site, (they lose track of your points), and also if the linked page is taken down, your image is lost. – user108787 Oct 3 '16 at 23:10