 # Relay lenses and optical formulas

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Relay lenses and optical formulas
1

Hi,
first of all, I'm not sure (but I hope it) that this is the correct section for this not usual argument:
If I want to use a relay system (composed by the primary lens and the relay lens), how can I calculate (which formula) the effective focal length (EFL), the depth of field (DOF) e the field of view (FOV) of the complete optical system?
I'm using a relay system to increase the image circle of a small CCTV primary lens to use it on a MFT body.
Ciao.
Toni Complain
Re: Relay lenses and optical formulas

First you need to discover the effective focal length, taking into account all the stuff you attached. Here are two methods; both involve knowing the magnification of the resulting image. Not as easy as you might think. The viewfinder view only displays about 95% of the actual view because of masking and a magnification of 0.82X. You can, however, take a picture of a metric ruler. You can use this image to obtain the actual image size and the resulting magnification.

1. This method depends on you achieving a magnification of 1 (i.e. life-size) often called unity or 1:1. Set up a target; best is a ruler with a millimeter scale. Adjust the distance, camera-to- ruler to achieve life-size (magnification 1). Accurately measure the distance from the ruler to the image plane. This camera features an image plane measuring point. This is the symbol adjacent to the LCD data display atop the camera body. Assuming you have precisely obtained magnification 1 and an accurate measure of object-to- image distance, divide this distance by 4. This will be the effective focal length.

2. This method depends on you achieving a magnification of 10 to 1 reduction. You position the distance camera-to- object to obtain a reduction 10 to 1. That will be a 100mm span on the ruler adjusted to occupy 10mm on the image. Having achieved this magnification, measure the distance from the object to a definite point on the lens barrel. Now repeat this procedure. This time we want a 20 to 1 reduction. A 400mm span occupies the same 10mm image span. Again measure this revised distance, ruler-to- the same reference point, on the lens barrel. Now we calculate focal length. The effective focal length is 10 times the difference between these two measurements. The formula behind this method ---If the distance between two camera positions is D, and difference between the two magnifications is M, then the focal length is D÷M.

Alan Marcus (alanmaxinemarcus@att.net)

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Re: Relay lenses and optical formulas

Thanks again.

Toni

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Re: Relay lenses and optical formulas

Glad I could help - once you figure out the focal length of this lash-up -- can help with angle of view etc.

Alan Marcus

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Re: Relay lenses and optical formulas
2

tonipuma wrote:

Hi,
first of all, I'm not sure (but I hope it) that this is the correct section for this not usual argument:
If I want to use a relay system (composed by the primary lens and the relay lens), how can I calculate (which formula) the effective focal length (EFL), the depth of field (DOF) e the field of view (FOV) of the complete optical system?
I'm using a relay system to increase the image circle of a small CCTV primary lens to use it on a MFT body.
Ciao.
Toni Looks like you have a 2.1 mm focal length f/1.6 M12 mount CCTV lens, and a reversed Canon 24 mm f/2.8 STM lens.

Note that Alan Marcus' formula for 1/4 subject-image distance at 1:1 magnification is only valid for a single thin lens. It won't work for the two compound lenses in your assembly.

In theory, you can use the lens equation to calculate image size from the subject and image distances, and from the separation of the two lenses. Calculate the position of the image of the subject formed by the first lens, then use this as the subject of the second lens. Unfortunately, for accurate results, you need to measure from the front and rear principal planes, rather than the front or rear elements, or the physical mid-point of the lens. Since both lenses are almost certainly retro-focal designs, the rear principal planes will be located somewhere outside the lens body, and you probably don't know exactly where.

Your reversed 24 mm lens focusses the image formed by the CCTV lens, and projects a magnified copy onto the sensor. My suggestion is to remove the 52 mm diameter adapter, and image a millimetre rule onto the sensor using only the reversed 24 mm lens. The exact magnification will change as you focus the lens, so record the maximum and minimum in-focus image size.

The EOS 760 D sensor is 22.3 x 14.9 mm. If you use a micro four thirds body, the sensor size is 18 x 13.5 mm. Compare this with the size of your subject, and calculate the magnification (or the range of magnification over the range of the lens' focussing motor).

My guess is that magnification using only the reversed 24 mm f/2.8 STM will be somewhere in the region of 4x to 6x.

With the CCTV lens attached and its position adjusted to focus a distant object at a fixed STM focus setting, then the effective focal length will be the product of CCTV lens focal length and STM lens magnification. If the CCTV lens focal length is 2.1 mm, and 24 mm relay magnification is 5x, then the effective focal length of the combination will be 10.5 mm. If you change the STM lens' focus, or the distance to the CCTV lens, the magnification and effective focal length will change.

Is the CCTV lens designed for a 1/2.5 inch sensor, or for something smaller, such as 1/3 inch? If designed for a relatively large sensor, the field of view may be limited by vignetting in the CCTV lens, especially with a MFT body. If designed for a smaller sensor, the DSLR sensor could capture the lens's nominal field of view.

Is the 2.1 mm lens designed for a rectilinear or a fish-eye projection? There will be a significant difference in field of view for the same effective focal length.

