CHAPTER IV. OPTICS
289
Solving these equations simultaneously and utilizing the lens formula we obtain 2a2
!=-=24cm ",3
563. a = l tan <p = 60 em. Note. The size of the object is determined by the minimum viewing angle of the eye. 564. At the principal focus of the lens .â&#x20AC;˘
Fig. 354 565. The image will be at a distance
a2
= 60 em on the same side
of the lens as the object (see Fig. 354).
Solution. In order to determine the position of the image produced by the system as a whole, calculate the positions of the images formed by the separate parts of the system consecutively along the path of the rays. " " F ._-~._. The lens for the object A will produce a virtual image B lying to the left of the lens at a distance
-.
a=
F at =30 em at- F
Fig. 355 The mirror having B as the object will form an image C lying at a distance a b = 45 em behind the mirror or, in other words, at a distance a' == a 2b = 60 em from the lens. On the right the lens recel ves rays from C as the object and th,
+
+