As long as the distance to the subject is small compared to the hyperfocal
distance, both before and after, the DOF will be halved. The doubling of the
f# willdouble the DOF, but the doubling of the focal length will quarter it,
for a net halving. For near hyperfocal cases, it's more complicated.
Details:
DOF = d**2/(h-d) + d**2/(h+d)
where h = F**2/(c*f) is hyperfocal distance
If d is small compared to h, then
DOF ~ 2*d**2/h = 2*c*f*d**2/F**2
leaving circle of confusion (c) and d unchanged,
DOF is proportional to f/F**2
Doubling f and focal length (F), halves DOF.
----- Original Message -----
From: "Richard F. Man" <richard@xxxxxxxxxxxxxx>
To: "oly" <olympus@xxxxxxxxxxxxxxx>
Sent: Sunday, July 28, 2002 6:44 PM
Subject: [OM] teleconverter and DOF
> I understand that for example, the 2X converter loses 2 stops of light. Q
> is whether that would increase the DOF accordingly. My intuition is that
it
> doesn't.
>
> OM content: this is about the Zuiko teleconverter, in case it matters :-)
>
> // richard http://www.imagecraft.com
> [ For technical support, please include all previous replies in your
msgs. ]
>
>
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