Optics Equations - to reduce height

In summary, the OP wants to reduce the distance between the camas and lens to still get the same focus. He wants to reduce the distance between the camas and lens and keep his same focus. He knows the simple lens equation and the ratio of image and object size being equal to the two distances. He knows diffraction is not a problem and f values around f8 will give least impairment. He knows f1.4 is a great advantage for low light or movement but not a 'preferred value' when a smaller stop will work.
  • #1
btb4198
572
10
Is there an optics equation that can take an existing optics set-up and tell you would components, you would need to reduce the distance and still get the same result?

I have a working system, but it is really larger and I would like to reduce the distance, the height.
I am using a light source to pre-analyzer filter to lens to camera, a basic Optical setup.

Is that any equation I can use ?
 
Science news on Phys.org
  • #2
That's all a bit too open ended for a proper answer, I'm afraid. What's the actual set-up and what do you actually need to know? A diagram is always worth the effort.
 
  • #3
I want to reduce the distance from the camas to the lens and keep my same focus.
 
  • #4
btb4198 said:
I want to reduce the distance from the camas to the lens and keep my same focus.
That still doesn't mean a lot to me, I'm afraid. You may know what you mean but you are leaving out a lot of stuff that's necessary for me to understand you.

I presume you know the simple lens equation (1/u +1/v = 1/f) and the ratio of image and object size being equal to the two distances. Have you read around about this or is this your first taste of optics? It isn't clear.
 
  • #5
Reducing the diameter and spacing of the elements without changing their power will do nothing but make your image worse. Optical systems are notoriously difficult to optimize, so I'm not sure there's a simple equation that would work. Unfortunately my 2 semesters of optics didn't give me enough knowledge to elaborate further or give a concrete answer.
 
  • Like
Likes sophiecentaur
  • #6
Drakkith said:
Reducing the diameter and spacing of the elements without changing their power will do nothing but make your image worse
Whilst this is sometimes the case, most camera lenses sit between diffraction and aberration problems. For a lot of photography, diffraction is not a problem and f values around f8 will give least impairment. f1.4 is a great advantage for low light or movement but not a 'preferred value' when a smaller stop will work.

The OP is far too vague for a good answer, as I have already said. If the requirement is very critical then any suitable (obviously multi-element) lens would be optimised for its size and you'd have to pay for it!

Some input from OP is required now.
 

1. What is the equation for calculating the reduced height in optics?

The equation for calculating the reduced height in optics is h' = h * (s / (s + f)), where h is the object height, s is the object distance, and f is the focal length of the lens.

2. How do I use the optics equation to reduce the height of an object?

To use the optics equation to reduce the height of an object, you need to know the object height, object distance, and focal length of the lens. Plug these values into the equation h' = h * (s / (s + f)) to calculate the reduced height.

3. Can the optics equation be used for any type of lens?

Yes, the optics equation can be used for any type of lens, as long as the focal length is known. It is a general equation that applies to both convex and concave lenses.

4. Is there a maximum or minimum height that can be achieved using the optics equation?

No, there is no maximum or minimum height that can be achieved using the optics equation. The reduced height will depend on the values of the object height, object distance, and focal length of the lens.

5. What are some real-life applications of the optics equation for reducing height?

The optics equation for reducing height is commonly used in the design and construction of optical instruments such as cameras, telescopes, and microscopes. It is also used in industries such as photography, astronomy, and medicine for precise measurements and imaging.

Similar threads

Replies
1
Views
704
Replies
13
Views
2K
Replies
22
Views
2K
Replies
8
Views
1K
Replies
1
Views
668
Replies
2
Views
668
Replies
1
Views
1K
Replies
3
Views
1K
Back
Top