# (Optics) Vignetting and field of view

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Aelo

## 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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## Answers and Replies

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.