Summarit 50/1.5 question

[thomas_krantz3]

Guys

 

Does anyone know if the Summarit is a "true 1.5"? Or is it more of a

1.8 or something? Reading optical theory it is clear that the step

from 2.0 to 1.4 is a huge one which introduces substantially more

abberation. The step from 1.5 to 1.4 was aparantly also quite an

effort since most lenses from Leica, Nikon and Canon started life as

1.5 and only later the 1.4 was developed.

 

Thomas Krantz

[al_kaplan1]

The f/1.5 Summarit started it's life before WWII as the 50mm f1.5 Scneider Xenon, an uncoated lens that was made in Leica screw mount and, I believe, marketed by Leitz. I've never owned one but did shoot a few rolls with a friend's Summarit back around 1970 to compare it to a 50mm f/1.4 Nikkor I was thinking of purchasing. Wide open there wasn't a significant variation in negative density between the two lenses, so I'm sure that the Summarit was a "true" f/1.5. The Nikkor had a bit less flare and was slightly sharper.

[albert_smith]

I would think that the Summarit was in fact f/1.5, and that the honesty was higher then. When the current 50mm f/1.4 Summilux was lab tested by Popular Photography Magazine back in 1994, the real aperture was listed as f/1.5 (see attached test snip). I guess these days, f/1.4 sounds sexier, than f/1.5.

 

Since the mechanical shutter of the M6 and earlier cameras is set in whole-stops, there would be no compensation via the shutter that you could do with that little .1 difference. If the exposure was f/2.0 at 1/60th, you couldn't go to f/1.5 at 1/119th. When I had my Summarit on my M3, I used a Sekonic meter and used f/1.4 to set the shutterspeed. Other than the soft and flarey look of the Summarit at full aperture, there was not obvious under exposure. It almost certainly was at least f/1.5.<div></div>

[mark_sampson]

I don't think the difference between f/1.5 and f/1.4 matters very much- tolerance in meter accuracy/calibration, shutter accuracy, film emulsion variations, process variablity, not to mention that bugaboo light falloff in the corners... all these factors suggest that the 1/6 stop difference will not show up in actual photography. That's .05 in density units... densotometers are usually accurate to .02...And people who are shooting in low light at f/1.4 or f/1.5 probably have other concerns, often relating to the subject.

[kelly_flanigan1]

A F 1.5 is alot easier to design than a F1.4 lens. In optics the abberations grow as the square and the CUBE of the aperture; depending on the abberation to be controlled. 1.5 divided by 1.4 doesnt sound like much. But when squared or cubed; it IS a really big deal. This lens was designed by hand; using many "human computers" that crunched the numbers for each ray; at each lens interface. To design a lens like this will consume several years of an optical teams energy in the pre WW2 era. In lens design; it is a big compromise. Many times the corners are purposely vignetted to radically reduce the off axis abberations when the lens is wide open. Layman think this is a lens "fault"; but using vignetting and adjusting the Max F number are tools that optical designers use.

[kelly_flanigan1]

The F1.5 lens is only a F1.5 at the central core ; 10mm? ??? due to vignetting. Also the lack of coating in pre WW2 lenses reduces the light transmission; so the on axis T number might be T1.8 ???? only a guess!

[victor_randin]

The �Leitz-Xenon 5cm F 1:1.5 Ernst Leitz Wetzlar� calculated by Schneider, made by Leitz is the predecessor of the Summarit 5cm F1:1.5.

 

I have both and hate them. Both are coated with bluish layers as most of Leitz�s early lenses, prone to flare and ghosts, soft wide open and low contrast, acceptable at F5.6 - 16. Very similar to the Summar, worse than the C.Z. Sonnar 5cm F 1:1.5, the Summitar. The difference between F1.4 and F1.5 is 0.538 f-stop.

[guy bennett]

I've got one in LTM. Use it on a IIIf. Image quality is PDG ("pretty darn good").

 

I just examined mine. No question: it is definitely a true 1.5. Says so on the lens barrel.