I suspect neither lens is rear-telecentric, so there is potential for some vignetting towards the edge of the field of view, depending on the locations of the exit pupils of the two lenses. In any case, the 24 mm lens will not accept the full f/1.6 aperture (0.31 NA) of the CCTV lens, though this won't necessarily truncate the field of view.

If you have this rig set up and producing reasonably sharp images, it may be simpler to measure the effective focal length directly - perhaps using a variation of Alan Marcus' second method, as discussed here.

Hope this helps.

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Alan Robinson

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Magnification and focal length
1

A Marcus wrote:

First you need to discover the effective focal length, taking into account all the stuff you attached. Here are two methods; both involve knowing the magnification of the resulting image. Not as easy as you might think. The viewfinder view only displays about 95% of the actual view because of masking and a magnification of 0.82X. You can, however, take a picture of a metric ruler. You can use this image to obtain the actual image size and the resulting magnification.

1. This method depends on you achieving a magnification of 1 (i.e. life-size) often called unity or 1:1. Set up a target; best is a ruler with a millimeter scale. Adjust the distance, camera-to- ruler to achieve life-size (magnification 1). Accurately measure the distance from the ruler to the image plane. This camera features an image plane measuring point. This is the symbol adjacent to the LCD data display atop the camera body. Assuming you have precisely obtained magnification 1 and an accurate measure of object-to- image distance, divide this distance by 4. This will be the effective focal length.

This only works for a thin lens. For a compound lens or relay scheme such as this, you need to measure from the lens principal planes. My guess is that these are separated by around 225 mm here, with an effective focal length perhaps 10 mm. You would estimate an effective focal length closer to 66 mm.

2. This method depends on you achieving a magnification of 10 to 1 reduction. You position the distance camera-to- object to obtain a reduction 10 to 1. That will be a 100mm span on the ruler adjusted to occupy 10mm on the image. Having achieved this magnification, measure the distance from the object to a definite point on the lens barrel. Now repeat this procedure. This time we want a 20 to 1 reduction. A 400mm span occupies the same 10mm image span. Again measure this revised distance, ruler-to- the same reference point, on the lens barrel. Now we calculate focal length. The effective focal length is 10 times the difference between these two measurements. The formula behind this method ---If the distance between two camera positions is D, and difference between the two magnifications is M, then the focal length is D÷M.

This formula is almost correct if the two magnifications are 1/10 and 1/20. In this case you require 200 mm span (not 400 mm) for the 1/20 x case, and the focal length is D/10 (i.e M x D).  This requires that the two magnifications differ by exactly a factor 2

More generally, if the subject distances are D1, D2, and the corresponding magnifications M1, M2, then we have: focal length f = (D1 - D2) M1 M2 / (M2 - M1)

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Alan Robinson

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Re: Relay lenses and optical formulas

tonipuma wrote:

Hi,
first of all, I'm not sure (but I hope it) that this is the correct section for this not usual argument:
If I want to use a relay system (composed by the primary lens and the relay lens), how can I calculate (which formula) the effective focal length (EFL), the depth of field (DOF) e the field of view (FOV) of the complete optical system?
I'm using a relay system to increase the image circle of a small CCTV primary lens to use it on a MFT body.
Ciao.
Toni To measure the focal length determine magnification with and without the 20mm extension tube.

focal length = delta_x / delta_magnification = 20mm / delta_magnification

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Bill ( Your trusted source for independent sensor data at PhotonsToPhotos )

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Re: Relay lenses and optical formulas

bclaff wrote:

tonipuma wrote:

Hi,
first of all, I'm not sure (but I hope it) that this is the correct section for this not usual argument:
If I want to use a relay system (composed by the primary lens and the relay lens), how can I calculate (which formula) the effective focal length (EFL), the depth of field (DOF) e the field of view (FOV) of the complete optical system?
I'm using a relay system to increase the image circle of a small CCTV primary lens to use it on a MFT body.
Ciao.
Toni To measure the focal length determine magnification with and without the 20mm extension tube.

focal length = delta_x / delta_magnification = 20mm / delta_magnification

This clearly works for a single thin lens. It also applies to a compound lens forming a real inverted image, provided the internal separations of the lens elements do not change.

I was mildly surprised to find that it also applies here too, where the first lens forms a real inverted image which the relay lens projects onto the sensor as a real non-inverted image. A caveat is that to form an in-focus image without the extension tube, we need to increase the separation between CCTV and relay lenses, which changes the effective focal length, compared with the infinite conjugation case.

If the relay lens operates with a magnification of 5x with the 20 mm extension tube in place, the relay magnification falls to 4.167x without the extension tube, and the location of the intermediate image shifts outwards by 0.96 mm with respect to the relay lens' principal plane. Focal length of relay lens is Δx / ΔM = 20 mm / 0.833 = 24 mm.

However, the 0.96 mm shift of intermediate image towards the first 2.1 mm focal length lens corresponds to a large shift in subject distance if the image is to remain focussed. If the lens separation is adjusted for infinity focus with the 20 mm extension tube removed, the effective focal length of the composite assembly is reduced by 15-20% compared with the case when infinity focus is achieved with the extension tube in place. With my estimated values, focal length is 8.75 mm compared with 10.5 mm.

