Amazing lens, incredible sharpness

I would immediately repurchase the EM-5 and the 75mm.  Outstanding lens...

khaledgawdat wrote:

Day light, olympus pen e-p5 75 mm 1.8

New to th micro 4/3, Coming from Pentax 645d, Nikon D4, D800 and skeptical of the smaller sensor size of the M4/3. I bought the olympus e-p5 with 5 primes for the size factor. the olympus 75mm 1.8 is simply amazing. Size, build quality are superb. Sharpness, bokhen and colors are unbelievable. of the 5 primes 12mm,17mm,45mm,75mm Olympus and 25 mm Leica, the 75 mm is my favourite of the group. 150 mm f 1.8 is a good portrait range with exceptional quality.

...it's either a 75mm f/1.8 or 150mm f/3.6 FF equivalent, but not a 150mm f/1.8 -- that would be a rather larger, heavier, and more expensive lens (think the Olympus 150 / 2).

Ironic, by the way, that your pic is at 75mm f/5, though. 

Do not hesitate to buy it.

It's a stellar lens for those that want a 75 / 1.8.

khaledgawdat wrote:

I bought the olympus e-p5 with 5 primes for the size factor. the olympus 75mm 1.8 is simply amazing. Size, build quality are superb. Sharpness, bokhen and colors are unbelievable.

Not to belittle the 75 f1.8, take a look at the new Sigma 60 f2.8.

It is better still - and 1/4 the price.

Allan

tinternaut wrote:

In terms of DoF, yes it's a 3.6 equivalent...

I'm glad you agree.

...but in terms of light gathering (i.e. the shutter speed it enables one to achieve at a given aperture for correct exposure)...

Anders explained it nicely in his post above, but I'll take a stab at it, too. The exposure is the amount of light per area that falls on the sensor, not the amount of light gathered. So, f/1.8 on mFT results in 4x as much light per area as f/3.6 on FF, but a FF sensor has 4x the area, so the two neatly cancel each other out, and the light gathering of f/1.8 on mFT is the same as the light gathering as f/3.6 on FF.

...surely 1.8 is 1.8.

Well, 75mm f/1.8 has an aperture (entrance pupil) diameter of 75mm / 1.8 = 42mm, and 150mm f/3.6 also has an aperture diameter of 150mm / 3.6 = 42mm.  So, f/1.8 on mFT is equivalent to f/3.6 on FF, where by "equivalent to", I mean results in the same diagonal angle of view, the same DOF for a given perspective, framing, and display size, and projects the same total amount of light on the sensor for a given shutter speed, resulting in the same noise for equally efficient sensors.

the 45mm is pretty damn close.

Great Bustard wrote:

tinternaut wrote:

In terms of DoF, yes it's a 3.6 equivalent...

I'm glad you agree.

...but in terms of light gathering (i.e. the shutter speed it enables one to achieve at a given aperture for correct exposure)...

Anders explained it nicely in his post above, but I'll take a stab at it, too. The exposure is the amount of light per area that falls on the sensor, not the amount of light gathered. So, f/1.8 on mFT results in 4x as much light per area as f/3.6 on FF, but a FF sensor has 4x the area, so the two neatly cancel each other out, and the light gathering of f/1.8 on mFT is the same as the light gathering as f/3.6 on FF.

I hear this often, but I don't think its correct. An imaging sensor needs to compute the difference in the number of photons hitting one site and the number hitting another site. The average is easily taken over the entire sensor. Having a 4 times larger sensing pixel does not give you 4 times better ability to compute that difference. Also its worth noting that a lot of image noise is not poisson photon counting; rather it is thermal noise related to electrical currents required to make the sensor work, and that does not scale in the same was as noise owing to pixel size.

Now I totally agree that all else equal, ff according to theory will have higher s/n than a sensor with smaller pixels. I just think the scaling with pixel size is complicated.

...surely 1.8 is 1.8.

Well, 75mm f/1.8 has an aperture (entrance pupil) diameter of 75mm / 1.8 = 42mm, and 150mm f/3.6 also has an aperture diameter of 150mm / 3.6 = 42mm. So, f/1.8 on mFT is equivalent to f/3.6 on FF, where by "equivalent to", I mean results in the same diagonal angle of view, the same DOF for a given perspective, framing, and display size, and projects the same total amount of light on the sensor for a given shutter speed, resulting in the same noise for equally efficient sensors.

thk0 wrote:

Great Bustard wrote:

tinternaut wrote:

In terms of DoF, yes it's a 3.6 equivalent...