 

;)

[kelly_flanigan1]

The difference between a F 1.400 and a F1.500 lens is 0.1992 stops; or 1/5 F-stop..........

[andy_piper2]

Also bear in mind that the Summarit started as a 1930's design (the Xenon) and in the thirties f/stops had not yet standardized on the numbers we now think of as full-stops (the nice round numbers that are exact powers of the square root of 2 (1.0-1.4-2.0-2.8) - there were f/6.8 lenses and f/2.2 lenses and so f/1.5 was not necessarily an oddball aperture the way it seems today in a world full of f/1.4s (so-called).

 

The uncoated Xenons that preceded the Summarit name probably had T/stops of close to f/1.8 - they were uncoated and lost more lgiht to reflection

[Rob F.]

Kelly, I think that's right. If I raise 2 to the 0.1992 power, I get 1.148, which is the ratio of light passed by an f/1.4 lens to that of an f/1.5 lens: (1.4/1.5) squared. I can see that the f-stop is the power to which you have to raise the base 2 to get the light ratio. Thus it's the logarithm to the base two of the light ratio. Beyond that, my brain froze. How did you work it backwards to get that power?

[victor_randin]

dF² =F1²- F2²

 

For example:

 

1) dF² (f2.0-f1.4) = 2.0²-1.4² = 4.0 � 1.96 =2.04

 

dF=1.428 (one f-stop)

 

2) dF²(f2.8-f2.0) = 2.8²-2.0²=7.84 � 4.0 = 3.84

 

dF=1.9595 (one f-stop)

 

3) dF²(f4.0-f2.8) = 4.0²-2.8²=16.0 - 7.84 = 8.16

 

dF=2.8565 (one f- stop) ��and so on��

 

 

So,

 

dF²(f1.5-f1.4) = 2.25 - 1.96 = 0.29

 

dF²=0.29

 

dF=0.538 (0.384 or 2/5 f-stop)

[victor_randin]

F2 : dF= 1.4 : 0.538 = 0.384

[victor_randin]

sorry,

dF: F2 = 0.538 : 1.4 = 0.384

[jsbc]

Sorry Victor, wrong equation. This is purely because we are concerned over the ratio of aperture so only multiplication and divison, no addition or subtraction.

A F1.4 lens is twice as fast as a F2 lens because it allows twice as much light to pass through the aperture (ie 1 EV more).

 

In terms of equation, the area of the aperture hole of a F1.4 lens is twice that of a F2 lens, ie for a 50mm lens, the aperture of a Summilux is 50/1.4 or 35.7cm whilst that of a summicron is 25cm.

 

Hence the area of aperture of a summilux is 3.14* (35.7)^2 while that of a summicron it is 3.14*(24)^2. So the ratio of their area is 2:1.

 

The ratio of the aperture area of a summilux to a Summarit is 1.15:1. Hence a true Summilux is 15% faster than a Summarit because it transmit 15% more light.

 

Actually one should take logs to get the answer but I think the example is more illustrative than equations.

[james_elwing]

Aargh; in short it is a 1.5; quod erat demonstrandum

[kelly_flanigan1]

<BR><BR>2**0.0 = 1.000 ie F 1.0<BR><BR>2**0.5 = 1.414 ie F 1.4<BR><BR>2**1.0 = 2.000 ie F2.8<BR><BR>2**1.5 = 2.828 ie F2.8<BR><BR>2**2.0 = 4.000 ie F4.0<BR><BR>The exponent changes by 0.5 for each full stop in the examples above<BR><BR>I used 1.400 and 1.500 as my F stops; to calculate the ratio...in fractional f stop:<BR><BR>2**X = 1.400 X= 0.4854<BR><BR>2**Y = 1.500 Y=0.5850<BR><BR> Y - X = 0.0996 ; remember a difference of 0.5 is a full stop<BR><BR>0.0996 / 0.5 = 0.1992 f stops or 1/5 stop <br><br>they would NOT allow me to take "advanced math" in 8th grade; because I didnt have a B average in ENGLISH...........Our educational system is run by a bunch of goofballs....