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Alan Robinson

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Re: Relay lenses and optical formulas

alanr0 wrote:

bclaff wrote:

tonipuma wrote:

Hi,
first of all, I'm not sure (but I hope it) that this is the correct section for this not usual argument:
If I want to use a relay system (composed by the primary lens and the relay lens), how can I calculate (which formula) the effective focal length (EFL), the depth of field (DOF) e the field of view (FOV) of the complete optical system?
I'm using a relay system to increase the image circle of a small CCTV primary lens to use it on a MFT body.
Ciao.
Toni To measure the focal length determine magnification with and without the 20mm extension tube.

focal length = delta_x / delta_magnification = 20mm / delta_magnification

This clearly works for a single thin lens. It also applies to a compound lens forming a real inverted image, provided the internal separations of the lens elements do not change.

...

No. It works for any lens.
I have used this technique many many times.

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Bill ( Your trusted source for independent sensor data at PhotonsToPhotos )

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Re: Focus breathing

bclaff wrote:

alanr0 wrote:

bclaff wrote:

tonipuma wrote:

Hi,
first of all, I'm not sure (but I hope it) that this is the correct section for this not usual argument:
If I want to use a relay system (composed by the primary lens and the relay lens), how can I calculate (which formula) the effective focal length (EFL), the depth of field (DOF) e the field of view (FOV) of the complete optical system?
I'm using a relay system to increase the image circle of a small CCTV primary lens to use it on a MFT body.
Ciao.
Toni To measure the focal length determine magnification with and without the 20mm extension tube.

focal length = delta_x / delta_magnification = 20mm / delta_magnification

This clearly works for a single thin lens. It also applies to a compound lens forming a real inverted image, provided the internal separations of the lens elements do not change.

...

No. It works for any lens.
I have used this technique many many times.

I don't doubt it. As I said, "it also applies here too".

The point I was trying to make is that removing the 20 mm extension tube is likely to push subject focus distance beyond infinity. If you adjust the inter-lens distance so that subject distances between 100 mm and infinity are focussed with the extension tube removed, then the subject distance will be much closer (between 6 and 7 mm from the CCTV lens' first principal plane) with the extension tube replaced.

The 24 mm relay lens' focus movement is not coupled to that of the primary CCTV lens. The composite assembly suffers from focus breathing. Changing the inter-lens separation changes the focal length of the assembly. What you will measure is the focal length when the assembly is configured to accommodate subjects less than 10 mm from the CCTV lens with the extension tube in place. The apparatus as described by the OP will have a longer focal length when focussing on subjects between 1 m and infinity.

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Alan Robinson

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Re: Focus breathing

alanr0 wrote:

bclaff wrote:

alanr0 wrote:

bclaff wrote:

tonipuma wrote:

Hi,
first of all, I'm not sure (but I hope it) that this is the correct section for this not usual argument:
If I want to use a relay system (composed by the primary lens and the relay lens), how can I calculate (which formula) the effective focal length (EFL), the depth of field (DOF) e the field of view (FOV) of the complete optical system?
I'm using a relay system to increase the image circle of a small CCTV primary lens to use it on a MFT body.
Ciao.
Toni To measure the focal length determine magnification with and without the 20mm extension tube.

focal length = delta_x / delta_magnification = 20mm / delta_magnification

This clearly works for a single thin lens. It also applies to a compound lens forming a real inverted image, provided the internal separations of the lens elements do not change.

...

No. It works for any lens.
I have used this technique many many times.

I don't doubt it. As I said, "it also applies here too".

The point I was trying to make is that removing the 20 mm extension tube is likely to push subject focus distance beyond infinity. If you adjust the inter-lens distance so that subject distances between 100 mm and infinity are focussed with the extension tube removed, then the subject distance will be much closer (between 6 and 7 mm from the CCTV lens' first principal plane) with the extension tube replaced.

The 24 mm relay lens' focus movement is not coupled to that of the primary CCTV lens. The composite assembly suffers from focus breathing. Changing the inter-lens separation changes the focal length of the assembly. What you will measure is the focal length when the assembly is configured to accommodate subjects less than 10 mm from the CCTV lens with the extension tube in place. The apparatus as described by the OP will have a longer focal length when focussing on subjects between 1 m and infinity.

Yeah, sorry; I read too quickly !

If removing the 20mm extension is a problem then I would just go the other way and add some extension.

Regards

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Bill ( Your trusted source for independent sensor data at PhotonsToPhotos )

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interesting ! - but does it make sense ?

The CCTV lens is designed for a small format sensor, with limited resolution. Maybe 2 Mpix. Relaying the intermediate image will not increase relative resolution. Relaying may degrade the image quality due to de deficiencies of the relay lens in this application. So the final image quality maybe lower than  2 Mpix worth.

The effective aperture of the relayed lens will be small (high aperture ratio). The relaying easily adds vignetting.

A quick estimate of the FL for the relayed compound lens system may be useful to find an equivalent original lens for the camera. The original lens is expected to offer better resolution across the image and an option to use lower aperture ratios like (f/1.4) - f/2.8 - f/8 .

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