Polarizing Filter advice

I assume its the "filter factor" for this filter - the amount of light it reduces going into the lens, 2.5 f-stops. Your built in meter should take the light reduction into account when metering scenes.

As noted, the filter can reduce light entering the lens by up to 2.5 f-stops. It will vary depending upon the direction the camera is pointing relative to the sun and the position of the filter as you turn it around to get the desired effect. Such filters offer their maximum effect when shooting at right angles to the sun - you can see the effect in the camera viewfinder.

If you're camera has an in-camera meter it will automatically take the light loss into account because the meter only sees the light that comes through filters and lens. If you are shooting in manual, things get a little trickier because you have to figure the light loss in when calculating shutter speed and the amount of loss depends on your angle to the sun. Practice is the only thing that will help in that situation.

Hi Tucker,

Mr. Bondrean is spot on with his advice.

When you mount a filter, the camera's automation will detect that a filter is attenuating (reducing) the exposing light energy. Now the camera using it's chip logic, will apply the necessary compensation and good exposure will be the result.

However, the notation 2.5 engraved on the filter is more likely a filter factor (FF) as this is the convention used to convey the required compensation needed for this specific filter. As you know, a filter modifies the exposing light in some way. In the process some light energy that otherwise would contribute to the exposure will be lost. We must compensate by opening up the lens's aperture or reducing the shutter speed. Because both the aperture and shutter speed are intertwined as to exposure, we can use a combination of both for the compensation. This is why a filter factor value is used to express the needed compensation.

Technically a filter factor is a multiplying factor. To use, we multiply the unfiltered shutter speed by the filter factor. This is easier said than done because typically the shutter speed will be stated as a fraction of a second such as 1/250.

Most of us will have difficulty multiplying the fractional shutter speed by the filter factor. In the case of your filter, assume the shutter speed was 1/250 without a filter. We mount the filter and apply the math: 1/250 X 2.5 = 1/100. Stated differently, if the shutter speed was 1/250, after mounting the filter with filter factor 2.5, the revised shutter speed is 1/100.

This method is easy if the shutter speed is long. Suppose you are making a time exposure 2 seconds long. You mount the filter with an FF or 2.5. The math is 2.5 x 2 =5 seconds.

Because multiplying a fractional shutter speed by a filter factor is difficult, most use a different method. We count on our fingers is powers of 2 thus 2 - 4 - 8 -16 - 32. Each finger represents 1 f/stop compensation. Thus if the filter factor was 8 we count 3 fingers and open up 3 f/stops. If the filter factor is 2 we count one finger and open up one f/stop. In the case of your filter, filter factor 2.5 we open up 1 1/2 f/stop.

Another way to handle the filter factor works with a hand-held meter. We divide the ISO by the filter factor. This forces the meter to do the math for you. This is a handy method that you should put in your bucket of photo knowledge. This simple method serves well when the camera is set on manual and an external meter is used.

So, let me make sure I have this right:
using a manuel FM10:
When I take light reading with the built in light meter,( with the filter on ) and it reads as a "perfect"exposure, I need to still open up the f/ stop 1.5 stops more and have a slightly "over exposed" reading on my meter? Is this the same for negative and slide film?

Tucker,

No. If you are using the camera meter, it will see everything as darker and automatically compensate. The information provided about is only needed if you are using an external meter to meter the scene. Try it. Just see what exposure your camera meter calls for of a scene without the filter and then add the filter and see what exposure the meter now calls for. It should be 2.5 stops more. You didn't have to do anything to adjust the exposure.

The FM10 contains an internal light meter that reads through the lens. Thus when a filter is mounted the density of the filter is detected and compensation is made. Stated another way, you need not take any action and the camera's metering system will consider the filter.

If you re-read my post, I advised manually applying a filter factor when a hand-held meter is being used. This is true because a meter external to the camera is unaware that a filter has been mounted. The best way to handle a detached meter and have it apply a filter factor is to set the meter using a revised ISO. We divide the ISO by the filter factor this forces the detached meter to give a reading that increases the exposure thus compensating for the filters density.

Also be advised that I believe the 2.5 is a filter factor. If this is true, you do not open up 2 1/2 stops to compensate as some have advised. The filter factor value is predicated on the f/stop which operates using a incremental change of 2X in exposure. Thus, filter factor 2 = 1 f/stop compensation. Filter factor 4 = 2 f/stops compensation. Filter factor 8 = 3 f/stops compensation. A filter factor of 2.5 requires just over 1 f/stops compensation.

All this stuff becomes moot if the camera sports a thru-the-lens metering system. Again you need not manually compensate when you mount a filter on a camera equipped with a thru-the-lens metering system.

Just checked the Heliopan SLIM Circular Polarizing filter specifications: They read
Filter Factor between 2.3 and 2.8 (approx. +1.3 stops).
So I was right.

thanks for the thorough responses, have some great contributors here, will be back with more questions