(For paraxial rays, MP can be considered to be a straight line
perpendicular to CP ) Therefore,
B A
B F
PM
FP
′
′
′
=
or 
B A
B F
BA
FP
′
′
′
=
 (∵PM = AB)
(9 4)
Since Ð APB = Ð A¢PB¢, the right angled triangles A¢B¢P and ABP are
also similar Therefore,
B A
B P
B A
B P
′
′
′
=
(9