How Does Immersing a Glass Rod in Water Affect Image Formation?

[Fizzicist]

Homework Statement

One end of a long glass rod is ground to a convex hemispherical shape. This glass has an index of refraction of 1.53. When a small leaf is placed 20.3 cm in front of the center of the hemisphere along the optic axis, an image is formed inside the glass 9.08 cm from the spherical surface.

Where would the image be formed if the glass were now immersed in water (refractive index 1.33), but nothing else were changed?

Homework Equations

f' = [n1(n2 -1)/(n2 - n1)]f, where n1 is the index of refraction of the medium (water) and n2 is the index of refraction of the lens (glass), f is the focal length of the glass in air, f' is the focal length in water.
1/f = 1/s + 1/s' where s is the object distance and s' is the image distance.

The Attempt at a Solution

I am having problems working this one out and I don't know why. I first used the lens equation to solve for the focal length in air. Then I used that focal length to find the focal length in water. Then I used that to get the new image distance. What am I doing wrong? It might be a sign problem, but I checked already and didn't seem to find one.

 

[Fizzicist]

For some reason I keep getting a large value for the image distance (250 cm). Anyone know why?

 

[Moetasim]

I would suggest checking your calculations and making sure you are using the correct values for the refractive indices and distances. It may also be helpful to draw a diagram to visualize the situation and double check your calculations. Additionally, make sure to use the correct signs for the object distance and image distance in the lens equation. If you are still having trouble, it may be helpful to consult with a colleague or reference a textbook for guidance.