Calculate the position of the image

[Pseudo Statistic]

A question:
"Suppose that a converging lens with focal length 30 cm in water is placed 20cm above a light source at the bottom of the pool. An image of the light source is formed by the lens.
a) Calculate the position of the image with respect to the bottom of the pool."
For a I did 1/f = 1/p + 1/q But got -60cm:
1/30 = 1/20 + 1/q
That doesn't sound right to me.. can anyone tell me what I'm doing wrong?

"b) If instead of water the pool were filled with a material with a different index of refraction, describe the effect, if any, on the image and its position in each of the following cases.
i) The index of refraction of the material is equal to that of the lens."
I'm going to guess there is no image?

"ii) The index of refraction of the material is greater than that of water but less than that of the lens."
No clue..

Can anyone clarify? Thank you.

 

[Doc Al]

A question:
"Suppose that a converging lens with focal length 30 cm in water is placed 20cm above a light source at the bottom of the pool. An image of the light source is formed by the lens.
a) Calculate the position of the image with respect to the bottom of the pool."
For a I did 1/f = 1/p + 1/q But got -60cm:
1/30 = 1/20 + 1/q
That doesn't sound right to me.. can anyone tell me what I'm doing wrong?

What doesn't sound right? (What's the position of the image with respect to the pool bottom?)

The focal length of a lens depends on the difference between the index of refraction of the lens and the index of refraction of the medium the lens is submerged in. (This should make intuitive sense, since without a difference in index of refraction, there would be no refraction and no lens.) For details, study the Lens Maker's formula. (Look here: http://hyperphysics.phy-astr.gsu.edu/Hbase/geoopt/lenmak.html#c1)