Recently I went out to shoot when it was snowing. I was playing around with different aperture sizes and exposure times. What I have noticed that the wider the aperture, the more snow I can capture compared with having a small aperture and longer exposure.

Why is this?

I've attached some examples (all with 100 ISO and 50mm):

f/3.2 @ 0.5sec

f/14 @ 4.0sec

  • 1
    You've changed not only aperture, but shutter as well. You need to try all combinations of aperture-shutter, and compensate with ISO. This question can be rephrased as "Why does faster shutter make snowflakes look bigger?" just as well.
    – Agent_L
    Mar 1, 2018 at 10:22

4 Answers 4

  1. The use of a large diameter, f/3.2 aperture allowed you to set the shutter at ½ second. You captured snow as it tumbled down. The shutter speed is insufficient to freeze the falling snow. The flakes you captured were moving, thus the camera recorded them as streaks. More importantly, the f/3.2 aperture yields a shallow depth-of-field. If you look closely, the snowflakes as well as the car grill are out-of-focus. The blurred snowflakes appear larger and indistinct because they are out of focus.

  2. The shot at f/14 has expanded depth-of-field. The snowflakes are rendered sharp. However, the shutter was open for 4 seconds. The snowflakes being in motion imaged as an elongated but in-focus streak. The fact that the streak is in-focus reduces the width of the band.

  • 3
    And the distance it travels during the exposure also reduces the intensity of the band. It would be useful to see this same experiment performed with ISO instead of exposure time.
    – dgatwood
    Feb 28, 2018 at 18:04
  • @ - dgatwood It is not the distance, it is exposure (dwell) time. More dwell time = more exposure and a wider streak. As to streak size: Out-of-focus objects yield larger images. True if the focus is too short or too long. This is because the image forming rays trace out a cone of light. Sharp focus occurs if the apex of the cone kisses the surface of film or digital sensor. If the cone of image forming rays fall too short or too long, the result is the same, the out-of-focus image will be slightly enlarged. Feb 28, 2018 at 18:39
  • 1
    I think you misunderstood my point. Because the speed of the snowflake is roughly constant, for any given pixel on the sensor, the snowflake blocks the bumper for a fixed amount of time. As the exposure increases, the percentage of the exposure during which each pixel's view of the car was occulted by the snowflake decreases, and the percentage of time in which the background was visible increases. Thus, there is higher signal from the bumper relative to the noise from the snowflakes, making the snowflakes less intense, in part, because they traveled farther during the (longer) exposure.
    – dgatwood
    Feb 28, 2018 at 19:13

The effect you see has nothing to do with the aperture.

The snow flakes aren’t larger. They are just more visible because they are less blurred due to the shorter exposure time.

In the second photo a longer exposure is used which blurs the movement more than in the first photo. If you went with even a slower shutter speed, you could almost make them disappear.

See also: How do moving objects disappear when exposed for a longer duration


A wider aperture lets more light in so snowflakes register faster on the sensor. When you close down the aperture, the exposure also gets longer, so each snowflake makes a longer path, in essence it is like spreading the snowflake on the sensor, the longer the spread the thinner it gets.

  • This can't be right. If you roll a ball past a camera that's making a long exposure, adjusting the exposure time, the speed of the ball or anything else will still give you a smeared track whose height is equal to the diameter of the ball. A snowflake is just a small, irregularly shaped ball. Feb 28, 2018 at 17:03
  • 1
    But the longer the exposure, the less pronounced that track is. Given a long enough exposure, that bit of noise from the rolling ball will be overwhelmed by the signal to the point that nobody will notice it. That's why people use extremely long exposures to make people disappear in landscape photos.
    – dgatwood
    Feb 28, 2018 at 17:58
  • @DavidRicherby - Yes it will give you a track the same size but it is dimmer which makes appear less and eventually even completely invisible, so some snowflakes would not appear at all while others will make a less impression in the sensor.
    – Itai
    Feb 28, 2018 at 21:14
  • @dgatwood But the ball will disappear by getting fainter not narrower. Feb 28, 2018 at 21:18
  • I kind of assumed that by "thinner", the original author meant "lighter". That said, the streaks will also be perceived as being more narrow, because an out-of-focus object looks brightest in the center, falling off more and more the farther you get into the blur region. So as you decrease the relative intensity of the snowflake, areas farther from the center will fall below the threshold of perception sooner than areas closer to the center.
    – dgatwood
    Feb 28, 2018 at 22:18

There's a couple of things going on here. One is a optical effect and the other is a illusory one:

  1. A larger aperture narrows your depth of field, blurring out-of-focus snowflakes and making them register across a larger area on the sensor (corresponding to the degree of blurring.)

  2. A larger aperture lets in more light, brightening the snowflake and making them appear larger; this however is an optical illusion and the actual width on the image is unchanged. Note that usually exposure time is used to counteract this brightness change, but it does not work for rapidly moving objects as the effective exposure is limited by the movement rather than the shutter (in other words, the light from a snowflake registers on a particular pixel for only a fraction of a second regardless of whether the shutter is open for 0.5 seconds or 4 seconds.)

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