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Dave Martindale 


 
Pat
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One would think that the use of 50 mm instead of 43 mm comes from a
combination of 3 things. First, it's just a heck of a lot easier to use/remember/etc. For example, would you then got to a 150 mm or a 129 mm. Second, you probably want a little extra light around the outside of the frame, so you round up a bit. Third, and most likely, when you are designing a lens, there's a lot of math going into the angles, etc. Back in the early days when the designers were using slide rules, it was probably easier to stick 50 into a formula than 43. But these are just guesses. Pat. Dave Martindale wrote: > "(EMail Removed)" <(EMail Removed)> writes: > > > >Somehow > >some manufacturer introduced the 50 mm as the normal lens and now it is > >general > >accepted as the normal lens. I believe that when people refers to 35 > >mm > >equivalent is based on 50 mm. So I used that as the reference. > > When you see lenses marked with a "35 mm equivalent" focal length, that > calculation depends only on the sensor dimensions. What's considered a > "normal" lens for the focal length has no part in it. > > For example, if you have a digital camera whose sensor is 7.2 mm wide, > that is 1/5 the width of a 35 film frame, and so all of the "35 > equivalent focal length" specification are 5 times the true focal > length. If the lens is actually 721 mm, it will be quoted as "35105 > equivalent". It simply doesn't matter whether you think "normal" is 40 > or 43 or 50 or 55 mm on a 35 camera. > > >I am referring to using the digital camera to take a picture through > >the > >eyepiece of a telescope or microscope. If the focal length of the > >eyepiece > >is 32 mm and using the maximum zoom of the CP995 (which is also 32 > >mm), > >the image you got is the same as formed by the objective of the > >telescope, > >i.e., the final magnification is 1. If the focal length of the eyepiece > >is 16, the > >final magnification is 2. Of course you have to make sure that the > >images > >match. > > That doesn't make sense to me at all. The magnfication of the > *telescope* is an angular magnification, determined by the ratio of the > objective and eyepiece focal lengths. The resulting virtual image is > projected at infinity (or so we assume, anyway). > > Then the camera uses its lens to capture that image. On a camera, the > lens focal length primarily determines angle of view, and you generally > want a rough match between the camera angle of view and the eyepiece > image size. That might mean using a 50 mm lens on a fullframe camera, > or a 10 mm lens on a P&S digicam, but the angle of view is the same for > both of these. And it doesn't matter whether the telescope eyepiece is > a 55 mm or a 9 mm. > > If you want to calculate the magnification of the digital image you get, > you do need to know the true focal length of the camera lens as well as > the overall magnification of the telescope. But you'll get the same > final magnification using a telescope with a 2000 mm objective and a 40 > mm eyepiece as you will with a 500 mm objective and a 10 mm eyepiece. > > Having the focal length of the camera lens match the focal length of > the eyepiece has no value at all that I'm aware of. They do two very > different jobs, and have entrance and exit pupils in very different > places. > > You *would* like to match the exit pupil of the eyepiece with the entrance > pupil of the camera lens, but that doesn't depend on the focal length, > and can vary greatly between zoom lenses that have the same focal > length. > > Dave 




