External focus in an optical system

In summary: If the source is a diffuse object, then the distance between the lens and the object will be greater than the focal length.Is the red rectangle a diffuse or point source?Ok, example:A light source is placed at the focal length of the paraxial lens #1. Lens 1 have an focal length of 40 mm. A second lens (the Eye) is placed 600 mm from the light source, with let say focal length 200 mm. My question is how I can determine the external focus from this system?I have found an external focus at about 294 mm from Lens 1
  • #1
nordmoon
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0
I feel a bit stuid for asking this.

I have an optical system conisting of a laser source, a lens with some focal length F and aperture stop `(with some diamater D), followed by another lens (with another focal length) f and aperture stop d at some distance r from the first aperture stop D. What is the external focus of the system?

Thanks in advance!
 
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  • #2
"External focus" often refers to the external focus control - some dial used to adjust the positions of the lenses.
 
  • #3
It's not clear what you are asking. If you have a laser "source" then that is collimated light going into your lens system. Is this a beam expander? What are the values of F, r and f?
 
  • #4
Ok, example:

A light source is placed at the focal length of the paraxial lens #1. Lens 1 have an focal length of 40 mm. A second lens (the Eye) is placed 600 mm from the light source, with let say focal length 200 mm. My question is how I can determine the external focus from this system?

I have found an external focus at about 294 mm from Lens 1 in ZEMAX, but to me it doesn't make sense.

Applying the lens formula on the first lens, 1/f = 1/a+1/b, with f = 40 mm, a = 40 mm: gives b = infinity.
And if a = 39 mm, then external focus should be at 1560 mm.

Is there something I am missing here? I just can't think of what could be wrong?

EDIT:My light source is a fiber, .. so the actual laser diode is about 60 mm from the fiber end, and is directly coupled into the fiber at the light source.
 
  • #5
Did you coin the term external focus?
 
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Likes davenn
  • #6
nordmoon said:
Ok, example:

A light source is placed at the focal length of the paraxial lens #1. Lens 1 have an focal length of 40 mm. A second lens (the Eye) is placed 600 mm from the light source, with let say focal length 200 mm. My question is how I can determine the external focus from this system?

I have found an external focus at about 294 mm from Lens 1 in ZEMAX, but to me it doesn't make sense.

So the source is 40mm from lens #1, Lens #2 is 600mm from the source. The image from lens #1 is at infinity, so lens #2 forms a real image 200mm away from it or 200mm + 600mm = 800mm from the source. Since that's not close to what you get, I am probably not envisioning the system correctly. A diagram would help, plus some information about the system - does it make a real or virtual image? Why do you say the second lens is the Eye?
 
  • #7
pixel said:
So the source is 40mm from lens #1, Lens #2 is 600mm from the source. The image from lens #1 is at infinity, so lens #2 forms a real image 200mm away from it or 200mm + 600mm = 800mm from the source. Since that's not close to what you get, I am probably not envisioning the system correctly. A diagram would help, plus some information about the system - does it make a real or virtual image? Why do you say the second lens is the Eye?

I am just trying to simulate the image that is obtained in the eye as a person looks into my objective lens with a light source. Simplied model of my setup from ZEMAX:

vpIu6eR.png


I am just trying to understand how to determine the external focus (which is the name I use for it).
I know its there in Zemax because I can see the rays converge at that point in space between the two lenses. The light source is a waveguide which have 4 fibers couple light into it. This waveguide end (if you couple light in the waveguide front side) is placed near 40 mm (could be 39 mm).

The first image must create an first image and the second lens creates an image in the eye. I thought the lens focal Point was 40 mm from the lens. I am just trying to udnerstand why its converging there and where the image should be located between the lenses.
 
  • #8
I think you need to tidy up both ends of that diagram. What's what?
For a simple lens, with a focal length of 40mm, an object, 40mm away will form an image at infinity. This may or may not be relevant, depending on what your ray diagram is actually depicting at the exit from the 'light source'. Is that red rectangle a diffuse source or is it just an exit pupil with a point source back inside the 'tube'? (In which case, the 40mm (a) is not really relevant in calculating the distance b. You need to specify the object distance from thatpoint source. If the source (laser) is producing a parallel beam, that would, i think, be focused at 40mm from the lens (what you call the External Focus)
Also, you seem to be ignoring the (your) eye lens, jumping straight from the 200mm lens to the retina.

What sort of optical instrument does this represent? What is it designed to achieve? Have you jumped into using a simulator, rather than using elementary (paper and pencil) optics and some thought?
 
  • #9
Well, in the beginning I have a laser. The light of the laser is coupled into a fiber, which in turn is coupled into a waveguide that has certain rectangular shape. The red rectangle in the image above was suppose to represent the fiber output. I made some additional simplification in ZEMAX, instead of the fibers I use a circular source with certain angles. It gives about the same result as with the complete modell using fibers.

It is the output of this waveguide I want to image onto the eye using the objective lens (lens f = 40 mm). I have also modeled the eye with a lens (lens2 F = 200 mm). I can calculate where the image is located using lens formula on each lens.

