(2010-11-26)
Matrix Methods in Paraxial Geometrical Optics
Each optical element acts on the distance and inclination of the ray.
Here, we are primarily concerned with optical systems endowed with
cylindrical symmetry around a straight line called the
optical axis (i.e., the optical system is unchanged in any rotation
around the optical axis).
A
meridional ray (or tangential ray is a ray that is contained
in a plane which includes the optical axis.
Other rays are called skew rays (this includes sagittal rays
whose direction is perpendicular to the optical axis but do not intersect it).
Wikipedia :
Ray
|
Optical axis
|
Paraxial approximation
|
Ray
transfer matrix analysis
Meghan (via Yahoo!
2011-01-05)
Crystal Ball
A solid sphere of glass (radius R, index n)
has focal length f = R/(2n-2)
There are several ways to obtain this result.
The easiest one is probably to notice that
the lensmaker's formula (intended for thin
lenses only) applies directly to this particular case of a
thick lens, because of the existence of an
optical center
(a point through which light rays are not deflected at all).
We may also do it the hard way,
without even using the small angle approximation:
For an incident ray at a distance u < R
from the center O of the sphere, we consider the plane xOy
where the x-axis is parallel to the ray
(whose direction is that of increasing values of x at a
constant value of y = u > 0).
See above figure.
The ray enters the sphere at point
I = ( x0 , y0 ) at an angle of incidence
denoted i
(that's the angle with respect to the normal to the surface).
x0 =
- ( R 2 - u 2 )½'
y0 = u = R sin i
The refracted ray emerges from I at an angle r
(with respect to the normal) whose sine is equal to u/nR
(according to Snell's law).
At this point, the ray's inclination with respect to the x-axis is
a (which is a negative angle).
Using a dummy parameter z, the equations of the ray inside the sphere are:
x = x0 + z cos a
&
y = y0 + z sin a
The exit point J corresponds to the nonzero value of z
for which x2 + y2 = R2 :
R 2 =
( x0 + z cos a )2 +
( y0 + z sin a )2
0 = z 2 +
2 z [
x0 cos a +
y0 sin a ]
So, we must plug z = -2 [
x0 cos a +
y0 sin a ]
into the previous parametric expressions
to obtain the coordinates
(x1 , y1 ) of the exit point J :
x1 = x0 - 2 [
x0 cos a +
y0 sin a ]
cos a
=
-
x0 cos 2a +
y0 sin 2a
y1 = y0 - 2 [
x0 cos a +
y0 sin a ]
sin a
=
-
x0 sin 2a +
y0 cos 2a
We could have obtained the same result geometrically...
Ransom
(2010-11-26)
Lensmaker's equation
Focal length of a lens with two concave faces of radii 0.300 & 0.970 m.
The following formula gives the focal length ( f )
in a medium of index 1 (air or vacuum)
for a thin lens consisting of a medium of index n enclosed between
two surfaces whose radii of curvature are respectively R1
and R2
Thin Lens Formula
| 1 |
= (n-1) [ |
1 |
+ |
1 |
] |
 |
|
|
| f |
R1 |
R2 |
|
The curvatures are counted positively when the surface bends toward the
denser medium and negatively otherwise. Similarly, the resulting
focal length is positive for a converging lens and negative for a diverging one.
In the above case of a plastic
biconcave lens (n = 1.44) the radii
of curvature are both negative (-0.300 and -0.970). So is the
focal length given by the above formula:
f = -0.521 m
(2010-11-25)
Radius of curvature of a concave mirror...
What's the curvature of a mirror magnifying 3 times an object at 22 cm?
(2008-10-26)
Shadow Hiding. The Opposition Effect.
The cause of extra brightness directed back to the source of illumination.
When illuminated, a smooth enough dull surface sends back
in all directions an intensity of light which is proportional to
its apparent area in the direction of the observer
(Lambert's Law).
However, some features of a rough surface may be large enough to cast shadows
on deeper patches which reduce the percentage of the surface that's illuminated.
This can reduce significantly the albedo of the surface of a rocky planet
whenever it's not pbserved directly at opposition.
