Bukhovtsev-et-al-Problems-in-Elementary-Physics

GEOMETRICAL OPTICS
399
When h ~ 2a, ex. and ~ are small. For small angles
H 01:: 15 h
a+b - · 2a
Upon excluding the unknown quantities ~
3
that 1=2: a.
and b from the equations, we find
Hence, R=2f=3a=6 metres.
719. The path of the ray is shown in Fig. 514. Let us continue AB up to
its intersection with the focal plane of lens NN. The beam of parallel
rays after refraction in the lens so travels that the continuations of the rays
should intersect at P'. Ray F'O is not refracted. Thus, ray C A passing to
point A is parallel to F'O up to the lens.
720. If A is the source and B is the image, then the lens will be convergent.
The position of the optical centre of the lens 0 and its foci F can be
found by construction as shown in Fig. 515.
If B is the source and A is the image, the lens is divergent. The respective
construction is illustrated in Fig. 516.
721. The centre of the lens 0 is the point of intersection of straight lines
SS' and NIN2. The foci can easily be found by constructing the rays parallel
to the major optical axis (Fig. 517).
722. Point 0, which is the optical centre of the lens, can be found by
dropping perpendicular 80 onto straight line N iN2 (Fig. 518). Let us draw
an auxiliary optical axis DO parallel to ray AB and extend straight line BC
until it intersects DO at point E lying in the focal plane. Let us drop a perpendicular
from E onto N IN 2 to find point F, one of the main foci of the
lens. By using the property of reversibility of the ray. we can find the other
main focus Ft.
723. The image S' may be real or virtual. In both cases let us draw an
arbitrary ray ADS' and auxiliary optical axis BOC parallel to it to find the
position of the source lFig. 519). By connecting the points of intersection B
and C (of the auxiliary axis with the focal planes) to point D by straight
lines, we can find the position of the source SI (if the image S' is real) and
S2 (if the image is virtual).
724. Since the ray incident on the mirror at its pole is reflected symmetrically
with respect to the major optical axis, let us plot point SI symmetrical
a
to S' and draw ray SSt until it in-
tersects the axis at point P (Fig. 520).
This point wilJ be the pole of the mir-
A
ror.
The optical centre C of the mirror
can obviously be found as the point of
JJ intersection of ray SS' with ax is N N' .
The focus can be found by the usual
____..o-.--;;:a~_o--_--_-construction of ray SM parallel to
Ai ~ the axis. The reflected ray must pass
through focus F (lying on the optical
ax is of the mirror) and through 8'.
725 .. (a) Let us construct, as in the
solution of Problem 724, the ray BAC
Fig. 518 and find point C (optical centre of the