β = y′/y = f/x = x′/f′,
(1)
and also
xx′ = ff′.
(2)
By differentiating equation (2) we obtain
dx′= −(ff′/x2) dx or dx′/dx = −ff′/x2.
(3)
The ratio of the displacement of the image dx′ to the displacement of the object dx is the axial magnification, and is denoted by α. Equation (3) gives important information on the displacement of the image when the object is moved. Since f and f′ always have contrary signs (as is proved below), the product −ff′ is invariably positive, and since x2 is positive for all values of x, it follows that dx and dx′ have the same sign, i.e. the object and image always move in the same direction, either both in the direction of the light, or both in the opposite direction. This is shown in fig. 3 by the object O3O2 and the image O′3O′2.
If two conjugate rays be drawn from two conjugate points on the axis, making angles u and u′ with the axis, as for example the rays OH1, O′H′1, in fig. 3, u is termed the “angular aperture for the object,” and u′ the “angular aperture for the image.” The ratio of the tangents of these angles is termed the “convergence” and is denoted by γ, thus γ = tan u′/tan u. Now tan u′= H′H′1/O′H′ = H′H′1/(O′F′ + F′H′) = H′H′1/(F′H′ − F′O′). Also tan u = HH1/OH = HH1/(OF + FH) = HH1/(FH − FO). Consequently γ = (FH − FO)/(F′H′ − F′O′), or, in our previous notation, γ = (f − x)/(f′ − x′).