Difference between revisions of "Lens"
m (1 revision)
Latest revision as of 22:54, 11 May 2013
Lenses are very useful for holography. Small lenses can be used to expand the beam, to change the diameter of the beam and to change the height to width ratio of the beam. The best lenses are Anti-Reflection coated for the frequency of laser/s in use.
Large lenses can be used to collimate a laser beam or to form a real image to make image planed holograms.
Choosing the Correct Lens
From Tom B.
Let beam diameter d be 4 mm and lens focal length f be 5 mm,
The beam is brought to a focal point f mm past the lens, and in cross section looks like an isoceles triangle with the acute angle A at the focus. The half-angle hA of the right-triangle part of the cone is given by
hA = arctan ((d/2)/f) = arctan (2/5) = arctan(0.4) = 21.8 degrees.
Past the focus, the beam diverges at 2 * hA. For a given distance past focus L, the beam radius R is L * tan(hA), or L = R / tan(hA)
For the given diameter D of 100 mm, R is 50 mm, and
L = 50 mm / tan(21.8) = 50/0.4 = 125 mm
So if we have an 8" collimation mirror, and we have a 4mm laser beam with no divergence and we want a colliated out beam then we would choose a lens with the focal length:
This is the most common shape for a lens. It has one or both surfaces ground to a spherical surface. It can either be concave or convex. They suffer from spherical aberrations but this is not important in many applications.
Plano-convex lenses have a positive focal length defined by f=R/(n-1) where R is the radius of curvature and n is the index of refraction of the glass used.
Plano-concave lenses have a negative focal length defined by f=-R/(n-1).
Bi-Convex lenses have a positive focal length defined by f=((2(n-1)/r)-(Tc(n-1)^2)/nR^2))^-1. Where Tc is the overall thickness of the lens.
Bi-Concave lenses have a negative focal length defined by f=-((2(n-1)/r)-(Tc(n-1)^2)/nR^2))^-1.
Cylindrical lenses are only ground on one axis. If placed into a laser beam they only expand one axis forming a line. With two lenses you can circularize an elliptical beam. If the ratio of the beam width to height is 3 to 1 then the focal lengths of the correcting lenses must be 3 to 1 (ie. 75mm and 25mm). The lenses are then placed at f1+f2 distance apart (ie. 100mm).
Cylindrical lenses are also useful before the collimating optic in order to make a slit reference beam. Holographic stereograms and Rainbow Holograms are examples of where a cylindrical lens could be useful.
Aspheric lenses are ground to a special figure that corrects for spherical aberrations.
Microscope objectives are often used in spatial filters. They work best if AR coated to the wavelength of the laser in use. You can convert from focal length to magnification with the following formula:
M = 250 / f
f = 250 / M
where M is the magnification of the objective and f is the focal length of the lens.
Industrial Optics makes a laser line series of objectives that have very low losses at 632nm.
Gradient Index Lenses have no figure cut into them. The focusing ability comes from changing the index of refraction across the surface of the lens. They are often used to focus light from a 808nm diode into the end of a YAG crystal for a DPSS laser.
They are produced by silver-ion exchange (or lithium-ion exchange for low numerical apertures) in a special glass. GRIN lenses come in two basic flavors: RADIAL or AXIAL which are sometimes referred to as RGRIN and AGRIN respectively. RGRINS are usually used where you want to add optical power to focus light. An RGRIN with flat surfaces can focus light just as a normal lens with curved surfaces does. Thin RGRIN lenses with flat surfaces are known as WOOD lenses, named after the American physicist R.W. Wood who did a lot of experimental work with radial gradients from about 1895 to 1905 and included descriptions of how to make them in his physics text book (available from OSA).
The Eikonal equation (Born & Wolf) says that d(nk)/ds = gradient n where nk is the ray vector, s is the distance along the direction of propagation, and n is the refractive index. So moving to finite differentials gives delta(nk) = delta(s) * gradient n
For Example, gradient n = (1.355 - 1.300) / 15 mm = 0.0037 per mm so delta(s) * gradient n = 5.5 mm * 0.0037 /mm = 0.0202 Now remember that nk is along the horizontal direction with a magnitude of 1.300 and delta(nk) is perpendicular to it, pointed toward the optical axis. Thus the vector sum of ray vector and change to the ray vector will have an angle of tangent (0.020/1.300) = 0.89 degrees in the glass In air, Snell's law says that this angle will increase to inverse sine(1.300*(sine(0.89 degrees)) = 1.15 degrees. So the ray exits at a height of about 15 mm with a slope of 1.15 degrees; this means it will strike the optical axis at a distance of 15 mm / tan(1.15 degrees), which is 747 mm. Which isn't the 409 mm you got