Calculating Chromatic Aberration

In summary, the speaker used the concept of convex lenses to solve for the focal lengths of red and blue light on the optical axis. However, their calculations did not result in the correct answer. They also express uncertainty about subtracting the radii of curvature correctly. After further consideration, the speaker was able to solve the problem.
  • #1
ab200
13
3
Homework Statement
The dependence of glass's index of refraction on wavelength causes imperfect images. A parallel beam of white light falls on a convex glass lens. Both surfaces of the lens have a 20 cm radius of curvature. For the glass, nred = 1.52 and nblue = 1.53.

What is the space between the points that red light and blue light focus on the optical axis, in cm?
Relevant Equations
(1/s) + (1/s') = 1/f = (n-1)[(1/R1) - (1/R2)]
Since the lens is convex, I figured that the points where the red and blue light focus on the optical axis would be equal to their respective focal lengths (f), given that the incoming rays are parallel to each other and perpendicular to the lens.

Solving this got me to 1/fred = (1.52 - 1)[(1/20) + (1/20)] = 0.013, so fred = 76.923 cm.
1/fblue = (1.53 - 1)[(1/20) + (1/20)] = 0.01325, so fblue = 75.472 cm.
Subtracting the two gives me 1.451 cm, but that isn't correct.

What I'm not sure about is whether I am subtracting R1 and R2 correctly. Both surfaces have the same radius of curvature, but I can't simply subtract them or I would end up with 0 on the right side of the equation, which doesn't make sense.

Reference: https://www.physicsforums.com/forums/introductory-physics-homework-help.153/post-thread
 
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  • #2
ab200 said:
(1.52 - 1)[(1/20) + (1/20)] = 0.013,
Really?
 
  • #3
haruspex said:
Really?
Alas, it is not. I figure it out though! Thank you for your help.
 
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