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

404 ANSWERS AND SOLUTIONS
~---d---~
Fig. 527
In our case the divergent lens is tightly fitted against the convergent one
(d=O) and therefore
I
I
I
II
~d
1 1 1 1 1
-+-=--+-=a1
b2 f 1 f 2 f
where f is the sought focal length of the system.
fIf
H t
2
ence, =f--f-'
1- 2
736. The light beam issuing from a point at a distance of a 2
= 5 em from
the second lens impinges on it. As follows from the formula of a lens, the
continuations of the rays refracted by this lens intersect at a distance of
b 2 =4 em from it (Fig. 530). This point coincides with the focus of the third
lens. For this reason the rays leaving the system will travel in a parallel
beam. The given system is telescopic.
737. Figure 531 illustrates the path of the ray through the plate from
point S of the object. As a result
/
/
of refraction of light by the plate
ray BE seems to be issuing from
I
/
point S' that is the virtual image
/A
of S in the plate. Thus, the distance
/
between the image of the object in
/
/
the plate and the lens is a' =a-SS'.
/ 8
The displacement SS'=AD=
=d-DC.
/
Assuming the angles of incidence
on the plate to be small, we have
DC=B.C =dr =~
t r n
Fig. 528
L
since - == n.
r