- #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

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