(Optics) Vignetting and field of view

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SUMMARY

The discussion focuses on calculating the unvignetted and fully vignetted fields of view for an 80 mm focal length thin lens with a 25 mm diameter and a 20 mm stop located 40 mm in front of the lens. The magnification is given as -1/2, leading to the determination of object and image distances as l' = -240 mm and l = 120 mm. The analysis involves deriving ray equations and applying vignetting conditions, ultimately concluding that a 5 mm object centered in the field of view remains unaffected by vignetting.

PREREQUISITES
  • Understanding of thin lens equations and magnification
  • Familiarity with vignetting concepts in optics
  • Knowledge of ray tracing techniques in optical systems
  • Ability to perform trigonometric calculations for field of view
NEXT STEPS
  • Study the derivation of vignetting equations in optical systems
  • Learn about ray tracing methods for complex optical setups
  • Explore the impact of lens diameter on field of view and vignetting
  • Investigate the relationship between object size and vignetting effects
USEFUL FOR

Optics students, optical engineers, and anyone involved in lens design and analysis will benefit from this discussion, particularly those focusing on vignetting effects in imaging systems.

Aelo
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Homework Statement


An 80 mm focal length thin lens is used to image an object with a magnification of -1/2.
The lens diameter is 25 mm and a stop of diameter 20 mm is located 40 mm in front of the lens.
How big is the unvignetted field of view [in terms of object size (in mm) and in terms of half field angle]?
How big is the fully vignetted field of view?


Homework Equations



Magnification = l'/l (coupling this with the focal length, we can find the object and image distances)
Vignetting equations attached

The Attempt at a Solution



I've drawn a picture and looked at http://spie.org/x32310.xml. I found that l' = -240 mm and l = 120 mm. I'm not sure where to go from here. Thanks in advance for help!
 

Attachments

  • vignetting.png
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The object/image distances are the wrong way round - as quoted they give a magnification of -2. Correcting that, I agree the values.

"Vignetting" is what happens when the bundle of rays that make it through one optical element don't all make it through the next.

I'm not sure from the description whether we've got object-stop-lens-image or object-lens-stop-image. I'm assuming the former, but extending the argument to the latter isn't hard. In this case, it's convenient to treat the object as positioned at x=0, the stop at x=200 and the lens at x=240. For a point on the object a distance ##h## off axis, it is easy to write down the equations of the rays that pass through the top and bottom of the stop:$$\begin{eqnarray*}
y_t&=&h-\frac{h-10}{200}x\\
y_b&=&h-\frac{h+10}{200}x
\end{eqnarray*}$$Then all you have to do is work out the height of the rays at the position of the lens (x=240):$$\begin{eqnarray*}
y_t&=&12-h/5\\
y_b&=&-12-h/5
\end{eqnarray*}$$and require that they pass through the 25mm diameter lens:$$\begin{array}{rcccl}
-12.5&\leq&12-h/5&\leq&12.5 \\
-12.5&\leq&-12-h/5&\leq&12.5
\end{array}$$
Which comes down to:$$\begin{array}{rcccl}
-0.5&\leq&h/5&\leq&24.5\\
-24.5&\leq&h/5&\leq&0.5
\end{array}$$We require all four of these conditions to be satisfied simultaneously, i.e. ##-0.5\leq h/5\leq 0.5##, or ##-2.5\leq h'\leq 2.5##. So a 5mm object centred in the field of view is not quite affected by vignetting.

In half-angle terms the field of view is simply ##\tan^{-1}(2.5/240)##, the angle subtended by ##h## at the center of the lens.
 

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