On my camera, between the mode dial and the flash, there is a white theta symbol. I've also seen it on review sites in pictures of other cameras, often in different locations but always near the pop-up flash. Why is it there, is it useful, and if so, how?
\$\begingroup\$
\$\endgroup\$
6
-
1\$\begingroup\$ That looks like a phi, not a theta. \$\endgroup\$– Konrad RudolphCommented Feb 19, 2011 at 12:11
-
9\$\begingroup\$ Actually it looks more like a plimsoll mark than any greek letter: en.wikipedia.org/wiki/Plimsoll_line#Load_line - When you put the camera in water, it should not sink below that line. ;) \$\endgroup\$– GuffaCommented Feb 19, 2011 at 13:03
-
5\$\begingroup\$ @Guffa: the camera does sink below that line. At least my EOS 40D did ;) \$\endgroup\$– Fredrik MörkCommented Feb 19, 2011 at 17:39
-
\$\begingroup\$ Or a diameter sign, ⌀ in unicode. fileformat.info/info/unicode/char/2300/index.htm \$\endgroup\$– mattdmCommented Apr 17, 2011 at 14:32
-
\$\begingroup\$ possible duplicate of What is an "image plane indicator"? \$\endgroup\$– mattdmCommented Mar 30, 2012 at 23:38
|
Show 1 more comment
1 Answer
\$\begingroup\$
\$\endgroup\$
3
This indicates the plane that the image sensor lies in, so you can use to calculate the exact distance between your subject and your "film".
-
2\$\begingroup\$ Or in other words, there the surface of your sensor is relative to the outside of the camera body for the reasons davr has given \$\endgroup\$– JamWheelCommented Feb 19, 2011 at 12:09
-
10\$\begingroup\$ This mark is used, by the way, for calculating mostly in macrophotography. The exact distance between the subject and the film plane/sensor is used to calculate the reproduction ratio (the relative size of the subject on the sensor) and exposure compensation. (The aperture you set on the lens is relative to the focal length of the lens, which is the length of the light path when the lens is focused at infinity. At macro distances for most lenses, the effective length of the lens is longer, so the effective aperture is smaller. If you are not metering through the lens, you need to compensate.) \$\endgroup\$– user2719Commented Feb 19, 2011 at 16:20
-
\$\begingroup\$ @Stan Rogers thanks for that great explanation, I was wondering who needed that precision when measuring the distance to a subject but never even thought of macro. \$\endgroup\$– geriksonCommented Feb 19, 2011 at 17:49