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title: Solutions for Image Formation
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# Exercises
## Equivalence of Points (Based on Exercise 3.1.)
Note that the thin lens model is similar since rays through the
centre of the lens (optical centre) are not deflected, and thus
serves the role of a pinhole.
## (Exercise 3.2)
This initial figure is slightly inaccurate.
Horizontal lines (object to lens, and lens to image point)
delimit the region of interest. In reality it is the aperture that
delimits the bundle of light that contributes to the image.
However it was my first impression of a possible solution, and
if we just image that the aperture just happens to extend exactly
to cover the rays drawn, then it is fine.
Let's redraw.
This figure shows the bundle of light emanating from the object point
The second model shows the bundle of light emanating from the object point
in the focus plane and passing through the apperture.
Assuming that the aperture is is circular, the bundle is a skew cone,
and the points contributing to the image point from the out-of-focus
plane is a section of the skew cone.
This may sound as if we have to dig deep into Geometry and not only
read the theory of conic sections but also the extensions to skew cones.
Luckily, it is a lot simpler. Let's redraw the figure, naming the points
for ease of discussion.
We see only a cross-section through the top $P$ and the centre
$O$ of the base, but the argument can be made with any such cross
section.
The sections $AO$ and $BO$ are equal to the radius of the aperture
The drawing of the skew cone shows cross-section through the top
$P$ and the centre $O$ of the base, but the argument can be made
with any such cross-section.
The line segments $AO$ and $BO$ are equal to the radius of the aperture
in every cross-section, and the aperture is a circle, so the radius
is always the same. Thus, even if the angle at $P$ is different,
the lengths $AO$ and $BO$ are constant.
We know that $A'B'$ is parallell to $AB$. Hence the triangles
$A'B'P$ and $ABP$ are equivalent. The same is the case for
$A'O'P$ versus $AOP$ and $B'O'P$ versus $BOP$.
It follows that the ratio of the heights is equal to the ratio
of the bases, that is:
$$\frac{h'}{h} = \frac{A'O'}{AO} = \frac{B'O'}{BO}$$
Since $AO=BO=r$ is the radius of the aperture, $A'O'=B'O'$
is the radius of the object. Hence the object forms a
circle with radius $r'=A'O'=B'O'$.
This gives
$$\frac{h'}{h} = \frac{r'}{r}$$
or
$$r'=\frac{h'}{h}\cdot r$$
Hence the object that contributes to the image of $P$ is a circle
with radius $r'=h'r/h$, where $h'$ is the distance between the
focus plane and the object plane.
(You can find the centre of the object using a similar argument.
In fact, that would similar to the argument you used in the previous
exercise. The new insight is in the part that we have solved.)
*I agree that this exercise is tricky, in the sense that the solution is
not at all obvious at the start.
If you expect to find a template solution you are stuck.
I had to make each of the figures in turn without knowing where
I would be heading next, just to earn a little more insight into
the problem each time.
To succeed, you have to focus on **understanding** the geometry,
rather than finding a computational procedure.*
## Scale Ambiguity (Exercise 3.8).
Use the figure from Exercise 3.1. Ask if it does not suffice.
## Exercise 3.3 Part 1-2.
