16th July 2006 - 11:31 PM
QUOTE (mr_homm+Jul 16 2006, 05:26 PM)
The formula for the focal length of a curved refracting surface is
1/f = (n2-n1)/R, where R is the radius of curvature of the surface and n1 is the refractive index on the side where the light is entering, and n2 is the refractive index on the exit side. The radius is positive if the surface bulges toward the direction of the incoming light, negative otherwise.
If the lenses are very thin, and the water between them is also very thin, then the focal lengths combine by a simple rule:
1/f_total = 1/f1 + 1/f2 + 1/f3 + ... .
It is hard to measure the radius of curvature of the lens surface, so suppose all you know is each lens's focal length in the ordinary situation with air on both sides, and the refractive index of the glass (n_g) and of air (n_a). If the lens is symmetrical (same curvature on both sides), then you get 1/f = (n_g - n_a)/R - (n_a - n_g)/(-R)
, by using the formulas for refraction at each surface and putting them together using the rule for combining focal lengths. This gives 1/f = 2(n_g - n_a)/R, so R = f*2(n_g - n_a).
The second lens will have the same formula, but with perhaps a different radius. Then when you combine them with water in between, you will get
1/f = (n_g - n_a)/(R1) + (n_w - n_g)/(-R1) + (n_g - n_w)/(R2) + (n_a - n_g)/(-R2)
=(n_g - n_a)(1/R1 - 1/R2) -(n_w - n_g)(1/R1 - 1/R2) =
(2n_g - n_w - n-a)(1/R1 - 1/R2).
Now put in the formulas for 1/R1 and 1/R2 to get
1/f =(1/2)*[(2n_g - n_w - n_a)/(n_g - n_a)]*(1/f1 + 1/f2).
I think this is what you wanted.
Hope this helps!
First of all thank you for your instant reply. Just a couple more questions.
a) I guess for a plano-convex lens the expression is:
1/f=(n_g - n_a)/R right?
Lets suppose that both lenses (a plano-conex and a bi-convex) have about 3mm thickness and their distance is about 5mm (filled with water). Is this configuration allows me to use this equation, for thin lenses and zerow distance or i have to use another expression including the distance?
Finally, is there any book you can suggest me to study these things in more depth?
17th July 2006 - 04:12 AM
You're right about the planoconvex lens. Basically, each surface where two different refractive materials meet contributes to the focal length.
As for the situation when there are distances between the lenses (or for thick lenses, when there is a non-negligible distance between the surfaces of the lens), you need a more advanced form of lens optics called "matrix optics." You might try searching the web for that phrase to find more information.
Briefly, you make a 2 component vector describing the height and angle of a light ray at a particular point, then you construct a matrix that changes the height and angle of a light ray in the proper way for each optical element, and then you multiply the matrices together onto the initial vector, in the same order the light ray encounters the lens surfaces.
Pick a plane to the left of all the lenses to be the "plane of entry," so the vector for a light ray is a column vector (h,a) where h is the height where the light ray crosses the plane of entry and a is the angle. As with all lens optics, you use the small angle approximation, so that tan(a) = a = sin(a).
Pick a second plane just to the left of the first surface, and let the distance between the two planes be d. Then the light ray strikes the second plane at the same angle as the first (since there's no refraction yet, this is just light traveling through space), but at a height h+ad (since the ray has had a slope a for a distance d). Then the matrix calculation for the light ray traveling from the first to the second plane is
| 1 d | | h | = | h+ad |
| 0 1 | | a | | a |
This matrix is called D(d)
When a light ray crosses a surface, you set up a plane of entry just to the left and a plane of exit just to the right of the surface, and then (since a lens will bring every horizontal ray to a point at its focus) the vector (h,0) will turn into (h,-h/f). The h stays h because there has been no horizontal distance traveled, so no space for the ray to descend. The angle becomes -h/f because this slope will bring the ray across the centerline at distance f. Likewise, the vector (0,a) stays (0,a) because all rays passing through the center of the lens do not bend. Therefore, the matrix calculation for passing through a lens surface is
| 1 0 | | h | = | h |
| -1/f 1 | | a | | a-h/f |
This matrix is called L(f).
Now what you do is compute the values of all the d and f entries, and then multiply all the matrices in order. If your plane of entry is just at the front surface of the first lens, and your plane of exit is just at the back surface of the second lens, then your matrix looks like
v_exit = L(f4)D(t2)L(f3)D(w)L(f2)D(t1)L(f1)*v_enter, where v_enter and v_exit are the light ray vectors, t1 and t2 are the thicknesses of the first and second lenses, f1 and f2 are the focal lengths of the front and back surfaces of the first lens, f3 and f4 are the same for the second lens, and w is the thickness of the water layer. Notice the order: the matrices are applied to v_enter in the same order the light ray would pass through the optical elements.
When you multiply out all these matrices, you will get one overall matrix that describes your lens system. HOWEVER, this may not look like a lens matrix. This means that there may not be an exact equivalent lens for your system. What this tells you is that you can use this matrix to compute what your system does to light rays, so you CAN solve any optical question you have about your system, but there is no single lens that does exactly what your system does.
If you really MUST have an equivalent lens, there is a method for representing your system as a single lens provided that you also use a "fake" distance. In other words, if your lens system is mounted in a tube, it can be turned into an equivalent lens, but you have to pretend that the tube is shorter (or maybe longer) than it really is, and then everything works out right. I mean that if the tube is 10cm long on the outside, you might have to pretend it's 12 cm long on the inside, or some such thing as that. I can give you the details if you want, but probably having the matrix is sufficient.
The place I learned about this is when I was helping a student with a university 4th year optics course through the electrical engineering department. I have never seen this matrix method described at a lower level than that, so if you want a book, you should be looking at 4th year engineering optics texts (expensive!). Unfortunately, I cannot remember the name of the book's authors. Oh, well, I think I did pretty well to remember this much detail a year later when I only got to see it for about an hour. It impressed me as a really cool method for analyzing optics, which is why it stuck in my mind.
Let me know if you need any more details, and good luck finding a book or web reference!