objective focal length
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I am confused with regard to the front focal length of objectives. I thought
that the magnification of an infinity corrected objective will be the distance between the tube lens and the intermediate image, divided by the front focal length of the objective (which is where I expect the object to be). Since the objective is not really a thin lens, I can understand that the actual working distance may be less than the focal length, since the focal length may need to be measured from inside the objective casing. However, I have a paper that describes a 25x/NA=0.5 air objective, which has an 11mm working distance, as having a 25.1mm focal length, while the distance from the tube lens to the intermediate image is 245mm. This has me confused, and I realize that I do not understand something fundamental here. So, I turn to the optics gurus on the list to clear this up, with many thanks in advance. --aryeh -- Aryeh Weiss School of Engineering Bar Ilan University Ramat Gan 52900 Israel Ph: 972-3-5317638 FAX: 972-3-7384050 |
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Hi Aryeh
Maybe the tube lens is not the original one, so that the "25x" objective really functions as a 10x in this setup? Can you give us more info about the setup. A 25x/0.5 air objective with a 11mm WD must be pretty big... Beat At 09:20 28-11-2008, you wrote: >I am confused with regard to the front focal length of objectives. I >thought that the magnification of an infinity corrected objective >will be the distance between the tube lens and the intermediate >image, divided by the front focal length of the objective (which is >where I expect the object to be). Since the objective is not really >a thin lens, I can understand that the actual working distance may >be less than the focal length, since the focal length may need to be >measured from inside the objective casing. > >However, I have a paper that describes a 25x/NA=0.5 air objective, >which has an 11mm working distance, as having a 25.1mm focal length, >while the distance from the tube lens to the intermediate image is >245mm. This has me confused, and I realize that I do not understand >something fundamental here. > >So, I turn to the optics gurus on the list to clear this up, with >many thanks in advance. > >--aryeh >-- >Aryeh Weiss >School of Engineering >Bar Ilan University >Ramat Gan 52900 Israel > >Ph: 972-3-5317638 >FAX: 972-3-7384050 |
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In reply to this post by Aryeh Weiss
I am absolutely not an optics guru!
But, as I understand it, an infinity corrected objective will form an image at 'infinity', so the distance from the principal plane to the object will be the focal length. But where the image is formed after the tube lens will depend on the focal length of the tube lens and we have to know this to make any useful calculation. It was all so much simpler in the days of 160mm tube length! Guy Optical Imaging Techniques in Cell Biology by Guy Cox CRC Press / Taylor & Francis http://www.guycox.com/optical.htm ______________________________________________ Associate Professor Guy Cox, MA, DPhil(Oxon) Electron Microscope Unit, Madsen Building F09, University of Sydney, NSW 2006 ______________________________________________ Phone +61 2 9351 3176 Fax +61 2 9351 7682 Mobile 0413 281 861 ______________________________________________ http://www.guycox.net -----Original Message----- From: Confocal Microscopy List [mailto:[hidden email]] On Behalf Of Aryeh Weiss Sent: Friday, 28 November 2008 7:20 PM To: [hidden email] Subject: objective focal length I am confused with regard to the front focal length of objectives. I thought that the magnification of an infinity corrected objective will be the distance between the tube lens and the intermediate image, divided by the front focal length of the objective (which is where I expect the object to be). Since the objective is not really a thin lens, I can understand that the actual working distance may be less than the focal length, since the focal length may need to be measured from inside the objective casing. However, I have a paper that describes a 25x/NA=0.5 air objective, which has an 11mm working distance, as having a 25.1mm focal length, while the distance from the tube lens to the intermediate image is 245mm. This has me confused, and I realize that I do not understand something fundamental here. So, I turn to the optics gurus on the list to clear this up, with many thanks in advance. --aryeh -- Aryeh Weiss School of Engineering Bar Ilan University Ramat Gan 52900 Israel Ph: 972-3-5317638 FAX: 972-3-7384050 No virus found in this incoming message. Checked by AVG. Version: 7.5.552 / Virus Database: 270.9.10/1815 - Release Date: 27/11/2008 9:02 AM No virus found in this outgoing message. Checked by AVG. Version: 7.5.552 / Virus Database: 270.9.10/1815 - Release Date: 27/11/2008 9:02 AM |
