OVF eyesight correction

[Jeff S]

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7 minutes ago, Rick said:

I don't believe the -0.5 diopter offset in the viewfinder has any effect on the  patch/lines.  The patchiness are reflected off the prism mirror in an other optical path, before the -0.5 diopter lens correction for the viewfinder.

The -0.5 correction seems to be incorporated in the viewfinder for other optical reasons.

So, you and jdlaing agree, while 01af and Jaap do not.  Forum life.

Jeff

[01af]

vor 30 Minuten schrieb Rick:

I don't believe the -0.5 diopter offset in the viewfinder has any effect on the  patch/lines.

Okay.

Can you read text in an 1 inch distance from your eye? No, you cannot. Then why can you clearly see framelines, LED digits, and a focus patch that are physically located approx. 1 inch from your eye?

The -0.5 dpt offset is what creates the virtual distance between those tiny viewfinder elements inside the rangefinder and your eye. It must not be confused with the additional (or subtractional) offset provided by correction lenses screwed into the viewfinder's eyepiece.

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[Rick]

6 minutes ago, Jeff S said:

So, you and jdlaing agree, while 01af and Jaap do not.  Forum life.

Jeff

No, I didn't say I agreed with anyone.  I said the -0.5 is not in the optical path of the path of the patch/lines and can therefore have no effect on forming a virtual image.  

You don't understand the optics of the rangefinder or don't seem to understand how a virtual image is formed.

Ask yourself this:  Why does the virtual image not change when you place a +0.5 diopter in the eyepiece?  It only changes the vergence(divergence) of the rays exiting the inner plus lens in the viewfinder.  If, the observer is young enough it is performed by the accommodation of the eye.

The -0.5 d seems have other optical design purposes in the rangefinder.

 

(Also, your last post adds nothing to the optical discussion, it is simply Forum Snark).  

[Rick]

 

1 hour ago, 01af said:

Okay.

Can you read text in an 1 inch distance from your eye? No, you cannot. Then why can you clearly see framelines, LED digits, and a focus patch that are physically located approx. 1 inch from your eye?

Great question.  In fact, I was thinking/hoping specifically you would reply.  🙂

The short answer is that the Frame lines and LEDs and Patch (FLP) are in a different optical path than the viewfinder.  The FLP is reflected by the prism mirror. The prism is only partially in the viewfinder optical path as the prism is in the middle of the viewfinder path.  More specifically, it is in the middle of a Reverse Galilean Telescope!  (more on  RGT later on)

The optical path of the FLP is such that it produces parallel rays out of the eyepiece. For an emmetropic eye the FLP will be in focus.

When you look into the eyepiece you are "seeing" the image of the FLP reflected by the prism mirror.  Big hint in that last sentence of how the virtual image is formed. 

The optics of the FLP are designed to produce parallel rays out the eyepiece.  An emmetropic is able to see the FLP clearly.  It has nothing to do with -0.5d as that is not in the FLP pathway.

 

1 hour ago, 01af said:

The -0.5 dpt offset is what creates the virtual distance between those tiny viewfinder elements inside the rangefinder and your eye. It must not be confused with the additional (or subtractional) offset provided by correction lenses screwed into the viewfinder's eyepiece.

I believe it is the optic design of the FLP optical path not the -0.5d that creates the virtual image.  The -0.5d does not to appear to me to be in the FLP optical path.  It is probably created in the RGT at the minus objective, possibly the eyepiece plus lens or likely just shortening the distance between the objective and eyepiece... which would make the viewfinder more compact.  

Edited January 4, 2023 by Rick

[Rick]

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Edited January 4, 2023 by Rick
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[Jeff S]

1 hour ago, Rick said:

No, I didn't say I agreed with anyone.  I said the -0.5 is not in the optical path of the path of the patch/lines and can therefore have no effect on forming a virtual image.  

You don't understand the optics of the rangefinder or don't seem to understand how a virtual image is formed.

Ask yourself this:  Why does the virtual image not change when you place a +0.5 diopter in the eyepiece?  It only changes the vergence(divergence) of the rays exiting the inner plus lens in the viewfinder.  If, the observer is young enough it is performed by the accommodation of the eye.

The -0.5 d seems have other optical design purposes in the rangefinder.

