If a beam of parallel rays enters an optical system the emergent beams will converge toward a point (or diverge away from a point for a negative system). This point will lie in the second focal plane. If the rays are parallel to the optic axis of a system the point is the second focal point. The focal plane passes through the focal point and is perpendicular to the optic axis. This is illustrated in Fig. 12.

We shall now determine the position of the second focal point,
**A**]_{OO '}.
To do this
let us take our second reference plane QQ to be the second focal
plane in Fig. 12. *y*
coordinate is, passes through the same point
*y* '*y* ',
' to *y*,
must be of the form

Thus, the lower diagonal element of the matrix of equation (22) must be zero so that,

(23) |

and thus

The position of the second focal point,

Rays emerging from a point in the first focal plane are illustrated
in Fig. 13. A parallel beam emerges from the optical system.
We can take the first reference plane PP through F and QQ the second
reference plane anywhere to the right of
*y*). Equation (22) gives
this relation if the upper diagonal element is zero, that is, if

(24) |

Here D is positive if F is to the left of O,
negative if F is to the right of
O. Then we have from equation (22) that

(25) |

Thus the focal points may be easily found from (23) and (24) once the matrix

From the discussion above, the matrix has a particularly simple form, since both diagonal elements are zero. Using (23) and (24) in (22), we obtain

(26) |

The significance of the two non-zero elements of the matrix connecting the two focal planes is not evident at this point. However, as will be shown later the elements are related to the first and second focal lengths,

(28) |

and

(29) |

A very general relationship
between *f* and *f* '

is unity and the determinant of the refraction matrix separating two media of index

Now if there is a series of refracting surfaces separating media with indices of refraction

ignoring all the translations as they involve multiplication by one. For the optical system

(30) |

The ratio of the first and second focal lengths is seen to be equal to the ratio of the index on the left of O to the index on the right of O '. Also

Equation (30) illustrates the power of the matrix method as the result is difficult to demonstrate in other ways for even a single thick lens and a general proof is not usually attempted for more complicated systems.

It is also perhaps well to note that the theorem
on matrices constitutes
a useful check on the accuracy of multiplication
in computing **A**]_{OO '}.*n* = *n* '

A thick biconvex lens in air has a radius of curvature of 10 cm at each surface and the surfaces are 10 cm apart. Find the focal points and the focal lengths of the lens if the index of the glass of the lens is 1.50.

Solution:

OF = (2/3)×(12)= 8 cm.

F is 8 cm to the
left of O,

Take the lens of the example to be 1 cm thick and show that

**Exercise 2**

Take the lens to be a biconcave thick lens
with radii of 10 cm, index 1.5,
thickness *f*' = *f* = -9.84 cm,

**Exercise 3**

A thick lens is concavo convex, the first radius
is *f*' = *f* = 36.36 cm,

**Exercise 4**

A biconvex lens of index 1.50 has radii *f*' = 24 cm,*f* = 18 cm,

**Exercise 5**

Find the matrix **A**]_{OO '},_{FF '}*n* on the left, *n* ' on the
right and the index of the lens, µ. Take
the radii of curvature as R_{1} and R_{2}.

The above exercises may be worked using the computer programme in the appendix.