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In reply to this post by Beat Ludin
Hi Beat,
The info is from a paper by Ovryn and Izen, JOSA A/vol 17, No 7, p. 1202 (july 2000), in section 3. A colleague asked me for some help, and since I dont understand it, I turned to the list. Best regards, --aryeh Beat Ludin wrote: > Hi Aryeh > > Maybe the tube lens is not the original one, so that the "25x" objective > really functions as a 10x in this setup? > Can you give us more info about the setup. A 25x/0.5 air objective with > a 11mm WD must be pretty big... > > Beat > > At 09:20 28-11-2008, you wrote: >> I am confused with regard to the front focal length of objectives. I >> thought that the magnification of an infinity corrected objective will >> be the distance between the tube lens and the intermediate image, >> divided by the front focal length of the objective (which is where I >> expect the object to be). Since the objective is not really a thin >> lens, I can understand that the actual working distance may be less >> than the focal length, since the focal length may need to be measured >> from inside the objective casing. >> >> However, I have a paper that describes a 25x/NA=0.5 air objective, >> which has an 11mm working distance, as having a 25.1mm focal length, >> while the distance from the tube lens to the intermediate image is >> 245mm. This has me confused, and I realize that I do not understand >> something fundamental here. >> >> So, I turn to the optics gurus on the list to clear this up, with many >> thanks in advance. >> >> --aryeh >> -- >> Aryeh Weiss >> School of Engineering >> Bar Ilan University >> Ramat Gan 52900 Israel >> >> Ph: 972-3-5317638 >> FAX: 972-3-7384050 > > |
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In reply to this post by Guy Cox
That is also how I understand it. So I assume that the distance between the tube
lens and the intermediate image plane is the focal length of the tube lens. Best regards, --aryeh Guy Cox wrote: > I am absolutely not an optics guru! > > But, as I understand it, an infinity corrected objective will form an image > at 'infinity', so the distance from the principal plane to the object will be > the focal length. But where the image is formed after the tube lens will > depend on the focal length of the tube lens and we have to know this to make > any useful calculation. > > It was all so much simpler in the days of 160mm tube length! > > Guy > > -----Original Message----- From: Confocal Microscopy List > [mailto:[hidden email]] On Behalf Of Aryeh Weiss Sent: > Friday, 28 November 2008 7:20 PM To: [hidden email] > Subject: objective focal length > > I am confused with regard to the front focal length of objectives. I thought > that the magnification of an infinity corrected objective will be the > distance between the tube lens and the intermediate image, divided by the > front focal length of the objective (which is where I expect the object to > be). Since the objective is not really a thin lens, I can understand that the > actual working distance may be less than the focal length, since the focal > length may need to be measured from inside the objective casing. > > However, I have a paper that describes a 25x/NA=0.5 air objective, which has > an 11mm working distance, as having a 25.1mm focal length, while the distance > from the tube lens to the intermediate image is 245mm. This has me confused, > and I realize that I do not understand something fundamental here. > > So, I turn to the optics gurus on the list to clear this up, with many thanks > in advance. > > --aryeh -- Aryeh Weiss School of Engineering Bar Ilan University Ramat Gan > 52900 Israel > > Ph: 972-3-5317638 FAX: 972-3-7384050 > > |
 
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Patrick Van Oostveldt |
Re: objective focal length
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Dear Arey,
Not being an expert I should say: if the tube lens forms an image at 245mm (=focal length) and if the lens magnification is 25x the objective should have a focal length of 9.8mm. The working distance is smaller than this, but is not directly predictable. However the magnifcation inscribed on the lens (25X) is dependent on a specific focal length of the tube lens. For Nikon objectives it is 200mm for Zeiss it is 164.5nmm. Probably this can be the point of confusion. Best regards Patrick Quoting Aryeh Weiss <[hidden email]>: > That is also how I understand it. So I assume that the distance between > the tube > lens and the intermediate image plane is the focal length of the tube lens. > > Best regards, > --aryeh > > Guy Cox wrote: >> I am absolutely not an optics guru! >> >> But, as I understand it, an infinity corrected