# Archived (Optics) Vignetting and field of view

1. Oct 19, 2013

### Aelo

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

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

3. 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!

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2. Sep 4, 2016

### Ibix

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.