 

(Also, your last post adds nothing to the optical discussion, it is simply Forum Snark).  

Absolutely no snark intended.  And I already wrote that I am neither an eye doc, nor a Leica expert.  That’s the exact reason I “invited” you, 01af and Jaap into this discussion via my mentions in post #14.  And as I wrote, “I’m happy to be corrected by any of these folks.”  So, thanks for weighing in.  I was hoping you would, and I appreciate it.  You’ve helped me understand a lot about Leica optical matters.
 

Forgive me if I jumped to the wrong conclusion about your agreement or disagreement with others; that’s merely the way I interpreted the respective posts.  And, yes, there are typically different views on forums like these. 
 

Jeff

[IkarusJohn]

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30 minutes ago, Rick said:

 

Great question.  In fact, I was thinking/hoping specifically you would reply.  🙂

The short answer is that the Frame lines and LEDs and Patch (FLP) are in a different optical path than the viewfinder.  The FLP is reflected by the prism mirror. The prism is only partially in the viewfinder optical path as the prism is in the middle of the viewfinder path.  More specifically, it is in the middle of a Reverse Galilean Telescope!  (more on  RGT later on)

The optical path of the FLP is such that it produces parallel rays out of the eyepiece. For an emmetropic eye the FLP will be in focus.

When you look into the eyepiece you are "seeing" the image of the FLP reflected by the prism mirror.  Big hint in that last sentence of how the virtual image is formed. 

The optics of the FLP are designed to produce parallel rays out the eyepiece.  An emmetropic is able to see the FLP clearly.  It has nothing to do with -0.5d as that is not in the FLP pathway.

 

I believe it is the optic design of the FLP optical path not the -0.5d that creates the virtual image.  The -0.5d does not to appear to me to be in the FLP optical path.  It is probably created in the RGT at the minus objective, possibly the eyepiece plus lens or likely just shortening the distance between the objective and eyepiece... which would make the viewfinder more compact.  

Thank you for the clarification, Rick. 

Can I revert to @jdlaing’s core point, before the micturating contest?  While technically interesting, for practical purposes the virtual 2m issue is largely irrelevant. 

Isn’t JD correct when he says the framelines are accurate only at 2m, but for actually focusing, the RF patch is accurate for the entire range of the lens?  The point, therefore, is when taking photos 2m is relevant only for frameline precise composition. 

I’ve never read anywhere that focusing is accurate only at 2m. I may be misinterpreting JD’s clarification, but that is what I understood him to say. 


 

[Rick]

3 minutes ago, IkarusJohn said:

Thank you for the clarification, Rick. 

Can I revert to @jdlaing’s core point, before the micturating contest?  While technically interesting, for practical purposes the virtual 2m issue is largely irrelevant. 

Correct.  And, I'm not even sure the distance we perceive the FLP is actually(optically) 2m.  It is simply a virtual image.

3 minutes ago, IkarusJohn said:

Isn’t JD correct when he says the framelines are accurate only at 2m, but for actually focusing, the RF patch is accurate for the entire range of the lens?  The point, therefore, is when taking photos 2m is relevant only for frameline precise composition. 

Correct.

3 minutes ago, IkarusJohn said:

I’ve never read anywhere that focusing is accurate only at 2m. I may be misinterpreting JD’s clarification, but that is what I understood him to say. 
 

Correct.  Focusing is accurate for the entire range 0.7 to infinity.  The frame line accuracy is most accurate and is set for 2m.

Frame line accuracy is best at 2m

Frame line virtual distance may appear to be at 2m, but I am not convinced for sure it is 2m.  I do know it is a virtual image by definition.  I highly suspect folks have come to believe it is 2m because that is the inverse f of a 0.5 diopter lens, and that is where that optical folklore started.  

But, the -0.5 diopter is not in the FLP pathway.  It can't play a role in determining the virtual distance of the FLP image, so the virtual distance could be virtually😁,  anything.

[Jeff S]

Rick…the only practical time I’ve considered the 2m issue, besides frame line accuracy (on models other than M8 or M9, which differ from 2m), has been when considering one’s proper diopter strength, or eyeglass prescription, for best M viewing/focusing. I had thought, possibly erroneously, that telling one’s optician that the display was set virtually to 2m was helpful, as was being able to see more distant subjects. In practice, however, I’ve let my eye doc figure out my normal distance correction, and have experimented with trial diopters over the eyepiece to see if they can additionally benefit, regardless of any calculations. 
 

