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C-82 Appendix C
Exercise 6 In this exercise we will use formula (11) to find the intersection of
a light ray and a lens surface. In your own sketch of Figure C introduce a coordinate
system such that the x-axis coincides with the axis of symmetry and
the y-axis goes through the point P 0 . Assume that P 0 is 1 cm above the x-axis,
so its coordinates are (0, 1), that the point V is on the x-axis 13 cm to the right
of the origin, that the radius of the first (spherical) surface of the lens is 5 cm,
and that the center of the sphere is on the axis of symmetry. If A 87, 19 is a
vector in the direction of the ray through P 0 and the top edge of the lens is 4 cm
above the x-axis, find the point P where the ray intersects the first surface of the
lens.
Hint: From the given information and your figure conclude that A 87, 19,
P 0 80, 19, r 5 and C 818, 09. Then find P 0 C, and use formula (11) to
show that the two values for t are t 2 or 3. Then use equation (1) to find the
points corresponding to the two values for t. Which, if any, is the point of intersection
of the ray with the first surface? What is the geometric significance
of the other point?
Real optical systems live in three-dimensional space, not in a two-dimensional
plane, and not all rays in a rotationally symmetric system lie in a plane containing
the axis of symmetry. It is surprising that our vector solution to the
problem of finding the intersection of a line and a circle in the plane is, through
the step-by-step derivation of equation (11), exactly the same as the vector solution
for the problem of the intersection of a line with a spherical surface in
three-dimensional space. In three dimensions points have three coordinates
and vectors have three components, but a line is still determined by a point it
passes through and a vector in the direction of the line. So equation (1) is also
the equation for a line in three dimensions. Every point on a spherical surface
is the same distance from the center of the sphere so equation (3) holds. The
rest is vector algebra, which works the same in any dimension: This is a major
advantage of a vector approach.