A camera is being used with the correct exposure at f/4 and a shutter speedof 1/

Question

A camera is being used with the correct exposure at f/4 and a shutter speedof 1/32s. In order to "stop" a fast moving subject, the shutter speed is changed to 1/256s. Find the new f stop that should be used to maintain satisfactory exposure, assuming no change in lighting conditions. The Answer is f/1.4.

I understand the first part. (1/256)/(1/32) so we know the exposure is reduced by a factor of 8 (8 times larger aperature).

I also know that this involves the f number = f/D formula.   D is the diameter of the lens. The rest of the teachers explaination don't make sense (he skips multiple steps a lot). So I don't understand how to take that I know the change will be 8 fold and produce the f/1.4 answer. Thanks for the help!

Explanation / Answer

Aperture openings are measured as fractions of the focal length of a lens. That is what the 'f' stands for in the aperture rating, 'focal length'. Assuming we have the epitome of lenses, the 50mm, with an aperture of f/2.8, we can determine the actual diameter of the aperture opening like so:

50mm / 2.8 = 17.85mm

If we open the aperture up to its maximum of, say, 1.4, we can measure that as well:

50mm / 1.4 = 35.71mm

The difference between an aperture of f/2.8 and an aperture of f/1.4 is a difference of four times as much light...or twostops. We know this because the area of the aperture opening itself is four times as large at f/1.4 (1001.54mm^2) as it is at f/2.8 (250.25mm^2). A stop in photography nomenclature means a difference of oneexposure value, which is the doubling, or halving, of the amount of light reaching the sensor.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.