Does minimum focus distance matter?

As far as I understand the physics of dome ports, lenses do not actually focus on the image per se, but rather on an image created by the dome. If the lense in question has a short enough minimum focus distance so focus is obtained on the domecreated image, everything is OK, otherwise a diopter is needed. However, does this mean that minimum focus distance only matters in relation to the domecreated image, that is short enough=OK, to long=diopter, or will there be advantages of even shorter minimum focus distances. To give an example: the Canon EF 15 mm fisheye has a mfd of 20 cm, the Sigma 15 mm fisheye a mfd of 15 cm. Both will focus without diopter in an Ikelite 8 inch dome. My question is (it took a while, didn´t it ), is it possible to do closer wa-close ups with the Sigma compared to the Canon, or is it the domecreated picture that gives the actaul closest possible wa-close up, so the Canon and the Sigma will render the same closest possible focus of an object.

Regards

Bent C

is it possible to do closer wa-close ups with the Sigma compared to the Canon

 

Regards

 

Bent C

Only if the focus subject is inside your dome. The only thing I've shot inside my dome is a dislodged zoom gear.

Going to chance my arm on an associated question.

How do you calculate the distance to the dome created image?

Paul C

I don't have either of those lenses (and use Nikon) but I believe you will be able to get closer to subjects with the Sigma and you will get greater magnification because you can get closer.

You are correct about the dome port creating an apparent image on which the lens focuses. That apparent image will be closer for close subjects and further away for more distant subjects. i.e. the distance to the apparent image is relative to the distance of the subject. I don't know how you can calculate the distance.

Even if you don't NEED a diopter lens to be able to focus on the apparent image, you can still benefit from using one for both edge sharpness and also reducing the minimum focusing distance of the whole rig resulting in an increase in the magnification possible.

Here is an example of what I mean although it isn't the difference between no diopter and using one, but rather the difference in the strength of the diopter, but the result is still the same.

Here is the result from my 18-55mm lens in a 6" dome port with a +4 diopter when focusing at the resulting minimum focusing distance:

Here is the same lens/dome port with a +5 diopter:

As you can see, increasing the strength of diopter results in higher magnification. The same thing will happen if you use a diopter even if you strictly don't need one. From my testing, I have determined (at least for a 6" dome) that a +4 diopter will give you the same magnification with the dome port underwater, as you get with the bare lens on land. Here is what I get from the same 18-55mm lens on land:

You'll notice that it matches the +4 diopter in a dome port.

So, basically, if you want to be able to get the same magnification/reproduction ratio of the lens as you can on land, you may want to consider using a +4 diopter even if you don't "need" one. Or you could go for a +2 which will improve the magnification somewhat.

With wide angle lenses, you do need to consider how close you can actually get as obviously the subject still has to be outside the dome. It is still worth the experiment as you may be able to shoot smaller subjects than you could without the diopter. If you use a quality diopter, you don't really lose that much if anything in image quality (in my experience).

Hi Andrew,

I only think that applies when you use the dome and lens and diopter on land. When you use it underwater, the dome port+water actually makes a convex lens - which means is 'de-magnifies' everything.

Cheers

James

James,

The photos with the black background were taken underwater and show what happens with the lens, dome and diopter are underwater. And yes, the dome port "demagnifies" everything which is exactly why you can benefit by using a diopter. As I said, in my experience with 6" dome ports, a +4 diopter negates the "demagnification" you get with the dome port.

Going to chance my arm on an associated question.

 

How do you calculate the distance to the dome created image?

 

Paul C

As a general rule of thumb, the virtual image is twice the dome's radius in front of the dome. So, with a 6" dome, the virtual image of an object at infinity will be ~6" in front of the dome. However, to find out exact distances, you need to account for the thickness of the dome as well as the refractive index of the material it's made of. If you feel like wading through the math, this link explains how to calculate the location of the virtual image based on the objects distance from the front of the port and takes into account the refractive index and port thickness.

Cheers,

Mike

As a general rule of thumb, the virtual image is twice the dome's radius in front of the dome. So, with a 6" dome, the virtual image of an object at infinity will be ~6" in front of the dome. However, to find out exact distances, you need to account for the thickness of the dome as well as the refractive index of the material it's made of. If you feel like wading through the math, this link explains how to calculate the location of the virtual image based on the objects distance from the front of the port and takes into account the refractive index and port thickness.

 

Cheers,

 

Mike

Thanks for that Mike - great stuff though I think I will stick with the general Rule!

It has a fine air of 'near enough is probably good enough' to it!

Thanks again.

Paul C

Thanks for that Mike - great stuff though I think I will stick with the general Rule!

 

It has a fine air of 'near enough is probably good enough' to it!

 

Thanks again.

 

Paul C

After I wrote that article referenced above, I realized it was too complicated to be of practical use, so I wrote another article where I derived a quite accurate but very simple "rule of thumb" for finding the dome's virtual image of a distant object. You can find it at http://www.scubageek.com/ , entitled "Secondary Focal Point of a Dome Port".

Les

I do realize that using a diopter will help in decreasing minimum focusing distance. However, at least on the Canon EF 15 mm fisheye, that is not an option due to the construction of the lens. In this thread the explicit answers to my question "is it possible to do closer wa-close ups with the Sigma compared to the Canon" has been two. One stating that there will be no difference, one the opposite. Already using the Canon fisheye, I guess I could buy the Sigma and simply do the tests underwater, but that would be quite an expensive way of knowing. Is there anyway of calculating the "nearest-object-to-the-dome-I-can-get-in-focus" given a minimum focus distance of the Sigma´s 15 vs the Canon´s 20 cm, both 15 mm fisheyes, both behind an 8'' dome on a full frame camera. It should be quite a difference in object size in the resulting picture if there was the same relative difference in mfd under water as above water on the two lenses.

