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Constant 2.8 is faster than the variable aperture lens you posted. It is equivalent to the GMaster 24-70 or the canon L 24-70 at less than half the price. Its

by adrr 6y ago

Constant 2.8 is faster than the variable aperture lens you posted. It is equivalent to the GMaster 24-70 or the canon L 24-70 at less than half the price. Its not on the same level as a Tamron, it’s made of machined aluminum and feels solid. Tamron is made out of plastic.

sudosysgen 6y ago
No, it isn't. Constant 2.8 at M43 is exactly equal to a constant f/5.6 on FF. Indeed, at the FoV equal to 50mm on FF, you would need 25mm on M43. At f/5.6, the aperture diameter of the 50mm FF lens is 50/5.6=8.93, and the M43 lens would have an aperture diameter of 25/2.8=8.93mm. This means that the lenses are equivalent. Indeed, the amount of light captured by a lens from a point source is proportional to the angle of the cone with its tip at the point and it's disk at the aperture. And so, since the FoV is the same, you will be at the same distance from the point for the same image, and since the diameter of the aperture is the same, the two cones will be identical, and thus the amount of light will be the same. For depth of field, the calculation is similar, and they are indeed also equal. And so, an f/5.6 aperture on FF is equal to an f/2.8 aperture on an M43 lens of the same FoV. As a conclusion, your 12-40mm f/2. 8 lens is exactly equivalent to a 24-80 f/5.6 lens on FF. And so, a 28-75 f/3.5-5.6 lens on FF is faster in average but has a slightly smaller zoom range. In the end, the variable aperture lens is optically superior, at a third of the price. As for the Tamron, it is indeed made of plastic, but I like this because it makes it lighter and barely less resistant as the G-Master. It is still capable of gathering four times more light. If you want to actually test this, we can print two test charts and put them one meter from the camera then one meter from each other in a line, and you can take a picture of them at 25mm f/2.8 on your Olympus, I'll take the same picture at 50mm f/5.6 on my A7ii, and you'll see that the depth of field will be exactly the same. If we had a way of controlling lighting, we would see that at the same shutter speed the noise levels of the two images will be very similar too.
- adrr 6y ago
  DOF is different than speed of the lens. F value for aperture is standard no matter the size of sensor and sensor will get the same amount of light if the shutter speed are the same.
  - sudosysgen 6y ago
    I explained to you exactly why this is incorrect. The amount of light the sensor catches from a point source depends on the diameter of the aperture and the distance from the point. So, at the same FoV, the diameter of the aperture is what matters. The light intensity at the same f number is indeed the case, however, since the image is smaller, then amount of light is four times smaller for an M4/3 lens at the same FoV. Another way to understand this is to realize that the focal length is smaller, therefore the image for the same object is smaller, therefore despite an equal light intensity the total light is the same. DoF is not actually any different from lens speed. There is a physically indestructible link where increasing the amount of light from the same object reduce the depth of field. DoF and amount of light are two faces of the same coin. I highly suggest actually doing the math. It's a very counter-intuitive concept. Take a point source of light, take some distance, then calculate the angle of the triangle formed by the point and the opposite points on the aperture. This angle is directly proportional to the light. You will see that the angle is equal in both. Then, to explain why the ISO is going to be four times as high in the FF camera, calculate the intensity of the image by taking the source to now be a sphere, and calculating the image size using the thin lens equation. You will se that the image will be twice the size with the 50mm FF lens, and thus the light intensity will be half that, which is why ISO is going to be four times higher in the FF camera. This is also why you can get away with ISOs four times on full-frame than on M4/3. Again, if you don't believe me, you really have to do the math in your own, failing that you should do an experiment and look at the noise levels at the same shutter speed at f/2.8 vs f/5.6 in the same speed with equivalent focal lengths. You will see.
    - deleted 6y ago
      [deleted]
    - colanderman 6y ago
      Parent (sudosysgen) is correct, I don't know why they're being downvoted. Another way to look at it is, if you have two setups with different sensor sizes, but the same relative aperture (i.e. f number), same ISO, same sensor resolution, and equivalent focal lengths (i.e. scaled by crop factor), you will have more noise in the image from the smaller sensor. As the parent stated, this is purely due to physics: the lens in the setup with the smaller sensor has a smaller absolute aperture; thus less total light is entering the camera.

