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

388 ANSWERS AND SOLUTIONS
Fig. 500
a
F'
J
I
J
I
I
I
\ I
\ I
\ 0
s
D
If n is odd (n=2i+ 1) it can
easily be seen that the i-th images
lie on the continuations of
the mirrors and th us coinci de
with the (i + 1) th and all the subsequent
images. For this reason
there will be 2i images, Le.,
n-l as before.
690. Using the solution of
Problem 689, let us plot consecutively
the first, second, third,
etc., images of source S in the
mirrors (Fig. 500). All of them
will lie on a circle with a radius
OS and the centre at pointo.
If a is an integer, the last i-th
images will either get onto poin ts
C and D at which the circle intersects
the continuations of the
mirrors, or will coincide with
point
P diametrally opposite to
the source. In both cases the
. number of images will be a-I.
If a is not an integer, for example a=2i ± ~, where ~ < 1, and i is an
integer, the last i-th images will lie on arc CPD that is behind both the
first and the second mirrors and there wiII be no more reflections. Thus, the
total number of images will be 2i.
691. Let us plot the image of point B in mirror bd (Fig. 501). Let us
then construct image B 1 in mirror cd. Also, B 3 is the image of B 2 In mirror
ac and 8 4 is the image of 8 3 in mirror abo
Let us connect points A and B 4 • Point C is the point of intersection of ab
with line AB 4 • Let us now draw line B 3C from B 3 , and connect point D at
which this line intersects ac with 8 2 • E with B 1 , and F with B.
It can be stated that broken line ACDEFB is the sought path of the
beam. Indeed, since B 3CB4 is an isosceles triangle, CD is the reflection of
beam AC.