In a rifle scope, why is the reticle at the focal point?

• I

In order for a conventional telescopic riflescope to function, it must invert the image two times.
As a result, the light-rays traveling through the tube must meet at two focal points.

With my limited knowledge of optics, i assume that parallell light going in through the objective will meet at a focal point (or two in this case).

As an example, if you point the scope at the sun, it should collect light at the focal points; And you could theoretically set something on fire by placing it at the focal point. (Please correct me if I am wrong!)

So the way I see it, if you place even the smallest reticle in this focal point, it will block most of the light of an image coming in. Also the size of the reticle should be magnified greatly.
Obviously I have misunderstood something. But what?

Also, why do they call it a "Focal Plane", and not a focal point?

Attachments

• Telescopic_sight_internals-600x195.png
17.2 KB · Views: 1,654

Homework Helper
Obviously I have misunderstood something. But what?
Only light coming in precisely parallel to the scope will land on the focal point. Light arriving from slightly off axis will land at an offset from the focal point.
Also, why do they call it a "Focal Plane", and not a focal point?
Because it is where the image of an object at infinity will form. Every point on the plane corresponds to an angle of incidence. More generally, for an object not at infinity, every point on the plane corresponds to a point on the object being viewed.

Only light coming in precisely parallel to the scope will land on the focal point. Light arriving from slightly off axis will land at an offset from the focal point.

Because it is where the image of an object at infinity will form. Every point on the plane corresponds to an angle of incidence. More generally, for an object not at infinity, every point on the plane corresponds to a point on the object being viewed.

Thank you!

I guess the lines in the picture are purely representational then.

Gold Member
I guess the lines in the picture are purely representational then.
There can be a whole family of lines from points on the object to different places on the focal plane. The reticle only blocks out parts of the image that it lays over.
The reticle has to be in the same plane as the intermediate image so that the eyepiece produces the image of the target for your eye to see at the same place as the reticle. We have all had the experience of a fuzzy reticle and a sharp target image and vice versa. The focal 'point' refers to a single point on the prime axis of the lens which is in the focal plane.

As an example, if you point the scope at the sun, it should collect light at the focal points; And you could theoretically set something on fire by placing it at the focal point. (Please correct me if I am wrong!)
A very bad idea to look at the Sun through almost any optics (except a solar scope or a regular scope with a proper Objective Filter. Lenses are fairly tolerant of high energy flux passing theory them but the reticle will absorb energy from the Sun, focussed on it and fry. It is vital that the Sun never gets a look in at the front end of any scope (even by chance if you leave the scope pointing at its future path in the day) It can score a burn down inside the tube when it is off axis even.
I guess the lines in the picture are purely representational then.
Yes. Showing all the possible rays would just be an unintelligible mush. Google 'Image formation convex lens' for a host of diagrams that will now make sense to you.

Google 'Image formation convex lens' for a host of diagrams that will now make sense to you.

Thank you. It makes slightly more sense now and you have pointed me in the right direction.

But I googled that and an interresting dillemma came up. See this video (Just turn the sound off and think of it as a series of powerpoint slides):

Evidently, the "image" will appear at different distances from the lens, based on distance to target. Is this adjusted for by moving the reticle inside the tube? I mean, the "focus" of the scope is usually at the ocular (the back) right? (and so cannot possibly affect distance betweeen objective and reticle).

Homework Helper
Gold Member
... Evidently, the "image" will appear at different distances from the lens, based on distance to target. Is this adjusted for by moving the reticle inside the tube? I mean, the "focus" of the scope is usually at the ocular (the back) right? (and so cannot possibly affect distance betweeen objective and reticle).
But the diffierence in distance from the lens is very small. For a 5cm focal length lens, an object at 10m distance forms an image only 0.01mm away from the focal plane where an image at infinity would form. You would hardly use a scope sight for such a short range and that difference in focal distance is negligible, less than the thickness of a wire graticule, or maybe comparable, if the line is etched on glass, but I've not seen that.

But the diffierence in distance from the lens is very small. For a 5cm focal length lens, an object at 10m distance forms an image only 0.01mm away from the focal plane where an image at infinity would form. You would hardly use a scope sight for such a short range and that difference in focal distance is negligible, less than the thickness of a wire graticule, or maybe comparable, if the line is etched on glass, but I've not seen that.

Thank you for the valuable insight.
How far away from the focal plane must the reticle be before I can notice that it becomes "fuzzy"?

Homework Helper
Gold Member
Depends on the focal length of the eyepiece.
If the eypiece is say a 1 cm focal length, the an object at 1.00 cm will appear at infinity and an object at 1.04 cm will appear at 25 cm (approx my near point.) So rather less depth of field than I'd thought, but enough to allow objects from 7 m to ∞ to stay in focus, if your eye can accommodate from 25 cm to ∞.

Image at 25 cm (1 cm Eyepiece ) 1.04 cm ( intermediate image ) 5 cm ( 5cm objective ) object at ∞
Image at ∞ (1 cm Eyepiece ) 1 cm ( intermediate image ) 5.04 cm ( 5cm objective ) object at 640 cm

E&OE, it's getting late here.

Depends on the focal length of the eyepiece.
If the eypiece is say a 1 cm focal length, the an object at 1.00 cm will appear at infinity and an object at 1.04 cm will appear at 25 cm (approx my near point.) So rather less depth of field than I'd thought, but enough to allow objects from 7 m to ∞ to stay in focus, if your eye can accommodate from 25 cm to ∞.

Image at 25 cm (1 cm Eyepiece ) 1.04 cm ( intermediate image ) 5 cm ( 5cm objective ) object at ∞
Image at ∞ (1 cm Eyepiece ) 1 cm ( intermediate image ) 5.04 cm ( 5cm objective ) object at 640 cm

E&OE, it's getting late here.

That makes sense, but have you factored in that both the reticle and the "target" should be in focus at the same time?

Homework Helper
Gold Member
No I haven't.
It did occur to me when reviewing the post, but I couldn't think of a quick fix, so didn't mention it!

It's something I haven't thought about before. Although I talk of people accommodating 25 cm to ∞, that involves refocusing the eye. I don't know how much depth of field the eye has to allow images at varying distances to appear in focus together. I expect this has been studied, so I may look into it when I've some time.

I'd guess the graticule is placed at the first image plane and at the focal length of the objective, so that for all reasonable objects the image and graticule are pretty much coincident. Then the user just adjusts the eyepiece to get the graticule comfortably in focus and accepts the image of the target as it comes.