What, if anything, can or should one say to their eye doc about the M viewing/focusing specs that might assist in determining best optical correction?

Jeff

[Rick]

6 minutes ago, Jeff S said:

Rick…the only practical time I’ve considered the 2m issue, besides frame line accuracy (on models other than M8 or M9, which differ from 2m), has been when considering one’s proper diopter strength, or eyeglass prescription, for best M viewing/focusing. I had thought, possibly erroneously, that telling one’s optician that the display was set virtually to 2m was helpful, as was being able to see more distant subjects. In practice, however, I’ve let my eye doc figure out my normal distance correction, and have experimented with trial diopters over the eyepiece to see if they can additionally benefit, regardless of any calculations. 
 

What, if anything, can or should one say to their eye doc about the M viewing/focusing specs that might assist in determining best optical correction?

Jeff

Your eye doctor would not be able to tell you the correct diopter without understanding the optics of the rangefinder.  You need trial lenses to be accurate.  

I can calculate it from your prescription but, it still may not be correct because your written prescription does not have to be correct, it is what the doctor finds to be the best correction for your visual system not just your optical system. I know it sounds crazy, but a lot of people are over corrected or under corrected or have under corrected astigmatism for reasons, and they don't know/understand that they are.

[Jeff S]

9 minutes ago, Rick said:

Your eye doctor would not be able to tell you the correct diopter without understanding the optics of the rangefinder.  You need trial lenses to be accurate.  

 

So it seems that there are no simple statements that a layperson can make about those rangefinder optics.  Seems also that the long-standing FAQ referencing the 2m virtual distance to the focus patch is wrong. Correct?

Jeff

Edited January 4, 2023 by Jeff S

[Rick]

21 minutes ago, Jeff S said:

So it seems that there are no simple statements that a layperson can make about those rangefinder optics. Seems also that the long-standing FAQ referencing the 2m virtual distance to the focus patch is wrong. Correct?

Jeff, the RF is a VERY complex optic.  There are actually 3 separate optical paths.  Heck, the optical path of the patch image is the third optical path which is different from the patch mask!  We haven't even touched on that.  So, no there is no simple statements.

I haven't read the FAQ in a long time, but again, as I said above, the 2m virtual distance may be correct but it has nothing to do with the -0.5d offset of the viewfinder image.  It could be anything. 

[Jeff S]

Thanks, Rick, the more I know, the less I know. Fortunately I can still see well enough to make pics with my M’s.🤓

Jeff

[Rick]

Here is an image that shows the optics of the RF.  The RF is an optic that is so impressively clever.  There are at least six different lenses, 2 prisms and a mirrored surface.  One element may be a compound lens?

The -0.5d correction, I measure to be in the viewfinder(blue) optical path.  The virtual image is formed in the FLP(red) optical path.

If, you read back through my previous posts you can reference the different optical paths I discuss in the image here.

Cheers.

 

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Edited January 5, 2023 by Rick

[jaapv]

On 1/4/2023 at 8:30 PM, Rick said:

Correct.  And, I'm not even sure the distance we perceive the FLP is actually(optically) 2m.  It is simply a virtual image.

Correct.

Correct.  Focusing is accurate for the entire range 0.7 to infinity.  The frame line accuracy is most accurate and is set for 2m.

Frame line accuracy is best at 2m

Frame line virtual distance may appear to be at 2m, but I am not convinced for sure it is 2m.  I do know it is a virtual image by definition.  I highly suspect folks have come to believe it is 2m because that is the inverse f of a 0.5 diopter lens, and that is where that optical folklore started.  

But, the -0.5 diopter is not in the FLP pathway.  It can't play a role in determining the virtual distance of the FLP image, so the virtual distance could be virtually😁,  anything.

The misconception here is that the frameline accuracy is somehow related to the optical properties of the viewfinder. The point is that the frameline accuracy means coincidence with the angle of view of the lens, which varies with the focus distance. Thus the frameline (size) accuracy is optimized for a lens focused at two meter. See Günther Osterloh, Advanced Photo School page 37 ff. Note that this is from the film ers, when the framelines were set for minimum focus distance.