Regards

Bent C

I do realize that using a diopter will help in decreasing minimum focusing distance. However, at least on the Canon EF 15 mm fisheye, that is not an option due to the construction of the lens. In this thread the explicit answers to my question "is it possible to do closer wa-close ups with the Sigma compared to the Canon" has been two. One stating that there will be no difference, one the opposite. Already using the Canon fisheye, I guess I could buy the Sigma and simply do the tests underwater, but that would be quite an expensive way of knowing. Is there anyway of calculating the "nearest-object-to-the-dome-I-can-get-in-focus" given a minimum focus distance of the Sigma´s 15 vs the Canon´s 20 cm, both 15 mm fisheyes, both behind an 8'' dome on a full frame camera. It should be quite a difference in object size in the resulting picture if there was the same relative difference in mfd under water as above water on the two lenses.

 

Regards

 

Bent C

Bent C,

The article I linked to above gives you all the math you need to calculate exactly where the virtual image will be given it's true location. However, as the author has already mentioned, the information can be a little thick (I thought it was an excellent article though Les). I had programmed the whole thing into an excel spreadsheet at one time, but I can't seem to find it anymore. If I ever track it down I'll provide a link. However, there are some calculations that have been done for the user for various dome radii that can be found at the bottom of the article.

Given that the two lenses you looking at are both fisheyes and both have incredible depth of field, I don't think you'll find any practical difference between the two lenses. Both lenses should be able to focus adequately at any usable working distance. You need to remember that you need to light your subject, and if you're only 10cm away from it, it's going to be very difficult to light your subject properly. I found myself in this situation with my tokina 10-17 (14cm minimum focus distance) recently. I actually had to back up a little because my dome shade was casting shadows on my subject.

Cheers,

Mike

Bent C,

 

The article I linked to above gives you all the math you need to calculate exactly where the virtual image will be given it's true location. However, as the author has already mentioned, the information can be a little thick (I thought it was an excellent article though Les). I had programmed the whole thing into an excel spreadsheet at one time, but I can't seem to find it anymore. If I ever track it down I'll provide a link. However, there are some calculations that have been done for the user for various dome radii that can be found at the bottom of the article.

 

Given that the two lenses you looking at are both fisheyes and both have incredible depth of field, I don't think you'll find any practical difference between the two lenses. Both lenses should be able to focus adequately at any usable working distance. You need to remember that you need to light your subject, and if you're only 10cm away from it, it's going to be very difficult to light your subject properly. I found myself in this situation with my tokina 10-17 (14cm minimum focus distance) recently. I actually had to back up a little because my dome shade was casting shadows on my subject.

 

Cheers,

 

Mike

As far as I can understand the article, nearer objects will map nearer virtual images in front of the dome (which of course isn´t that surprising, but other articles I have read have stated differently). In that case, there will obviously in principle be situations were the Sigma can focus on the virtual image while the Canon will not. But, thinking about it, I guess that you are entirely correct in your view that the practical consequences of the difference will be very hard to find in real world situations. So, thanks a lot, I just saved my self a bunch of $ and will keep the Canon, which I like a lot for over water pictures.

Regards

Bent

I do realize that using a diopter will help in decreasing minimum focusing distance. However, at least on the Canon EF 15 mm fisheye, that is not an option due to the construction of the lens. In this thread the explicit answers to my question "is it possible to do closer wa-close ups with the Sigma compared to the Canon" has been two. One stating that there will be no difference, one the opposite. Already using the Canon fisheye, I guess I could buy the Sigma and simply do the tests underwater, but that would be quite an expensive way of knowing. Is there anyway of calculating the "nearest-object-to-the-dome-I-can-get-in-focus" given a minimum focus distance of the Sigma´s 15 vs the Canon´s 20 cm, both 15 mm fisheyes, both behind an 8'' dome on a full frame camera. It should be quite a difference in object size in the resulting picture if there was the same relative difference in mfd under water as above water on the two lenses.

 

Regards

 

Bent C

Hi Bent;

I guess it's academic now, since you're sticking with the Canon, but there is a general procedure for figuring out how to achieve the minimum dome-to-subject distance with any lens, with any camera, with any dome. Here it is:

1. Assemble your rig. If you can focus closer than the front of your dome, go to step 5. Otherwise go to step 2.

2. Determine the distance from the front of your lens (filter mount) to the front of the dome.

3. Determine the distance from the front of the dome to the "infinity" virtual image: It is 3R-4t, where R=dome radius and t=dome thickness.

4. Add the above two distances, convert to meters, and take the inverse. Round this down to the nearest integer to get the maximum power of the diopter you can use. This diopter will allow you to achieve the closest possible dome-to-subject distance with your setup while still allowing you to focus on infinity.

5. You're done.

If you do the above steps for the Canon lens, you can adjust the calculations for any other lens (like the Sigma) simply by going to a camera store where you can measure the length of the lens (mounting flange to filter mount).

If this sounds too complicated, the alternative is to buy 3 diopters --- +2, +3, +4. The highest-powered one that still gives you a sharp background when the lens is set to closest focus is the one that you want.

Les

Hi Les

Thanks for the input. It really isn´t purely academic for me, as I also use the 17-40, which I certainly need a diopter for. On the Canon fisheye (and I assume the Sigma as well) as far as I know, I can not add any diopter, as the front glass bulges out and there is no filter thread on the lens.

Bent