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

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The discussion revolves around calculating the image formation of a convex hemispherical glass rod when immersed in water. The initial setup involves a glass rod with a refractive index of 1.53 and an object placed 20.3 cm from its center, resulting in an image 9.08 cm from the surface. The user attempts to find the new image distance after immersing the rod in water, using the lens equation and the modified focal length. However, they encounter difficulties, particularly with an unexpectedly large image distance of 250 cm, suggesting a potential error in their calculations or sign conventions. The thread seeks clarification on the correct approach to resolve these issues.
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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.
 
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For some reason I keep getting a large value for the image distance (250 cm). Anyone know why?
 
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