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

420 ANSWERS AND SOLUTIONS
from the equations:
d 2 tan cp d 2 q>
---~-
(x 2 + d 2 )- 1 X 2 - Xz
(considering triangles EOP, E AB ~nd OPO I ) and
1 1 1
-+-=d
2 .x?, /2
(considering E as the image of°1 ) . ' Hence DC}. ~ 0.011 em.
Fig. 549
In the third arrangement (d 3=5 em) the path of the rays will be somewhat
different (Fig. 549) than that ShO\\'D in Fig. 548.
The equations for 'D 8 can be written similar to the preceding cases as
follows:
Ds ds tan q> e: dscp and ..!.._..!..=..!..
(l- d s ) + Xs Xs Xs ' da Xa fI
Hence, Ds=O.18 em.
762. It follows from the formula of a lens
1 1 1
(i+1J=Fob
that the magnification of the objective is
b' Fob
k1=(i=;-="P o b
where b is the distance from the image to the objective.
The real inverse magnification of the object produced by the objective can
be viewed through the eyepiece as through a magnifying glass, the virtual
image produced by this magnl fying glass being arranged from the eye at the
distance of best vision D=25 em.
According to the formula ~f a magnifying glass
1 1 1
al -D = F ey e
where al is the distance from the image produced by the objecti ve to the
eyepiece.
1 he magnification of the eyepiece is
D D+Feye
k,,=-=----
at F ey e
The total magnification of the microscope is
Fob (D+Feye)
k=ktk"
180 times
(a-Fob) Feye