Why are objects far away inverted through a lens but not through the viewfinder?

Question

When I look through a lens the image of objects far away are inverted but when looking the the viewfinder on my camera they are not. Why is this?

I am having a hard time understanding why objects far away are inverted in the first place.

Could anyone provide an explanation or ray diagrams (Preferably using a point source on an object and including the lens in the human eye)?

EDIT: Thanks everyone I now understand why objects far away from a lens appear inverted. But can anyone now explain how the camera elements make far inverted objects appear right way up without also making close normal objects appear upside down?

EDIT 2: I can't provide an image right now because I am at school but you know how when you look through a magnifying glass and far objects will be inverted and blurry but close objects will be sharp and erect (normal)?

That is what is happening when I look through my camera lenses while they are not attached to the camera, but when they are attached to the camera and I look through the viewfinder (or at processed film) the objects in the image produced are all of the same orientation.

Does this mean that lens doesn't actually produce images like a magnifying glass would because the objects on the images produced on film are all of the same orientation? Or does this mean a magnifying glass doesn't actually produce objects with different orientations?? If a magnifying glass doesn't, then why does it look like it does and are the convex lens diagrams wrong (they show a virtual image upright for close objects and real upside-down images for far objects)? Isn't a magnifying glass just a convex lens?

It DOES look like a magnifying glass when I look through the lens. That's why I thought that then lens was producing objects with different orientations. This also goes with the convex lens diagrams below that show objects with different orientations.

So does the lens produce objects with different orientations or doesn't it??? If not why does it look like it does when I look through the lens, and also based on the convex lens diagrams it seems like it should. If it doesn't then how do the other lenses in a camera lens attachment correct the convex lens. And if it does then why does film and the viewfinder show objects with the same orientation?

Sorry for asking so much. This is just so confusing!

EDIT 3: This is how I thought a camera lens would work:

I forgot to mention in EDIT 2 that it seems that close objects shouldn't even appear on film based on the diagrams.

I still don't understand... =(

EDIT 4: So objects really close to the camera lens should not appear on film, correct?

So...Why do all objects in the viewfinder appear upright??? Since my eye is reviving both the light rays from close objects (virtual upright images) and far objects (real inverted images) shouldn't really close objects and objects farther away have different orientations? Just like looking through the lens directly? How does the viewfinder change anything?

EDIT 5: Thanks so much everyone. Thanks for the help.

"Anything close enough to form a virtual image is not focused onto the focusing screen"

So lets say I put a pen right in front of the lens and look through it directly. The image I see is upright so this means it is a virtual image. Now lets say I attach the lens to the camera and look through the viewfinder. I can still see the pen but it is blurry (because the focal length is longer, right?). The lens forms a virtual image of the pen but I can still see it in the viewfinder. Why is this? If the viewfinder shows me exactly what would be on the film it should not show the pen at all (based on the diagrams in the image above) should it?

EDIT 6: Maybe it should form a blurry image. Like a pin hole camera or something. In any case thanks for all the help everyone. I know it can be frustrating trying to teach me. I can be pretty dense sometimes.

Answer 1

It is "easy enough" to explain the basic question with a ray diagram or similar means - see below
BUT it is important to realize that the answer to why viewfinder or human eye images are not inverted is "by design" or "because" (choose one, both the same essentially). That is to say, the system requires the outcome to be a certain way, so whatever steps are required to implement the outcome are provided.

In the case of a viewfinder, extra lenses, mirrors or prisms (or a combination of these) are added as required to achieve the final result. The real question becomes not "why is this this way up" but how is this done.

In the case of the human eye, the image on the Retina IS inverted and the brain looks at it the "right way up" as far as the viewer is concerned.

The information below from this excellent site shows how basic inversion works.

Also see -> More on Ray Diagrams

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In the case of the eye, the image is inverted:{ From here - low tech but interesting }

http://www.quantumtheatre.co.uk/Lights%20&%20Sounds%20notes%20Key%20Stage%202_files/image022.jpg

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IMPORTANT:

Note that while the above image grabs your attention as it demonstrates inversion, it actually does a very bad job of showing how the eye lens works. As the eye lens is increasingly embedded in the cornea, the air-cornea interaction does most of the 'lensing' while the cornea-lens interface only manages about 10% of the total bending.

An excellent discussion of this is available here - see the In your eye and a reasonably correct image of how light is actually bent by the eye is shown below.

Answer 2

This link provides a good (sometime complex) answer to your question.

In short:

• One normal lens element magnifies but is limited to a certain magnification factor
• Complex Lens-combinations can get more magnification and may invert the picture due to the different attributes of the lenses and their use in the objective. (You might catch the beam of a lens behind its focal point and refocus it with another lens-> inverts the image)
• A pentaprism inverts the image when it is sent to the viewfinder, so you have the final "look" of the image and can work with it.

Answer 3

"Anti-focusing" of objects closer than 1 focal length to lens:

• This addresses one of the questions which arose during your "Edits" but is not implied in your subject line.

The Question: This is a precis of your question - all text is yours.

