topic Re: Close focus distance for RF100mm F2.8 L MACRO IS USM in EF & RF Lenses

Close focus distance for RF100mm F2.8 L MACRO IS USM

wes1790 — Mon, 01 Jan 2024 19:57:19 GMT

What is the distance between the subject and lens front [filter ring] at 1.4x maximum magnification for

RF100mm F2.8 L MACRO IS USM?

Does anyone have a chart showing subject size at various lens-to-subject distances?

Re: Close focus distance for RF100mm F2.8 L MACRO IS USM

rs-eos — Mon, 01 Jan 2024 20:26:31 GMT

Note that focusing distances are not typically computed from the front of the lens, but instead from the sensor. Your camera should have a symbol of a circle with a line running through it (typically near the viewfinder).

Pages 13 and 14 of the User Manual should include all the information you're after (to include distances from the from the front of the lens).

Re: Close focus distance for RF100mm F2.8 L MACRO IS USM

wes1790 — Mon, 01 Jan 2024 20:53:31 GMT

I don't care how far it is from the sensor to the subject; what counts is the working distance from the front of the lens to the subject: 0.28' or 3.36" according to the user Manual. Thanks. A chart of subject size and distance from the lens would still be helpful.

Re: Close focus distance for RF100mm F2.8 L MACRO IS USM

shadowsports — Tue, 02 Jan 2024 00:19:06 GMT

Greetings,

I found the information Ricky provided to be helpful. MFD is an industry specification. Since MFD is calculated from the subject to the sensor, this is the information provided by the manufacturer.

The lens is approx 5.8 in. Add a filter its approx 5.825

Mag 1.4x is 10.32 in

Mag 1.0x is 11.04 in

Mag .5x is 15 in

Lens measurement is from the front of the lens to the flange.

Flange distance on an RF lens is 20mm or .787 in

So if you take the lens and filter (if applicable) 5.825 + the distance from the flange to the sensor .787 = 6.612 in

Mag 1.4x is 10.32 in - 6.612 in = 3.708 in

Mag 1.0x is 11.04 in - 6.612 in = 4.428 in

Mag .5x is 15 in - 6.612 in = 8.388 in

The MFD at 1.4x is 10.32. The lens + filter is 5.825 in = 4.495 in. - (minus) the flange distance .787 in = 3.708 in

Someone should probably verify my calculations 🤣