Exposure via a black glass

[berci]

Hey Everyone and Everyone else,

If I set the aperture to f22 and put a dark (density 3.0, 10 f stops) ND filter in front of it what f number do I get?

Is it f22 + 10 f stops = f720?

 

[ClaremontPhoto]

As I understand it ND 3x is three f stops. So f22 on the lens becomes f64 effectively, although a through the lens metering system will see the ND filter and need no compensation.

If you're after long exposure times you can add more ND filters or use a pinhole camera. My pinhole is from Zero Image and is f138.

Edit: ND 3x may be eight f stops. Sorry for any confusion.

 

[EmilGil]

Each stop represent half the light or dividing the area of the open aperture by 1.4 (sqrt(2)). 10 stops is then (sqrt(2))^10=2^5=32. Multiplying f/22 with 32 yields 704 so f/720 sounds reasonable as the f-stops don't follow the powers of 1.4 perfectly.

 

[vicmortelmans]

density is in 10-base logarithm, so density 3 means that a fraction of 1/1000 of the light passes through. That's more or lesse equivalent to 10 f-stops, being actually a fraction of 1/1024.

And if I calculate right, f22 + 10 stops = f704

Groeten,

Vic

 

[MadMan2k]

Aperture affects depth of field and the rendering of pin light sources, so it's still f/22, but with the EV of the light entering the lens reduced that number of stops.

 

[sheepdog]

I suppose that one could say that the effective transmission of the collective optic is now f22 [T704], using the T-stop designation popularised among cinematographers to describe the amount of light hitting the film..

Kind regards
Kjetil