Pat 


 
ih@ece.udel.edu
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Dave Martindale wrote: > "(EMail Removed)" <(EMail Removed)> writes: > > When you see lenses marked with a "35 mm equivalent" focal length, that > calculation depends only on the sensor dimensions. What's considered a > "normal" lens for the focal length has no part in it. > > For example, if you have a digital camera whose sensor is 7.2 mm wide, > that is 1/5 the width of a 35 film frame, and so all of the "35 > equivalent focal length" specification are 5 times the true focal > length. If the lens is actually 721 mm, it will be quoted as "35105 > equivalent". It simply doesn't matter whether you think "normal" is 40 > or 43 or 50 or 55 mm on a 35 camera. > You have used a legitimate method to calculate the 35 mm equivalent focal lengths by linear proportions. By using linear proportion, all the "form" factors are automatically included. This method is good enough for all practical purposes. However, strickly speaking this method is correct if the aspect ratio is the same and the normal focal length is equal to the diagonal. We know the diagonal the 35 mm (double frame) film format is 43. Its aspect ratio is 3:2 and that of S&P digicam is 4:3. 50 mm is now generally accepted as the normal length. Because ' of these factors, the calculation may produce slight different values. However, it is of no important concerns. As I said above this method is good enough for all purposes. Let us go through the "formal" exercise in calculating the 35 mm equivalent. The diogonal of the sensor of 7.2 mm width is 9 mm. The normal focal length of this sensor is 9 mm (strickly by definition). 7/9 = .7778 and 21/9 = 2.3333. So the 35 mm equivalent is 39  117 mm based on 50 mm and 33  100 based on 43 (which nobody uses now). You can see that what normal focal lengths to use does matter if you follow the "formal" procedure. Is this important? probably not. I don't know which of the three is more right than others. I probably chosen either 35  105 or 39  117. The normal focal length difinition has been used only as a guideline or reference. > >I am referring to using the digital camera to take a picture through > >the > >eyepiece of a telescope or microscope. If the focal length of the > >eyepiece > >is 32 mm and using the maximum zoom of the CP995 (which is also 32 > >mm), > >the image you got is the same as formed by the objective of the > >telescope, > >i.e., the final magnification is 1. If the focal length of the eyepiece > >is 16, the > >final magnification is 2. Of course you have to make sure that the > >images > >match. > > That doesn't make sense to me at all. The magnfication of the > *telescope* is an angular magnification, determined by the ratio of the > objective and eyepiece focal lengths. The resulting virtual image is > projected at infinity (or so we assume, anyway). > > Then the camera uses its lens to capture that image. On a camera, the > lens focal length primarily determines angle of view, and you generally > want a rough match between the camera angle of view and the eyepiece > image size. That might mean using a 50 mm lens on a fullframe camera, > or a 10 mm lens on a P&S digicam, but the angle of view is the same for > both of these. And it doesn't matter whether the telescope eyepiece is > a 55 mm or a 9 mm. > > If you want to calculate the magnification of the digital image you get, > you do need to know the true focal length of the camera lens as well as > the overall magnification of the telescope. But you'll get the same > final magnification using a telescope with a 2000 mm objective and a 40 > mm eyepiece as you will with a 500 mm objective and a 10 mm eyepiece. > > Having the focal length of the camera lens match the focal length of > the eyepiece has no value at all that I'm aware of. They do two very > different jobs, and have entrance and exit pupils in very different > places. I never said that. > > You *would* like to match the exit pupil of the eyepiece with the entrance > pupil of the camera lens, but that doesn't depend on the focal length, > and can vary greatly between zoom lenses that have the same focal > length. > > Dave What you said is generally correct and you try to give an overall description of the whole operation. In my post I only try to describe the image "relay" operation. Two things can be discussed separately. You said the image is projected at infinite. That means the image is formed right at the focal plane of the eyepiece. Then you said that the camera uses its to capture that image. The lens must be set at infinite in order to capture an image at infinite. This means that the image is formed at the focal plane of the camera. So the real operation is that the image is formed at focal plane of the eyepiece and transferred (or "relay") to the focal plane of the camera. This is the relay operation I was talking about. Please see below for a more detailed explanation. I also generally agree with you about the pupils match. Since I was only talking about the relay, I don't need to discuss that. The best way to record the telescope image is to directly expose it on the sensor of a DSLR. Since we cannot remove the lense of a P&S digicam, one way is to "relay" the image to the sensor by using a "relay lense" formed by the eyspiece and the lens of the camera. This is normally done. In doing so there is a "final" magnification/demagnification depending on the focal ratio of the two. It is best to see this by the following experiment (theoretical or physical). Align two lenses of equal focal length on their optical axis. If you put an image at the first lens and an image of equal size (inverted) is formed at the focal plane of the second. If you double (or halve) the focal length of the second lens, the image is doubled (or halved). I did not discuss the many technical details which you discussed some. The operation the relay lens is just like this. Your description of the visual operation of the telescaope and many technical aspects of the operations are correct. We are in general agreement except that you are looking at the big picture and I was talking only about a small part of it. Charles S. Ih 