I'm glad you agree.

...but in terms of light gathering (i.e. the shutter speed it enables one to achieve at a given aperture for correct exposure)...

Anders explained it nicely in his post above, but I'll take a stab at it, too. The exposure is the amount of light per area that falls on the sensor, not the amount of light gathered. So, f/1.8 on mFT results in 4x as much light per area as f/3.6 on FF, but a FF sensor has 4x the area, so the two neatly cancel each other out, and the light gathering of f/1.8 on mFT is the same as the light gathering as f/3.6 on FF.

I hear this often, but I don't think its correct.

It is. The photos here demonstrate it fairly well:

http://www.dpreview.com/forums/post/51782704

An imaging sensor needs to compute the difference in the number of photons hitting one site and the number hitting another site. The average is easily taken over the entire sensor. Having a 4 times larger sensing pixel does not give you 4 times better ability to compute that difference.

Pixel size has a secondary, and minor, effect on image noise, as a comparison between the D600 and D800 will clearly demonstrate.

Also its worth noting that a lot of image noise is not poisson photon counting; rather it is thermal noise related to electrical currents required to make the sensor work, and that does not scale in the same was as noise owing to pixel size.

If you mean read noise, yes, that absolutely has an effect (more so at higher ISOs such as those in the link above), but this all comes under the "equally efficient sensors" clause for noise equivalence.

By the way, we would all agree that a 75 / 1.8 on an EM5 is equivalent to a 75 / 1.8 on the more noisy G1, right?  That is, we wouldn't say that f/1.8 on an EM5 is "equivalent to" f/1.4 on a G1, or something like that simply because the noise levels were not exactly the same, would we?

Now I totally agree that all else equal, ff according to theory will have higher s/n than a sensor with smaller pixels.

For a given exposure, but the same noise for a given DOF and shutter speed (all else equal, of course).

I just think the scaling with pixel size is complicated.

Not complicated, but largely irrelevant, again, as a comparison between the D600 and D800 will clearly demonstrate.

David Kieltyka wrote:

Just to note...the lens doesn't give a rat's dupa what the characteristics are of the sensor or film frame it's mounted in front of.

Correct.  But the visual properties of the resulting photo sure as heck do.

If the ratio of focal length to max. aperture diaphragm size is 1.8:1 then the lens is a 1.8. Period (allowing for manufacturing tolerances and lens maker rounding/exaggerating).

Except what does that have to do with the visual properties of the resulting photo?

It's certainly a good thing to know that larger photosites tend to be more efficient both at photon collection and in turning those photons into legit image-generating electrons.

Except that's not true, as the D600 vs D800 clearly demonstrates.

It's also a good thing to know that using a larger sensor/film frame yields less DOF for a particular field-of-view & f-ratio compared to a smaller sensor/film frame.

And perspective and display size.  But the same DOF for the same perspective, framing, aperture (entrance pupil) diameter, and display size.

You know how they compute the f-ratio for a lens, right?  They take the quotient of the focal length and aperture diameter.  For example, the max aperture diameter for the 75 / 1.8 is 42mm, so the computed f-ratio is 75mm / 42mm = 1.8.

(Whether or not this second feature is an advantage or disadvantage is up to the individual photographer.) But when taking photos with an m43 camera/lens combo it's also irrelevant. The format is what it is...you go with it and work within its boundaries, same as any other format including 35mm.

Sure.  But when comparing different formats, it sure helps to understand the meaning of the numbers and how they relate to the visual properties of the photo.

David Kieltyka wrote:

Great Bustard wrote:

David Kieltyka wrote:

Just to note...the lens doesn't give a rat's dupa what the characteristics are of the sensor or film frame it's mounted in front of.

Correct. But the visual properties of the resulting photo sure as heck do.

No argument.

If the ratio of focal length to max. aperture diaphragm size is 1.8:1 then the lens is a 1.8. Period (allowing for manufacturing tolerances and lens maker rounding/exaggerating).

Except what does that have to do with the visual properties of the resulting photo?

Geez, it has to do with the properties of the lens. You know...the subject of my post.

