- #1
StonieJ
- 29
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A friend and I are getting conflicting answers for a focal length/power question. He says he gets 22.5 diopters, whereas I get 5.1 diopters, so I was wondering who's screwing up. Here is the question and my answer:
Q: If a +20 diopter lens is a symmetric double convex lens of index 1.50, what is its focal length and power when placed in a tank of water with index 1.33?
A:
A priori, it seems to make sense that the focal length would increase and the power would decrease. Since the change of light going from water to the lens is less than the change of light going from air to the lens (in terms of indices of refraction), it seems that the light should be less bent in water and therefore travel further before converging to an image.
Q: If a +20 diopter lens is a symmetric double convex lens of index 1.50, what is its focal length and power when placed in a tank of water with index 1.33?
A:
Code:
f = 1/d = 1/20 = 0.05 m
Using lens maker's equation:
1/f = (n1/n2 - 1) (1/r1 - 1/r2)
Let (1/r1 - 1/r2) = K.
Solve for K in air (which will be the same in water).
1/0.05 m = (1.50 / 1.00 - 1)K
20 = (0.5)K
K = 40
Now plug in values for water:
1/f = (1.50/1.33 - 1) (40)
1/f = 5.1 diopters
f = 0.20 m
A priori, it seems to make sense that the focal length would increase and the power would decrease. Since the change of light going from water to the lens is less than the change of light going from air to the lens (in terms of indices of refraction), it seems that the light should be less bent in water and therefore travel further before converging to an image.
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