In vacuum (and approximately in air) the field of view has angular width in radians
theta[air]=2 arctan(L/2 f)
where f is the focal length and L is the width of the sensor, and theta is in radians. But, usually field of view is quoted for the diagonal, and in that case L=sqrt(w^2+h^2), where w is width and h is height.
If the lens is directed through a flat port into water, then the field of view has angular width given by Snell's Law:
theta[water] = 2 arcsin( sin( theta[air] /2)/1.333 )
where 1.333 is the index of refraction of water.
For example:
The D70 sensor is 23.7 by 15.6 mm, so it measures 28.37mm diagonally. Then, an 18mm lens has field of view in air of
theta[D70,air] =2 arctan(28.37/2*18) = 1.335 radians
Convert radians to degrees by multiplying by (180/pi) = 57.29:
theta[D70,air] = 1.335 * 57.29 = 76 degrees :: is the field of view of a D70 with 18mm lens in air.
A full-frame sensor is 36x24 mm, so 43.27mm diagonally. Then, a 27mm lens would have field of view
theta[FF,air] =1.34 radians = 77 degrees :: is the field of view of a FF sensor with 18mm lens in air.
This is almost the same as the 18mm lens for the D70. The ratio. (27 mm/18 mm) is the "Nikon factor" of 1.5, to account for the smaller sensor.
In water, the D70 with a flat port would have field of view
theta[D70,water] = 2 arcsin( sin( 1.335 /2) /1.333) = 0.965 radians = 55 deg :: FOV of a D70 w/ 18mm lens in water.
With a little algebra, you can show that this is the same field of view as a 41mm lens with a full-frame sensor in air, or with your D70 in air with a 27mm lens.
I hope this helps!