objective will form an image >> at 'infinity', so the distance from the principal plane to the >> object will be >> the focal length. But where the image is formed after the tube lens will >> depend on the focal length of the tube lens and we have to know this to make >> any useful calculation. >> >> It was all so much simpler in the days of 160mm tube length! >> >> Guy >> >> -----Original Message----- From: Confocal Microscopy List >> [mailto:[hidden email]] On Behalf Of Aryeh Weiss Sent: >> Friday, 28 November 2008 7:20 PM To: >> [hidden email] Subject: objective focal length >> >> I am confused with regard to the front focal length of objectives. I thought >> that the magnification of an infinity corrected objective will be the >> distance between the tube lens and the intermediate image, divided by the >> front focal length of the objective (which is where I expect the object to >> be). Since the objective is not really a thin lens, I can >> understand that the >> actual working distance may be less than the focal length, since the focal >> length may need to be measured from inside the objective casing. >> >> However, I have a paper that describes a 25x/NA=0.5 air objective, which has >> an 11mm working distance, as having a 25.1mm focal length, while >> the distance >> from the tube lens to the intermediate image is 245mm. This has me confused, >> and I realize that I do not understand something fundamental here. >> >> So, I turn to the optics gurus on the list to clear this up, with >> many thanks >> in advance. >> >> --aryeh -- Aryeh Weiss School of Engineering Bar Ilan University Ramat Gan >> 52900 Israel >> >> Ph: 972-3-5317638 FAX: 972-3-7384050 >> >> -- Dep. Moleculaire Biotechnologie Coupure links 653 B 9000 GENT tel 09 264 5969 fax 09 264 6219 |
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The problem is that an infinity corrected 25x lens with a 25mm focal length
should require a tube lens with a 625mm focal length (25x25). Does such a microscope exist? --aryeh Patrick Van Oostveldt wrote: > Dear Arey, > > Not being an expert I should say: if the tube lens forms an image at > 245mm (=focal length) and if the lens magnification is 25x the objective > should have a focal length of 9.8mm. The working distance is smaller > than this, but is not directly predictable. > > However the magnifcation inscribed on the lens (25X) is dependent on a > specific focal length of the tube lens. For Nikon objectives it is 200mm > for Zeiss it is 164.5nmm. Probably this can be the point of confusion. > > Best regards > > Patrick > > Quoting Aryeh Weiss <[hidden email]>: > >> That is also how I understand it. So I assume that the distance between >> the tube >> lens and the intermediate image plane is the focal length of the tube >> lens. >> >> Best regards, >> --aryeh >> >> Guy Cox wrote: >>> I am absolutely not an optics guru! >>> >>> But, as I understand it, an infinity corrected objective will form an >>> image >>> at 'infinity', so the distance from the principal plane to the >>> object will be >>> the focal length. But where the image is formed after the tube lens >>> will >>> depend on the focal length of the tube lens and we have to know this >>> to make >>> any useful calculation. >>> >>> It was all so much simpler in the days of 160mm tube length! >>> >>> Guy >>> >>> -----Original Message----- From: Confocal Microscopy List >>> [mailto:[hidden email]] On Behalf Of Aryeh Weiss Sent: >>> Friday, 28 November 2008 7:20 PM To: >>> [hidden email] Subject: objective focal length >>> >>> I am confused with regard to the front focal length of objectives. I >>> thought >>> that the magnification of an infinity corrected objective will be the >>> distance between the tube lens and the intermediate image, divided by >>> the >>> front focal length of the objective (which is where I expect the >>> object to >>> be). Since the objective is not really a thin lens, I can understand >>> that the >>> actual working distance may be less than the focal length, since the >>> focal >>> length may need to be measured from inside the objective casing. >>> >>> However, I have a paper that describes a 25x/NA=0.5 air objective, >>> which has >>> an 11mm working distance, as having a 25.1mm focal length, while the >>> distance >>> from the tube lens to the intermediate image is 245mm. This has me >>> confused, >>> and I realize that I do not understand something fundamental here. >>> >>> So, I turn to the optics gurus on the list to clear this up, with >>> many thanks >>> in advance. >>> >>> --aryeh -- Aryeh Weiss School of Engineering Bar Ilan University >>> Ramat Gan >>> 52900 Israel >>> >>> Ph: 972-3-5317638 FAX: 972-3-7384050 >>> >>> > > > -- Aryeh Weiss School of Engineering Bar Ilan University Ramat Gan 52900 Israel Ph: 972-3-5317638 FAX: 972-3-7384050 |
