70
Dioptrics of the Eye
[51, 52.
Now by the conditions as to the size of the object, the angle a must be
very small, and by putting cosa = l, we shall neglect only a small
quantity of the second order of smallness ; and then we may write :
If g>> denotes the distance from a of the image of the axial point p,
then
and hence
9"
= 1,
X = grr
(5)
That is, the point r at the foot of the perpendicular tr is the image of p.
Accordingly, the images of all points lying in a plane perpendicular
to the axis at p are found to be approximately in a plane perpendicular
to the axis through the image of p. Thus having first located the
point r which is the image of the axial point p, the image of any other
point in the object may be found by simply drawing a line from the
given point through the centre of the spherical refracting surface, and
the point where it crosses the transversal image-plane at r will be the
required image. It follows from the principles of elementary geometry
that the image and the object are similar.
It is easy to find the ratio of the corresponding linear dimensions of
the object and its image. For example, let the height of the object ps
be denoted by ß>, and the corresponding length in the image rt by ß>>;
then
where the minus sign has to be inserted in the formula so that it will
apply to the cases of both erect and inverted images. Combining this
or
(2c), (3a),
(3b) and (3c), we find:
0"
G..
Grr-grr
(6a)
ß>
~G'-gr
G'
ß"
Fr
F"—f"
, (6b)
ß'
F•>
When the refracting surface is plane, the focal lengths become
infinite, and for this case equation (6b) becomes
t
ß'
= 1
(6c)
Accordingly, the image in a plane refracting surface is exactly as large
as the object.