Consider a simple diopter consisting of one bi-convex element with radius of curvature 'r' and index of refraction n(g). When surrounded by a medium having index of refraction 'n', its focal length 'f' is given by the lens maker's formula:

So, if n(w) and n(a) are the indices of refraction of water and air, respectively, the corresponding in-water and in-air focal lengths are

Dividing and simplifying, we get

Since n(a)=1.0 and n(w)=1.33, this becomes

As long as the index of refraction n(g) of the lens is greater than that of water, the lens' in-water focal length will always be **greater** than its in-air focal length. When used underwater it will therefore cause less bending of light rays and will require increased lens-to-subject distances. Since lens power varies inversely as the focal length, the ratio of in-water power P(w) to the in-air power P(a) is

The following table shows the results corresponding to some typical values of the lens index of refraction.

n(g) f(w)/f(a) P(w)/P(a)---------------------------- 1.5 4.0 .25 1.6 3.0 .33 1.7 2.5 .40 1.8 2.3 .43