edit: sorry, forgot to specify this is a macro question

I ordered a D850 so that I can print larger images at high ppi, but I am wondering if I am going to see a significant reduction in reproduction ratio by doing this, or will I just lose working distance? I haven't used a full frame before, and the D7500 is a DX. Anyone have any insight?


There is no significant difference between the two camera's you mentioned when it comes to catching detail. The D7500 should be a bit better at it, but the difference is only about 3%.

The reproduction ratio does not change. It is a lens property, so wit the same lens, it will always stay the same. The sensor of the D850 is much bigger, but fortunately it also has a lot more pixels. The amount of pixels that are used by your subject will be about the same.

Suppose you are taking a picture of a fruit fly that is about 4 mm long. At a 1:1 reproduction ratio, it will also be 4 mm long on the sensor. On the D7500 sensor it will be roughly 1030 pixels long, and on the D850 it will be about 1000 pixels long. Not a real difference. But the D7500 sensor is 5,568 pixels long, so you have 4,565 pixels before or after the fly. The D850 sensor is 8,256 pixels long, so you have 7,256 pixels before or after the fly. With both camera's you can print the fly with the same size and quality, but if you do that, the D850 will give a larger photo which shows much more of the surroundings of the fly. You are not losing anything, you just get more of the surrounding with the D850.

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    The difference is only about 1.15%. The D850 has about 229.3333 pixels per millimeter of sensor width, the D7500 has about 232 pixels per millimeter of sensor width. – Michael C Nov 6 '17 at 10:02
  • Right which isn't a huge issue, my main goal was to be able to print 36x24 at a higher ppi than the d7500 can (154), 229 isn't bad per se. Thank you for the answer, Rick, very straight and to the point. The "...4mm long on the sensor..." really helped with the perspective. – d4rk s1gm Nov 10 '17 at 19:30

The lens projects the same image size regardless of the sensor behind it. The size of the sensor determines how much of the lens' image circle is being captured in the photograph. At the minimum focus distance the size of an object in focus will be projected the same size on either sensor. If the reproduction ratio of the lens at MFD is 1:1 (the same thing as a 1.0X MM), a 10mm long object will have an image projected on the sensor that is 10mm long. With an APS-C sensor it will be projected 10mm long onto a sensor that is 24x16mm in width and height. With a FF sensor it will be projected 10mm long onto a sensor that is 36x24mm in width and height.

Where the difference lies is when one displays an image from a smaller sensor and an image from a larger sensor at the same display size.

  • If both images are displayed at 8x12 inches, the image from the crop sensor is enlarged by a (1.5X) greater ratio than the image from the FF sensor.
  • If, on the other hand, one enlarges both images by the same amount the image from the crop sensor would be displayed at 8x12 inches while the image from the FF sensor would be displayed at 12x18 inches (1.5X larger). If one then looked at the central 8x12 inches of the FF image currently displayed at 12x18 inches, one would see the subject at the same size as one sees it in the APS-C image.

When viewing an image at 100% (one image pixel per one screen pixel) on a computer monitor, pixel density will come into play because the number of pixels per inch or mm on the sensor will determine the enlargement ratio when it is displayed on a monitor with a specific pixel pitch. In this regard you will lose a small amount of resolution going from the D7500 sensor that is 5,568 pixels long in 24mm versus the D850 that is 8,256 pixels long in 36mm. The D7500 has 232 pixels per millimeter. The D850 has 229.3333 pixels per millimeter. Thus the 10mm long object will be ≈1.15% smaller when displayed on the same monitor at 100% from an image taken with the D850 than with an image taken with the D7500.

If the images from each camera are displayed so that the full width and height of the image from each camera fits on the monitor's full screen, then the higher resolution (8,256 pixels wide) image from the D850 will make the 10mm object appear to be 1.5X smaller than the smaller resolution (5,568 pixels wide) image from the D7500.


Going from APS-C (DX) to Full-Frame (FX), your sensor is going to be bigger, so at the same focus distance your magnification ratio logically will be lower.

The only thing you can do is get closer but lenses have a minimum focus distance, so that is only possible up to a point. If you were already at the minimum focus distance for your lens, then you will need to get a higher magnification lens to compensate.

  • Thanks, I thought that might be the case. I think currently my effective ratio is 1.63:1 (100mm FX on DX = 150 + 93mm tubes). If I use the name setup on FX, (100mm FX + 93 mm tubes) I would get 1.93:1. Would it still appear smaller, or am I using misguided thought. – d4rk s1gm Nov 3 '17 at 17:30
  • Yes, your subject would cover a smaller portion of the frame that is what the lower magnification means. – Itai Nov 3 '17 at 17:41
  • Ah okay. I redid my math and found that right now I am at 2.43, but would be at 1.93 with full frame. Isn't the reproduction ratio the size of the subject on the sensor, though?Edit: hit enter on accident. – d4rk s1gm Nov 3 '17 at 17:41
  • Magnification is the ratio between the size of the subject on the sensor and the actual subject size. – Itai Nov 3 '17 at 18:00

You said macro, and speaking of macro, ratios like 1:1 are exactly the SAME size and SAME reproduction ratio on any size sensor. 1:1 is 1 to 1. 1:1 specifies the ratio, the object is same real life size in the sensor image, on both sensors if both are at 1:1 (a U.S. Penny coin will be 3/4 inch diameter on both sensors). The lens set to 1:1 does 1:1 regardless of whatever camera body is attached. The larger sensor will simply show a larger view around it (not cropped by sensor).

Speaking of a normal photo scene (not macro), then if with the same lens from the same location spot and distance (one way to compare them), again, the same lens does the same thing. The larger sensor does see a larger wider view (a larger frame), but within that frame view, the objects are the same size (same lens at same spot). Because the same lens does the same thing. The ratio of the frames is larger, but the ratio of the scene objects are exactly the same (with the same lens and distance).

But in the normal world, if trying to get the "same picture" to compare, we use a longer lens on a larger sensor from same spot, or stand back farther with the smaller sensor and same lens, to get a similar view (to compare the entire frame). Both focal length or distance are a multiple of the crop factor. That's so they both show the same field of view, the same picture. Then in THAT case, the larger sensor showing same scene would show smaller objects in a larger frame (of same scene). In THAT case, the larger sensor would see a smaller reproduction ratio. A large factor is that the lenses and/or distances are not the same then, which may be chosen due to the sensor size, but the difference is not caused by the sensor.

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