In Zemax this intersection of rays is located at 294 mm from lens 1. Could this be the location of the image from lens 1? Which source is it then imaging and what location would the source then be? Would it be the waveguide output (my source I want to image onto the eye). I know that when applying the lens formula on a lens I can find the location of the "image" of the object, provided I know the focal length of the lens and the location of the object relative to the lens. I am uncertain becuase I have waveguides and fibers involved.

The application is laser saftey. This is just to find worst case locations using simulation. I am going to do RL measurments later.
 
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  • #10
So you have an array of fibres, approximately in phase (?). I cannot tell whether that would represent a unidirectional source (object at infinity) or a diffuse image, each part at distance a.
nordmoon said:
I have also modeled the eye with a lens (lens2 F = 200 mm).
That's pretty massive for a human eye!
Google eye optics for a better estimate of effective focal length and how we focus on objects / images.
nordmoon said:
In Zemax this intersection of rays is located at 294 mm from lens 1.
Zemax is clearly making an assumption that the 'object is not at 40mm but at a greater distance (that's why it's 'pulled in' the intermediate image)
You are making many assumptions about the eye's optics. I would imagine that, with your eye at a reasonable distance from that middle image (a bit further away than the near point of the observer's vision) then the observer should 'see it' as a real image at that location. Use a realistic eye model, for a start - but is it really necessary? What are you trying to find out with this exercise? If you produce a real image at a suitable place, the eye should be able to 'see' it and it takes care of its own focusing arrangement.
BTW, is that laser source safe for someone to look straight into?
 
  • #11
I changed my last question a bit because I realized that the intersection of the rays in ZEMAX must represent the image. But I am still uncertain why it is at 294 mm and what it is imaging (the fiber output, the waveguide output or my laser source)? In my simplification I removed the fiber (it took so long time to ray-trace - I usually never use Zemax for waveguide studies) and the original light source. So the source my ZEMAX sees is the fiber output, modeled as circular point sources with divergence at two angles. However the location of the intersection point did not change.

Yeah, I realize it is a large focal length. My understanding is that the eye has changing focal length. It's change depends on what you are focus on. This particular case study is suppose to be with an eye located 6000 mm from the objective lens with an eye focused on the external focus.

I still don't get what the 'external focus' is..., this isn't named by me. But I can see it as interecting beams in Zemax.
 
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  • #12
nordmoon said:
I changed my last question a bit because I realized that the intersection of the rays in ZEMAX must represent the image. But I am still uncertain why it is at 294 mm and what it is imaging (the fiber output, the waveguide output or my laser source)? In my simplification I removed the fiber (it took so long time to ray-trace - I usually never use Zemax for waveguide studies) and the original light source. So the source my ZEMAX sees is the fiber output, modeled as circular point sources with divergence at two angles. However the location of the intersection point did not change.

Yeah, I realize it is a large focal length. My understanding is that the eye has changing focal length. It's change depends on what you are focus on. This particular case study is suppose to be with an eye located 6000 mm from the objective lens with an eye focused on the external focus.

I still don't get what the 'external focus' is..., this isn't named by me. But I can see it as interecting beams in Zemax.

Yes, The fibers is an array of fibers along the waveguide.
PF seems just as perplexed about the term 'external focus'.
You are doing a good job of demonstrating the perils of using a simulator when things it's doing are uncertain. If it is telling you the image distance then you can tell where it is assuming the object is. No idea about that.
The idea of an eye with focal length of 200mm is just a joke.
The original question just doesn't make sense as it stands. It is incomplete. If you place your head at a reasonable distance then your eye will see that image. From the geometry, it looks like the image will appear 'smaller' than the end of the waveguide.
If this is an exercise then perhaps you should ask for clarification. If it is just a 'worked example' then kick it into touch and try another one.
 

Related to External focus in an optical system

1. What is the concept of external focus in an optical system?

The concept of external focus in an optical system refers to the location of the focal point or image plane outside of the optical system. This means that the light rays from an object are focused by the optical system to form an image that is located outside of the system.

2. How does external focus differ from internal focus in an optical system?

In internal focus, the focal point or image plane is located within the optical system itself, meaning that the light rays are focused to form an image that is within the system. This is typically achieved through the use of lenses or mirrors to manipulate the path of light rays. In contrast, external focus involves the light rays being focused to form an image outside of the optical system.

3. What are the advantages of using external focus in an optical system?

One advantage of external focus is that it allows for a longer working distance, meaning that the object being viewed can be placed further away from the optical system. This can be useful in applications where there is limited space or when working with delicate objects. External focus also reduces the risk of damage to the optical system, as the object is not physically close to the system.

4. Can external focus be achieved with all types of optical systems?

Yes, external focus can be achieved with various types of optical systems, including lenses, mirrors, and prisms. However, the specific design and components of the system may affect the ease and precision of achieving external focus. Some systems may be designed specifically for external focus, while others may require additional adjustments or attachments to achieve it.

5. How does external focus impact the quality of the image produced by an optical system?

The impact of external focus on image quality depends on various factors, such as the type and quality of the optical system, the distance between the object and the system, and the properties of the object being viewed. Generally, external focus can result in a clearer and sharper image, as the object is further away from the system and there is less risk of distortion or aberrations. However, this may not always be the case and the overall image quality will also depend on other factors such as the resolution and sensitivity of the system.

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