ih@ece.udel.edu 
Dave Martindale
Guest
Posts: n/a

"(EMail Removed)" <(EMail Removed)> writes:
>You have used a legitimate method to calculate the >35 mm equivalent focal lengths by linear proportions. >By using linear proportion, all the "form" factors >are automatically included. This method is good enough >for all practical purposes. However, strickly speaking >this method is correct if the aspect ratio is the same >and the normal focal length is equal to the diagonal. No, it doesn't matter what the normal focal length is considered to be, since that doesn't need to enter the calculations at all. If the aspect ratios of the two formats are different, then you have to make it clear whether you are calculating "equivalent" focal lengths by making the horizontal FOV, the vertical FOV, or the diagonal FOV the same. There are 3 different answers depending on which you choose to make equal. >Let us go through the "formal" exercise in calculating >the 35 mm equivalent. The diogonal of the sensor of 7.2 mm >width is 9 mm. The normal focal length of this sensor is >9 mm (strickly by definition). 7/9 = .7778 and 21/9 = 2.3333. >So the 35 mm equivalent is 39  117 mm based on 50 mm and 33 > 100 based on 43 (which nobody uses now). You can see that >what normal focal lengths to use does matter if you follow >the "formal" procedure. I've never seen that particular procedure used before. You're converting every focal length into a ratio against the normal lens for the one format, then multiplying that by the normal lens for the other format. That does give you a sort of "comparative" focal length, but it doesn't give you the same field of view unless the two normal FLs are carefully matched! Instead, here is the usual method of focal length conversion: scale focal lengths proportional to sensor dimensions to preserve FOV. For example, suppose we want to preserve horizontal FOV. The sensor width is 7.2 mm, while full frame width is 36 mm. So focal lengths of 7 and 21 mm on the digital camera are equivalent to 7 * 36/7.2 = 35 and 21 * 36/7.2 = 105 mm. These are exactly equivalent, in terms of matching horizontal FOV. If you want to match diagonal FOV instead, the diagonal of a 7.2 mm wide 1.333 aspect sensor is 9 mm, the diagonal of a 36 mm wide 1.5 aspect ratio 35 frame is 43.27 mm, and the multiplier to convert from true FL to "35 equivalent" FL is 4.8 instead of 5.0. But none of these calculations depend on what is considered "normal" for either format. >The best way to record the telescope image is to directly >expose it on the sensor of a DSLR. Since we cannot remove >the lense of a P&S digicam, one way is to "relay" the >image to the sensor by using a "relay lense" formed by the >eyspiece and the lens of the camera. This is normally done. >In doing so there is a "final" magnification/demagnification >depending on the focal ratio of the two. It is best to see this >by the following experiment (theoretical or physical). Align >two lenses of equal focal length on their optical axis. If you >put an image at the first lens and an image of equal size >(inverted) is formed at the focal plane of the second. If you >double (or halve) the focal length of the second lens, the image >is doubled (or halved). I did not discuss the many technical >details which you discussed some. The operation the relay lens >is just like this. Ok. But, in general, it's not desirable to make the FL of the camera lens equal to the FL of the telescope eyepiece. That *is* true if you're trying to simulate the effect of removing the lens from the camera, removing the eyepiece from the telescope, mounting the camera directly on the telescope, and doing "prime focus" photography. But if you leave the eyepiece and camera lens in place, you generally do not want 1:1 relaying. You want to relay most of the image that the eyepiece "sees", whose real size is set by the field stop inside the eyepiece, onto the sensor in the camera. For a given eyepiece and field diameter, this means that the camera lens FL should be proportional to the size of the camera sensor. Put another way, you want the camera to capture a particular field of view, no matter whether that's a 14 mm FL on a small digicam or a 70 mm FL on a fullframe SLR. So you're right: the magnification of the "relay optics" is determined by the ratio of their focal lengths. Doing calculations with these, you need to determine how the objective's FL converts an angle at infinity into a distance in the primary focal plane, and then use the relay optics magnification to convert that to a distance on the sensor, which ultimately becomes a distance on the print given the printing magnification. In contrast, I was keeping things in the angular domain as long as possible. The telescope objective and eyepiece together provide an angular magnification, and this angle is a certain portion of the FOV of the camera. The two methods are equivalent and should give the same ultimate answer. Dave 