by adrr 6y ago

sudosysgen 6y ago

No, it isn't. Constant 2.8 at M43 is exactly equal to a constant f/5.6 on FF. Indeed, at the FoV equal to 50mm on FF, you would need 25mm on M43. At f/5.6, the aperture diameter of the 50mm FF lens is 50/5.6=8.93, and the M43 lens would have an aperture diameter of 25/2.8=8.93mm. This means that the lenses are equivalent. Indeed, the amount of light captured by a lens from a point source is proportional to the angle of the cone with its tip at the point and it's disk at the aperture. And so, since the FoV is the same, you will be at the same distance from the point for the same image, and since the diameter of the aperture is the same, the two cones will be identical, and thus the amount of light will be the same. For depth of field, the calculation is similar, and they are indeed also equal. And so, an f/5.6 aperture on FF is equal to an f/2.8 aperture on an M43 lens of the same FoV. As a conclusion, your 12-40mm f/2. 8 lens is exactly equivalent to a 24-80 f/5.6 lens on FF. And so, a 28-75 f/3.5-5.6 lens on FF is faster in average but has a slightly smaller zoom range. In the end, the variable aperture lens is optically superior, at a third of the price. As for the Tamron, it is indeed made of plastic, but I like this because it makes it lighter and barely less resistant as the G-Master. It is still capable of gathering four times more light. If you want to actually test this, we can print two test charts and put them one meter from the camera then one meter from each other in a line, and you can take a picture of them at 25mm f/2.8 on your Olympus, I'll take the same picture at 50mm f/5.6 on my A7ii, and you'll see that the depth of field will be exactly the same. If we had a way of controlling lighting, we would see that at the same shutter speed the noise levels of the two images will be very similar too.

adrr 6y ago
DOF is different than speed of the lens. F value for aperture is standard no matter the size of sensor and sensor will get the same amount of light if the shutter speed are the same.
- sudosysgen 6y ago
  I explained to you exactly why this is incorrect. The amount of light the sensor catches from a point source depends on the diameter of the aperture and the distance from the point. So, at the same FoV, the diameter of the aperture is what matters. The light intensity at the same f number is indeed the case, however, since the image is smaller, then amount of light is four times smaller for an M4/3 lens at the same FoV. Another way to understand this is to realize that the focal length is smaller, therefore the image for the same object is smaller, therefore despite an equal light intensity the total light is the same. DoF is not actually any different from lens speed. There is a physically indestructible link where increasing the amount of light from the same object reduce the depth of field. DoF and amount of light are two faces of the same coin. I highly suggest actually doing the math. It's a very counter-intuitive concept. Take a point source of light, take some distance, then calculate the angle of the triangle formed by the point and the opposite points on the aperture. This angle is directly proportional to the light. You will see that the angle is equal in both. Then, to explain why the ISO is going to be four times as high in the FF camera, calculate the intensity of the image by taking the source to now be a sphere, and calculating the image size using the thin lens equation. You will se that the image will be twice the size with the 50mm FF lens, and thus the light intensity will be half that, which is why ISO is going to be four times higher in the FF camera. This is also why you can get away with ISOs four times on full-frame than on M4/3. Again, if you don't believe me, you really have to do the math in your own, failing that you should do an experiment and look at the noise levels at the same shutter speed at f/2.8 vs f/5.6 in the same speed with equivalent focal lengths. You will see.
  - deleted 6y ago
    [deleted]
  - colanderman 6y ago
    Parent (sudosysgen) is correct, I don't know why they're being downvoted. Another way to look at it is, if you have two setups with different sensor sizes, but the same relative aperture (i.e. f number), same ISO, same sensor resolution, and equivalent focal lengths (i.e. scaled by crop factor), you will have more noise in the image from the smaller sensor. As the parent stated, this is purely due to physics: the lens in the setup with the smaller sensor has a smaller absolute aperture; thus less total light is entering the camera.