 

 

images under Zitatrecht I'll rotate them tomorrow.

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[jaapv]

4 hours ago, Jeff S said:

So it seems that there are no simple statements that a layperson can make about those rangefinder optics.  Seems also that the long-standing FAQ referencing the 2m virtual distance to the focus patch is wrong. Correct?

Jeff

I don’t see a contradiction here; Rick tells us that an optometrist or ophthalmologist needs specialized knowledge of the RF to calculate the diopter which is true, and the FAQ states that it is most practical to use an empirical method. In fact, that is what Leica used in the past. They provided their dealers with a wheel containing all diopter strengths that went into the camera shoe and covered the ocular, so the customer could spin it and find the correct strength.
As for the focus patch, the point is that it is projected into the optical path of the Albada viewfinder by a semi-silvered  mirror. That makes the virtual image of the patch part of the optical system of the viewfinder. It would make little sense for Leica to create an RF viewfinder where either the framelines or the RF patch were unsharp. I think the explanation by 01AF that the purpose of the -0.5 diopter strength is to help the eye with normal vision accommodate from the overlay patch at 2m to the background at infinity ( this is a telescope system) which is obviously needed to obtain coincidence, is correct. 
I would add that it makes little sense for optical experts like Leica to arbitrarily introduce a diopter strength for no reason. 

[Rick]

32 minutes ago, jaapv said:

The misconception here is that the frameline accuracy is somehow related to the optical properties of the viewfinder. The point is that the frameline accuracy means coincidence with the angle of view of the lens, which varies with the focus distance. Thus the frameline (size) accuracy is optimized for a lens focused at two meter. See Günther Osterloh, Advanced Photo School page 37 ff. Note that this is from the film ers, when the framelines were set for minimum focus distance.

Correct.  

It is nice to see #15 labeled as an achromatic element.  As I described above I believed there must be a compound element as well as the simple lenses.  My image sort of shows it is a glued doublet.  I figured it was achromatic and maybe field flattening element.  Lot's chromatic aberration in that design of all those simple lenses.

Now, if you really want to go down the rabbit hole, there is another lens at #18.  The image it projects through the mask is NOT 0.5d vergence exiting the eyepiece.  It is plano(0.0 diopters).

[jaapv]

3 minutes ago, Rick said:

Correct.  

It is nice to see #15 labeled as an achromatic element.  As I described above I believed there must be a compound element as well as the simple lenses.  My image sort of shows it is a glued doublet.  I figured it was achromatic and maybe field flattening element.  Lot's chromatic aberration in that design of all those simple lenses.

Now, if you really want to go down the rabbit hole, there is another lens at #18.  The image it projects through the mask is NOT 0.5d vergence exiting the eyepiece.  It is plano(0.0 diopters).

I think that we can agree that this is a highly complicated system that lends itself to interesting discussion.  This is an older design ( M4 I believe) and Leica has refined it quite a bit since then. 

[Jeff S]

11 minutes ago, jaapv said:

I don’t see a contradiction here; Rick tells us that an optometrist or ophthalmologist needs specialized knowledge of the RF to calculate the diopter which is true, and the FAQ states that it is most practical to use an empirical method. In fact, that is what Leica used in the past. They provided their dealers with a wheel containing all diopter strengths that went into the camera shoe and covered the ocular, so the customer could spin it and find the correct strength. 

Yep, I’ve seen pics of the device. As I wrote, I use my own similarly practical method, i.e., my optician’s set of trial diopters.  I use glasses to correct for distance (and astigmatism), but now, with aging eyes,  benefit from an additional +.5 diopter.  
 

When folks here ask how to determine diopter strength, I generally advise to ignore the math (unless they have Rick’s optical expertise) and just experiment with diopters.  But I also have mentioned that the RF patch is set at a virtual distance of 2m.  jdlaing disagreed with this statement, in fact with the whole premise of a set virtual distance.  I was trying to determine if Rick agreed or disagreed.

Jeff

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[Jeff S]

7 minutes ago, jaapv said:

I think that we can agree that this is a highly complicated system that lends itself to interesting discussion.  This is an older design ( M4 I believe) and Leica has refined it quite a bit since then. 

More complicated than I thought before this discussion.

And, separately, I think the frame lines on an M4 are accurate at 1m. 🫣

Jeff