Chapter 5

Image Formation

The Positive Nodal Points The positive nodal points, N and N ', are two points lying on the optic axis which have the property that a ray incident on N emerges from N ' in a direction parallel to the incident ray. Thus the angular magnification \alpha ' = 1 for these two points.

To determine the position of the nodal points, we again use (36) but with the conditions now that for y = y ' = 0, \alpha ' = alpha. Equation (36) states that for y ' = 0, alpha = (-z '/f) alpha ' and therefore
$$Z'=-f \ \ \{\rm or} \ \ \
F'N'=-f\eqno(51)$$ (51)


Also (36) states that for y ' = y = 0, 0 = (zz '/f - f ') \alpha ' = alpha which combined with equation (51) gives
$$z = -f' \ \ \ {\rm or}\ \ \  FN  =  -f'. \eqno(52) $$ (52)

Figure 20.

Thus for a positive system the first nodal point is a distance f ' to the right of F and the second nodal point is a distance f to the left of F '. The use of the nodal points in a ray diagram is illustrated in Fig. 20. The distances NN ' and HH ' are obviously equal. If f = f ', N and H are coincident, as are N ' and H '. Equations (51) and (52) together with (36) give the transformation from one nodal plane to the other as
$$\bmatrix{y \cr \alpha}_{N}=
\bmatrix{f'/f & 0 \cr 1/f & 1}_{NN'}
\bmatrix{y' \cr \alpha'}_{N'} . \eqno(53)$$ (53)

Again the two planes through N and N ' are recognized to be object-image planes as there is a zero in the upper right hand corner of the matrix.

The linear magnification for an object at the first nodal plane of an image at the second nodal plane is seen to be m = y '/y = f/f ' = n/n '.

It is left to the reader to show that the equations for image formation may be expressed in terms of the object distance v from N and the image distance v ' from N ' as
$$f'/v+f/v'=1\eqno(54)$$ (54)

and
$$m=-v'/v . \eqno(55)$$ (55)

The points N and N ' may lie outside the optical system OO ' and also the rays may not in fact pass through the nodal points.


Example 4

Compute the nodal points of the lens of example 2.

Figure 21.

Solution: We had O 'F ' = 2.167 cm and OF = 4.167 cm. The second nodal point is 3.333 cm to the left of F ' and the first nodal point 4.333 cm to the right of F as illustrated in Fig. 21. A ray incident towards N emerges parallel to the incident ray as if it came from N '.

Negative Nodal Points

The two negative nodal points, N(-) and N '(-), are two points on the optic axis such that a ray incident on the first point at an angle, $\theta$, say, emerges from the second point at an angle of -$-\theta$. The angular magnification for these points is then alpha '/alpha = -1. With this definition it is left to the reader to show that
$$N (-) F  =  f' , \eqno(56)$$ (56)

$$ N'(-) F'=  f , \eqno(57)$$ (57)

and that
$$\bmatrix{y \cr \alpha}_{N(-)}=
\bmatrix{-f'/f & 0 \cr 1/f & -1}_{N(-)N'(-)}
\bmatrix{y' \cr \alpha'}_{N'(-)} . \eqno(58)$$ (58)

It is further left as an exercise to show that the equations for image formation may be expressed in terms of the object distance, w, from N(-) and the image distance, w ', from N '(-) as
$$f'/w+f/w'=-1 , \eqno(59)$$ (59)

and
$$m=w'/w . \eqno(60)$$ (60)

Figure 22.

The ray diagram of Fig. 22 illustrates the use of the negative nodal points. The parallel rays through F and N(-) are brought to a focus in the second focal plane while the ray through F is rendered parallel to the optic axis.

Exercise 2

Compute the positions of all the cardinal points of the lens in the example of chapter 4.

Exercise 3

Compute the positions of all the cardinal points of the lens in exercise 4.