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Hi
calculate the back vertex power of a lens made from glass n = 1.5. with an axial thickness of 9mm if the front surface has a power of +10.76D and the back surface has a power of -3.00D?
so far I have:
n= 1.5
t= 9mm
F1= +10.76D
F2= -3.00D
F= +7.76D
I used this formula: t/n which is 9/1.5 = 6mm
if someone could point out some guidance, that would be great. thanks
I suggest you purchase Ellen Stoner's excellent book 'Optical Formulas'.
Start by removing the thickness from your equation and creating a thin lens equation, to do this convert the front surface to the effective power at the back surface.Quote:
Originally Posted by londongirl312
Convert front to focal distance:
1.5 / +10.76D = 0.1394m
Now since you are traveling 9mm through the lens remove that:
0.1394m - 0.009m = 0.1304m
Now convert back to diopters:
1.5 / 0.1304m = +11.50D
Now you have a thin lens equivalent so just plug that into the thin lens equation:
+11.50D + (-3.00D) = +8.50D
The formula you are using is converting the thickness to the air equivalent so 9mm / 1.5 = 6mm, which now if you used would work the same as the steps above but instead of using the index of the material to convert the front surface you would use the index of air since you compensated out the thickness:
1 / +10.76D = 0.0929m
Subtract the air equivalent thickness:
0.0929m - 0.006m = 0.0869m
Now convert back to diopters:
1 / 0.0869m = +11.50D
Now plug that into the thin lens equation:
+11.50 + (-3.00D) = +8.50D
For more information look up Swiane's step along method, and vergence formulas to gain a better understanding. Or you could just use a formula which I believe in the book mentioned above is an approximation.