Focal length based on light color

[Seadragon]

Homework Statement

A thin lens has radii of curvature R1=-8.00 cm and R2=-6.00 cm, indices of refraction
n(red)=1.50 and n(violet)=1.53 for red and violet light, and is surrounded by air. An object at
distance s=20.0 cm from the lens has an image at distance s' from the lens. Is there a
second lens with n(red), n(violet), and R2 as stated above, but a different R1 so that s' due to
violet light and the second lens equals that due to red light and the first lens? Explain
your result.

Homework Equations

1/f = (n-1)(1/R1 - 1/R2)

1/s + 1/s' = 1/f

The Attempt at a Solution

In general:

1/s + 1/s' = (n-1)(1/R1 - 1/R2)

1/s' is negative in this case due to a negative radius of curvature (diverging lens)?? But then f is also negative...

1/s - 1/s' = - 1/f

1/f = 1/s' - 1/s

1/s' - 1/s= (n-1)(1/R1 - 1/R2)

1/s' = (n-1)(1/R1 - 1/R2) + 1/s

We need to have 1/s' equal for two lenses, the first one (given) and the second one (to solve for). Let x be R2 of the second lens:

1/s' = (n(red)-1)(1/R1 - 1/R2) + 1/s = (n(violet)-1)(1/R1 - 1/x) + 1/s

Subtract 1/s from both sides:

(n(red)-1)(1/R1 - 1/R2) = (n(violet)-1)(1/x - 1/R2)

[(n(red)-1)(1/R1 - 1/R2) / [(n(violet)-1)] ] + 1/R2 = 1/x

x = {[(n(red)-1)(1/R1 - 1/R2) / [(n(violet)-1)] ] + 1/R2}^-1

So the second lens would have a second radius of curvature x.

There're so many points to make mistakes... is this somewhat correct? Thanks!

Michael

 

[anathema]

Hello Michael,

Your solution is mostly correct, but there are a few things that need to be clarified.

Firstly, your calculation for x is correct, but it should be noted that x is the radius of curvature of the second lens, not R2. R2 is the radius of curvature of the first lens.

Also, your approach assumes that the two lenses are placed in series, with the second lens immediately after the first one. This may not always be the case, as the lenses could be placed at different distances from each other.

Another point to consider is that the focal length of the second lens may not necessarily be the same as the focal length of the first lens. This means that the image distance, s', due to violet light and the second lens may not be exactly equal to the image distance due to red light and the first lens. However, it is possible to choose a value for x that would make the image distances very close to each other.

Overall, your solution is a good approach to solving this problem. Just make sure to clarify the points mentioned above and consider the possibility of the lenses being placed at different distances from each other.

Best of luck with your research!