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The paper says it is a Zeiss Planachromat 25x/0.5 objective. I
searched for such an objective and the only I found was the Zeiss Jena GF-Planachromat shown at http://cgi.ebay.de/Zeiss-Jena-GF-Planachromat-Phv-25x%2F0,50-Phako--neu-_W0QQitemZ380085053453QQcmdZViewItemQQimsxZ20081127?IMSfp=TL0811271110001r39141 It's easy to see by the diameter of the front lens (<5mm) alone that this objective cannot possibly have a working distance of 11m at an NA of 0.5. So it seems very likely that the specs given in the paper are simply wrong. Beat At 18:39 29-11-2008, Aryeh Weiss wrote: >Content-Transfer-Encoding: 7bit > >The problem is that an infinity corrected 25x lens with a 25mm focal >length should require a tube lens with a 625mm focal length (25x25). >Does such a microscope exist? > >--aryeh > > >Patrick Van Oostveldt wrote: >>Dear Arey, >>Not being an expert I should say: if the tube lens forms an image >>at 245mm (=focal length) and if the lens magnification is 25x the >>objective should have a focal length of 9.8mm. The working distance >>is smaller than this, but is not directly predictable. >>However the magnifcation inscribed on the lens (25X) is dependent >>on a specific focal length of the tube lens. For Nikon objectives >>it is 200mm for Zeiss it is 164.5nmm. Probably this can be the >>point of confusion. >>Best regards >>Patrick >>Quoting Aryeh Weiss <[hidden email]>: >> >>>That is also how I understand it. So I assume that the distance between >>>the tube >>>lens and the intermediate image plane is the focal length of the tube lens. >>> >>>Best regards, >>>--aryeh >>> >>>Guy Cox wrote: >>>>I am absolutely not an optics guru! >>>> >>>>But, as I understand it, an infinity corrected objective will form an image >>>>at 'infinity', so the distance from the principal plane to the >>>>object will be >>>>the focal length. But where the image is formed after the tube lens will >>>>depend on the focal length of the tube lens and we have to know >>>>this to make >>>>any useful calculation. >>>> >>>>It was all so much simpler in the days of 160mm tube length! >>>> >>>>Guy >>>> >>>>-----Original Message----- From: Confocal Microscopy List >>>>[mailto:[hidden email]] On Behalf Of Aryeh Weiss Sent: >>>>Friday, 28 November 2008 7:20 PM To: >>>>[hidden email] Subject: objective focal length >>>> >>>>I am confused with regard to the front focal length of >>>>objectives. I thought >>>>that the magnification of an infinity corrected objective will be the >>>>distance between the tube lens and the intermediate image, divided by the >>>>front focal length of the objective (which is where I expect the object to >>>>be). Since the objective is not really a thin lens, I >>>>can understand that the >>>>actual working distance may be less than the focal length, since the focal >>>>length may need to be measured from inside the objective casing. >>>> >>>>However, I have a paper that describes a 25x/NA=0.5 air >>>>objective, which has >>>>an 11mm working distance, as having a 25.1mm focal length, >>>>while the distance >>>>from the tube lens to the intermediate image is 245mm. This has >>>>me confused, >>>>and I realize that I do not understand something fundamental here. >>>> >>>>So, I turn to the optics gurus on the list to clear this up, with >>>>many thanks >>>>in advance. >>>> >>>>--aryeh -- Aryeh Weiss School of Engineering Bar Ilan University Ramat Gan >>>>52900 Israel >>>> >>>>Ph: 972-3-5317638 FAX: 972-3-7384050 >>>> >> > > >-- >Aryeh Weiss >School of Engineering >Bar Ilan University >Ramat Gan 52900 Israel > >Ph: 972-3-5317638 >FAX: 972-3-7384050 |
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Thank you for that pointer. I agree, and the consensus seems to be that
something has not been reported properly in the paper. When I am puzzled about a paper, I assume that I probably made a mistake. However, all of the responses that I received support the idea that the paper has a mistake. Best regards, --aryeh Beat Ludin wrote: > The paper says it is a Zeiss Planachromat 25x/0.5 objective. I searched > for such an objective and the only I found was the Zeiss Jena > GF-Planachromat shown at > http://cgi.ebay.de/Zeiss-Jena-GF-Planachromat-Phv-25x%2F0,50-Phako--neu-_W0QQitemZ380085053453QQcmdZViewItemQQimsxZ20081127?IMSfp=TL0811271110001r39141 > > > It's easy to see by the diameter of the front lens (<5mm) alone that > this objective cannot possibly have a working distance of 11m at an NA > of 0.5. So it seems very likely that the specs given in the paper are > simply wrong. > > Beat > > > > At 18:39 29-11-2008, Aryeh Weiss wrote: >> Content-Transfer-Encoding: 7bit >> >> The problem is that an infinity corrected 25x lens with a 25mm focal >> length should require a tube lens with a 625mm focal length (25x25). >> Does such a microscope exist? >> >> --aryeh >> >> >> Patrick Van Oostveldt wrote: >>> Dear Arey, >>> Not being an expert I should say: if the tube lens forms an image at >>> 245mm (=focal length) and if the lens magnification is 25x the >>> objective should have a focal length of 9.8mm. The working distance >>> is smaller than this, but is not directly predictable. >>> However the magnifcation inscribed on the lens (25X) is dependent on >>> a specific focal length of the tube lens. For Nikon objectives it is >>> 200mm for Zeiss it is 164.5nmm. Probably this can be the point of >>> confusion. >>> Best regards >>> Patrick >>> Quoting Aryeh Weiss <[hidden email]>: >>> >>>> That is also how I understand it. So I assume that the distance between >>>> the tube >>>> lens and the intermediate image plane is the focal length of the >>>> tube lens. >>>> >>>> Best regards, >>>> --aryeh >>>> >>>> Guy Cox wrote: >>>>> I am absolutely not an optics guru! >>>>> >>>>> But, as I understand it, an infinity corrected objective will form >>>>> an image >>>>> at 'infinity', so the distance from the principal plane to the >>>>> object will be >>>>> the focal length. But where the image is formed after the tube >>>>> lens will >>>>> depend on the focal length of the tube lens and we have to know >>>>> this to make >>>>> any useful calculation. >>>>> >>>>> It was all so much simpler in the days of 160mm tube length! >>>>> >>>>> Guy >>>>> >>>>> -----Original Message----- From: Confocal Microscopy List >>>>> [mailto:[hidden email]] On Behalf Of Aryeh Weiss >>>>> Sent: >>>>> Friday, 28 November 2008 7:20 PM To: >>>>> [hidden email] Subject: objective focal length >>>>> >>>>> I am confused with regard to the front focal length of objectives. >>>>> I thought >>>>> that the magnification of an infinity corrected objective will be the >>>>> distance between the tube lens and the intermediate image, divided >>>>> by the >>>>> front focal length of the objective (which is where I expect the >>>>> object to >>>>> be). Since the objective is not really a thin lens, I can >>>>> understand that the >>>>> actual working distance may be less than the focal length, since >>>>> the focal >>>>> length may need to be measured from inside the objective casing. >>>>> >>>>> However, I have a paper that describes a 25x/NA=0.5 air objective, >>>>> which has >>>>> an 11mm working distance, as having a 25.1mm focal length, while >>>>> the distance >>>>> from the tube lens to the intermediate image is 245mm. This has me >>>>> confused, >>>>> and I realize that I do not understand something fundamental here. >>>>> >>>>> So, I turn to the optics gurus on the list to clear this up, with >>>>> many thanks >>>>> in advance. >>>>> >>>>> --aryeh -- Aryeh Weiss School of Engineering Bar Ilan University >>>>> Ramat Gan >>>>> 52900 Israel >>>>> >>>>> Ph: 972-3-5317638 FAX: 972-3-7384050 >>>>> >>> >> >> >> -- >> Aryeh Weiss >> School of Engineering >> Bar Ilan University >> Ramat Gan 52900 Israel >> >> Ph: 972-3-5317638 >> FAX: 972-3-7384050 > > -- Aryeh Weiss School of Engineering Bar Ilan University Ramat Gan 52900 Israel Ph: 972-3-5317638 FAX: 972-3-7384050 |
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B. Prabhakar Pandian |
Hypoxia/ pH assay in viable cells
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In reply to this post by Beat Ludin
Hello,
Can someone tell me what fluroescent asssays are currently available for studying pH and hypoxia changes in a cell culture real-time. Thanks, -Prabhakar |
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Farid Jalali |
Re: Hypoxia/ pH assay in viable cells
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Hello Prabhakar,
For the case of hypoxia, I have used the Oxylite system from Oxford Optronics to measure O2 concentration in vitro during live-cell imaging. The system is quite easy to use and the probes themselves are about $200-$300. They are fibre-optic and quite delicate. Alternatively, something that we have wanted to try is to use the bioreduced 2-nitroimidizoles EF5 or pimonidazole and stain indirectly using immunofluorescence (there is a Cy3-conjugated ab available to EF5). I know its not real time, but you may be able to develop a calibration curve for future experiments. See any publication by CJ Koch for the use of EF5. Best Farid On Tue, Dec 2, 2008 at 4:25 PM, B. Prabhakar Pandian <[hidden email]> wrote: Hello, |
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Stephen Cody-2 |
Re: Hypoxia/ pH assay in viable cells
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Dear Prabhaker,
We had great success with SNARF-1/AM for measuring intracellular pH. It relatively easy to do this quantitatively too.
See:
http://dx.doi.org/10.1016/0968-4328(93)90035-Y Cheers
Stephen Cody
2008/12/3 Farid Jalali <[hidden email]> Hello Prabhakar, -- Stephen Cody |
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