That's correct.

You know the apparent depth di (the depth of the "image") is given by
di = - do (1/n2)

(this assumes we are looking from the air with n1 = 1.0, which is usually the case!)

Now we can just "plug in" the numbers,

- 0.75 m = - do (1/1.33)

The image distance di is negative, di = - 0.75 m, because this is a "virtual image". The light appears to come from there but it does not actually pass through that point.

- 0.75 m = - do (1/1.33)

do = (0.75 m) (1.33)

do = 1.00 m

That is, the "object", your toes, is(/are) 1.0 m below the water's surface.

(c) 2002, Doug Davis; all rights reserved.