Is there any lens filter or any other technique, that I can use to photograph lakes, oceans or any water source, without the reflections on it?

Every photo of a water source can be seen with a reflection of any objects or the reflection of sky. I just need to capture the lake without any reflections on water.

What is the technique?

  • \$\begingroup\$ do you have an example of what you're expecting the image to look like? \$\endgroup\$ Jul 12, 2017 at 7:01

3 Answers 3


Use a polarizing filter, which will filter out light polarized in a certain direction (turn the filter until the desired effect is achieved).

This works, because reflected light is polarized in the plane of the surface it is reflected from.

Note that the sky's blue is also polarized, you'll see a distinct graduation when you look at different angles to the sun.

Here you can see two pictures i recently shot (smartphone through sunglasses, please don't judge me) to test the effect.

non polarized

The first picture shows reflections in the water and the window; the polarizer is turned in the wrong direction.


In this picture, the filter is aligned with the reflections from the water, which are greatly reduced.

  • \$\begingroup\$ Polarizing filter will remove scattered light to some extent. However, if you take a picture of a still pond w/ the reflection of the far shore , a polarizing filter will do just about bupkis for you. What you've eliminated here is glint, not resolved reflections of distant objects. \$\endgroup\$ Jul 12, 2017 at 11:19
  • \$\begingroup\$ @CarlWitthoft how do you explain that en.wikipedia.org/wiki/Brewster%27s_angle#/media/… then? (well, it's a "polorizer" according to the filename, but still) \$\endgroup\$
    – ths
    Jul 12, 2017 at 12:04
  • \$\begingroup\$ I'm well aware of Brewster's Angle rules as well as the fact that it only indirectly applies here. So long as the far object (shoreline or whatever) is in a position that a mirror would relay it to the camera, the surface of the lake will do so as well. The preferential polarization of the reflection is not strong enough to allow a polarizer to remove the image. \$\endgroup\$ Jul 12, 2017 at 12:17
  • \$\begingroup\$ @CarlWitthoft why does the image of the tree in the window disappear then? \$\endgroup\$
    – ths
    Jul 12, 2017 at 12:22
  • 1
    \$\begingroup\$ Your eyes must be better than mine :-) as I can't see a tree-ish thing in either image. It's possible that the ghost image of the windows is a back-surface reflection, in which case the light which passes the front surface is certainly polarization (and angle) dependent. \$\endgroup\$ Jul 12, 2017 at 12:26

If you want to get the effect without photoshop trickery there are couple of factors to think about. Generally reflections off of medium boundary (like water) are governed by Frensell equations that predict that the amount of reflected light is dependent on the orientation of the reflective surface and the polarization of the incoming light.

  1. The effect of the angle of incidence

When photographing water some light is always reflected, but the amount of reflected light is very low below the Brewster angle (53 degrees) and rises quite rapidly above that. What this translates to is that water photographed from above at no more than a 45 degree angle from shooting vertically will have much less reflections. So get up on a high vantage point and shoot down.

Secondly when the water is still it will keep a consistent angle of water toward the camera, a smaller angle (shooting more at a downward angle) is needed for a wavy sea than a still lake.

  1. The effect of polarization

Light polarized at different angles to the plane of incidence reflects in different proportions, generally all light reflects at the same rate when shooting vertically, and then one polarization decreases with increasing angle, until it is not reflected at all at the Brewster angle, after which it starts to be reflected again. Light from the sky is also polarized, most highly at a 90 degree angle from the sun, which affects the amount of reflecting sky at different angles towards the sun.

The first consequence is that a polarizing filter can partially cut out one polarization from the reflection, leaving you with less reflected light, especially at moderate reflection angles and at 90 degree angles relative to the sun (so something like mid-late afternoon and shooting with the sun to the side is usually good enough). This is widely used to get clear shots of people through the front or side windows and decrease the amount of reflected light off of a water surface. Though it will generally not eliminate all reflections, if done correctly they will be imperceptible.

The second consequence is that usually clouds will usually be reflected more than the sky (because they scatter light without polarizing it), and a polarizing filter enhances this effect, so shooting during a cloudless day will give less reflections.

Additional considerations

The last thing to note is that it is easier to achieve all this with longer focal lengths, and positioning the frame in landscape orientation because the angle of incidence to the water will change less across the frame (this can be quite a big factor as if you were to position a typical phone camera in portrait mode the difference in the angle across the frame would be 54 degrees, so typically you go from almost no reflections at the bottom of the frame to a lot of reflections at the top).

  • 1
    \$\begingroup\$ Good to know the scientific aspect of the problem, thanks for mentioning the details. \$\endgroup\$
    – inckka
    Jul 16, 2017 at 17:07

A polarizing filter will do the trick for you. It removes the reflections you mentioned.

Browse online for examples.


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