As a photographer, I tend to judge lenses in the context of the photos they produce with the cameras they are used with.  For example, if I just had a lens, no matter how amazing that lens was, it would be useless without a camera to go with it.

It's certainly a good thing to know that larger photosites tend to be more efficient both at photon collection and in turning those photons into legit image-generating electrons.

Except that's not true, as the D600 vs D800 clearly demonstrates.

I did say "tend to be." There are no absolutes in this stuff.

Doesn't even "tend to be".  In fact, what "tends to be" is that pixel size plays little to no role at all.

It's also a good thing to know that using a larger sensor/film frame yields less DOF for a particular field-of-view & f-ratio compared to a smaller sensor/film frame.

And perspective and display size. But the same DOF for the same perspective, framing, aperture (entrance pupil) diameter, and display size.

No argument here either...but when taking photographs I don't care about the entrance pupil diameter. I care about the f-ratio.

That's because when you're taking photographs, you're not comparing your equipment to another format.

You know how they compute the f-ratio for a lens, right? They take the quotient of the focal length and aperture diameter. For example, the max aperture diameter for the 75 / 1.8 is 42mm, so the computed f-ratio is 75mm / 42mm = 1.8.

Doesn't "ratio of focal length to max. aperture diaphragm size" (see above) cover this?

Yes.

(Whether or not this second feature is an advantage or disadvantage is up to the individual photographer.) But when taking photos with an m43 camera/lens combo it's also irrelevant. The format is what it is...you go with it and work within its boundaries, same as any other format including 35mm.

Sure. But when comparing different formats, it sure helps to understand the meaning of the numbers and how they relate to the visual properties of the photo.

A shame the original poster even mentioned the dreaded 35mm format "equivalent." More info is a good thing. But so is less pedantry.

Over and most emphatically out.

I always think that understanding the context of the numbers, in connection to the visual properties of the resulting photo, is a good thing.  Can't speak for others on that point, though.

PaulM2 wrote:

David Kieltyka wrote:

Just to note...the lens doesn't give a rat's dupa what the characteristics are of the sensor or film frame it's mounted in front of. If the ratio of focal length to max. aperture diaphragm size is 1.8:1 then the lens is a 1.8. Period (allowing for manufacturing tolerances and lens maker rounding/exaggerating).

It's certainly a good thing to know that larger photosites tend to be more efficient both at photon collection and in turning those photons into legit image-generating electrons. It's also a good thing to know that using a larger sensor/film frame yields less DOF for a particular field-of-view & f-ratio compared to a smaller sensor/film frame. (Whether or not this second feature is an advantage or disadvantage is up to the individual photographer.) But when taking photos with an m43 camera/lens combo it's also irrelevant. The format is what it is...you go with it and work within its boundaries, same as any other format including 35mm.

-Dave-

With no intent to hi-jack the original post, my 2 cents worth.

I play with old legacy glass from my old film cameras and the above discussion, I have read with great interest.

I have the adapter that allows me to mount, say my 45mm/f2.0 Minolta film lens, to my PEN. With the 2x crop factor in mind, my angle of view or should I say DOF, is equivalent to a 90mm on a 35mm film camera. But the f stop is still f/2.0, as I did not change any of the physical aspects of the lens itself? Its still the same old lens I used many years ago. (the focal length of the lens and the physical size of the aperture determines the "f" number)

Indeed, the lens did not change.  That is, it's still a 45 / 2 -- it didn't change into a 90 / 2 or a 90 / 4.  However, photos at 45mm f/2 on mFT will look remarkably similar to photos at 90mm f/4 on FF -- much more similar than they will to photos at 90mm f/2 on FF.

I fought with this concept in my mind for some time. But when I consider that I must shoot this old stuff in manual mode, set the focus, aperture, shutter speed, ISO, I am lucky to get the first exposure even close.

The exposure will be the same on all formats at all focal lengths for a given f-ratio and shutter speed (ignoring differences in t-stops, of course).  However, the total amount of light falling on the sensor will be very different (as will the DOF).  Of course, if you are not comparing formats, this doesn't matter.

And as a added note, this old lens on the PEN does not work as well as it does on my DSLR. Could be a problem with the adapter? But it was fun to try.

Does not work as well how?

Again, thanks to all who have posted. Very interesting.

Glad to hear it!