Dave Martindale 
ih@ece.udel.edu
Guest
Posts: n/a

Dave Martindale wrote: > "(EMail Removed)" <(EMail Removed)> writes: > > >You have used a legitimate method to calculate the > >35 mm equivalent focal lengths by linear proportions. > >By using linear proportion, all the "form" factors > >are automatically included. This method is good enough > >for all practical purposes. However, strickly speaking > >this method is correct if the aspect ratio is the same > >and the normal focal length is equal to the diagonal. > > No, it doesn't matter what the normal focal length is considered to be, > since that doesn't need to enter the calculations at all. > > If the aspect ratios of the two formats are different, then you have to > make it clear whether you are calculating "equivalent" focal lengths by > making the horizontal FOV, the vertical FOV, or the diagonal FOV the > same. There are 3 different answers depending on which you choose to > make equal. > > >Let us go through the "formal" exercise in calculating > >the 35 mm equivalent. The diogonal of the sensor of 7.2 mm > >width is 9 mm. The normal focal length of this sensor is > >9 mm (strickly by definition). 7/9 = .7778 and 21/9 = 2.3333. > >So the 35 mm equivalent is 39  117 mm based on 50 mm and 33 > > 100 based on 43 (which nobody uses now). You can see that > >what normal focal lengths to use does matter if you follow > >the "formal" procedure. > > I've never seen that particular procedure used before. You're > converting every focal length into a ratio against the normal lens for > the one format, then multiplying that by the normal lens for the other > format. That does give you a sort of "comparative" focal length, but it > doesn't give you the same field of view unless the two normal FLs are > carefully matched! > > Instead, here is the usual method of focal length conversion: scale > focal lengths proportional to sensor dimensions to preserve FOV. > For example, suppose we want to preserve horizontal FOV. The sensor > width is 7.2 mm, while full frame width is 36 mm. So focal lengths of 7 > and 21 mm on the digital camera are equivalent to 7 * 36/7.2 = 35 and > 21 * 36/7.2 = 105 mm. These are exactly equivalent, in terms of > matching horizontal FOV. > > If you want to match diagonal FOV instead, the diagonal of a 7.2 mm wide > 1.333 aspect sensor is 9 mm, the diagonal of a 36 mm wide 1.5 aspect > ratio 35 frame is 43.27 mm, and the multiplier to convert from true FL > to "35 equivalent" FL is 4.8 instead of 5.0. > > But none of these calculations depend on what is considered "normal" for > either format. The way I described is also OK. Let me give you a real example, Canon S400. The specifications are the following. Focal lelngth: 7.4  22.2 (35mm film equivalent 36  108 mm) CCD Size: 1/1.8" The CCD size follows the TV Vidicon convention. The diagonal of the image on the 1" tube is 16 mm. Thus the diagonal of the 1/1.8" CCD is 16/1.8 = 8.88889. 7.4/8.8889 = .8325 22.2/8.8889 = 2.4975 The true diagonal of the 35 mm film is 43.2666153. If you multiple this number to the above number, you get 36  108 mm exactly as given by the manufacturer. If you use 50 mm, you get 41  125 mm. Use your method, you get 37.46  112.38 mm. Apparently the manufacturers use the true diaponal to calculate the equivalent. So the 35 mm equivalent given by the manufacturers is as good you can get. From the very beginning I said your method is legitimate. The slightly different values are due to the different aspect ratio of the two format and I also mentioned that. If the aspect ratio is the same, the two methods will give the same results. Conventionally people try to match the diagonal as normally given by the manufacturers. If you are interested in matching horizonal FOV, your calculation is right because you started by matching the horizonal width. If you started by matching the diagonal, you wil get the same answer as given by the manufacturer. Both method are right and legitimate. It happens that one method needs the diagonal value and the other doesn't. There is nothing wrong with this. > > > Ok. But, in general, it's not desirable to make the FL of the camera > lens equal to the FL of the telescope eyepiece. That *is* true if > you're trying to simulate the effect of removing the lens from the > camera, removing the eyepiece from the telescope, mounting the camera > directly on the telescope, and doing "prime focus" photography. > But if you leave the eyepiece and camera lens in place, you generally do > not want 1:1 relaying. You want to relay most of the image that the > > So you're right: the magnification of the "relay optics" is determined > by the ratio of their focal lengths. That's all I said and never mentioned to match the focal lengths or making them 1:1. I never intended to discuss all the technical details. Charles S. Ih > 




ih@ece.udel.edu 
Dave Martindale
Guest
Posts: n/a

"(EMail Removed)" <(EMail Removed)> writes:
>The way I described is also OK. Let me give you a real >example, Canon S400. The specifications are the following. >Focal lelngth: 7.4  22.2 (35mm film equivalent 36  108 mm) >CCD Size: 1/1.8" >The CCD size follows the TV Vidicon convention. The diagonal >of the image on the 1" tube is 16 mm. Thus the diagonal of the >1/1.8" CCD is 16/1.8 = 8.88889. >7.4/8.8889 = .8325 >22.2/8.8889 = 2.4975 >The true diagonal of the 35 mm film is 43.2666153. >If you multiple this number to the above number, you get >36  108 mm exactly as given by the manufacturer. Right. That's perfectly legitimate, when you use the true diagonal. Note that you didn't need to assume a "normal" focal length for either format. >If you use 50 mm, you get 41  125 mm. Which I argue is the wrong answer, since it doesn't match *any* field of view  neither horizontal, vertical, or diagonal. >Use your method, >you get 37.46  112.38 mm. The method I described gives 3 different answers, depending on whether you decide you care most about horizontal, vertical, or diagonal FOV. If you pick diagonal, it is exactly the method you used above. >Both method are right and legitimate. It >happens that one method needs the diagonal value and the >other doesn't. There is nothing wrong with this. But notice that neither method needs to make any assumption about the normal lens focal length. That was my original point. Dave 




Dave Martindale 


 
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