• I forgot to mention in EDIT 2 that it seems that close objects shouldn't even appear on film based on the diagrams.

• EDIT 4: So objects really close to the camera lens should not appear on film, correct?

• "Anything close enough to form a virtual image is not focused onto the focusing screen"

• So lets say I put a pen right in front of the lens and look through it directly. The image I see is upright so this means it is a virtual image. Now lets say I attach the lens to the camera and look through the viewfinder. I can still see the pen but it is blurry (because the focal length is longer, right?). The lens forms a virtual image of the pen but I can still see it in the viewfinder. Why is this? If the viewfinder shows me exactly what would be on the film it should not show the pen at all (based on the diagrams in the image above) should it?

• EDIT 6: Maybe it should form a blurry image. Like a pin hole camera or something.

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What you are describing is exactly what happens, but because the defocusing of objects closer than a focal length from the lens is progressive as distance inside the focal point increases - just as your diagram suggests - they do not just "vanish" as they come inside the critical distance - rather they become progressively more indistinct the closer they get to the lens face.

The pictures below show reasonably extreme examples of this 'feature' being used to good effect to nearly completely remove closeground items from the photo - in this case vertical bars and a reasonably heavy mesh are nicely "vanished by being defocused and spread so widely as not to be noticed.

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Foreground objects (in this case a heavy mesh and cage bars) which are closer to the lens than its focal length are "anti-focused" to the point of near invisibility.

Your diagram 3 with cage bars added:

This is one of my standard "tricks" for photographing objects in cages and similar environments where there is an incomplete obscuring layer that you can get right up against. An extremely useful "trick".

In this photo there are cage bars very close to the front element of the lens - as close as I could get them. I use this method to sucessfully "drop out" even quite solid bars. In this case it's normal thickness cage bars. Distance to front element is under 50mm and it's a 50 mm f1.8 lens. There ARE some optical effects present but they are not normally noticed by most viewers. Higher res version of this is here and click download icon 2nd from right at top of photo. This gives a much better look at what you CAN'T see.

CAGE BARS BETWEEN BIRD & VIEWER

This is an even better example, in that there is a small pitch very thick square mesh between the camera and the subject (I think not more than 20mm squares - I can check other photos). This was using an 18-250 lens at 18mm, f6.3 * See photos showing mesh that was present in 2nd photo below. Visually the mesh ruins the bird's presentation and the camera "sees" the bird far better than the eye can.
Same photo on facebook here

VERY THICK & UGLY SQUARE MESH BETWEEN BIRD & VIEWER

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(*) I originally said that this was taken with a 50mm f1.8 lens but after checking the original I have changed the details, as above.

Answer 4

If a converging lens has focal length f, a subject which is position p relative to the lens will generate an image at location q=f/(f/p-1) [the fundamental equation is f/p+f/q=-1]; the ratio of image sizes will be p:q. When p and q have the same sign, the image will be on the same side of the lens as the object, and the size ratio will be positive. When p and q have opposite sign, the image will be on the opposite side of the lens, the size ratio will be negative (implying an inverted image).

Note also that if an image formed by one lens is used as the "subject" for a second, that second lens won't won't "care" on which side of the first lens the image appears, nor even which side of the second lens the image appears; the same position and size formula will apply. The distinction between virtual and real images is only relevant when trying to place a target (like a sheet of film) on the focal plane, and can most simply be expressed by observing that lenses can't do anything if they're not located between the real subject and the intended focal target; if the final lens would present a virtual image, that would imply the target would have to be between the real subject and the final lens, rendering that final lens irrelevant.

A telescopes or other similar instruments will use a lens or sequence of lenses to focus an image, then use another lens or sequence of lenses which is "looking" at that image to focus another image, etc. The first lens will produce an image which is at least a focal length away from it. In a telescope, the second lens is placed such that the image will always be on the same side as the viewer. In such a situation, the second lens will focus an image to be less than a focal length away. Subjects which were infinitesimally far from the first lens will be focused almost infinitely far behind the second, which makes the f/p term approach zero, thus yielding an image one focal length behind the second. Subjects which are infinitely far from the first lens will be focused a discrete distance behind the second, and will yield an image whose distance from the lens is even shorter. The net effect is that regardless of the location of the original image, the second lens will produce an image which is between zero and one focal lengths away. Since the first lens produced an image on the side opposite the original subject, it will invert the image; since the second lens produced an image on the same side as its "subject" [the subject was an image located on the same side as the viewer) it will not invert the image.

Many kinds of telescopic apparatus intended for eye viewing add a third lens which is located such that the image from the second lens will always be significantly more than a focal length in front of it. This lens will thus refocus the image formed by the first two lenses to form a second image which is on the opposite side of e third lens from the second. Because that image and its subject will be on opposite sides of the third lens, the third lens will cause a second inversion, thus flipping the image upright.

With regard to the original question, the reason that a telescopic viewfinder always shows objects upright is that while excessively-close subjects may cause the primary lens to generate an image which is nearly infinitely far past the second lens, the second lens always produces an image which is between zero and one focal lengths past it, such that the final viewing lens will never see an object so close as to change its inverting behavior.